Ordinal Arithmetic and the First Uncountable Ordinal — Examples

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$1 + \omega = \omega$ and $\omega + 1 > \omega$ , computed both from the recursion and as order types

Example

The two smallest infinite sums behave differently:

$1 + \omega = \omega, \qquad \omega + 1 = \omega^{+} > \omega.$

Each is computed twice below: once from the recursive clauses of Ordinal addition $\alpha + \beta$ , and once from the order-type description of $\alpha + \beta$ is the order type of $\alpha$ followed by $\beta$ , where $\alpha + \beta$ is the order type of a copy of $\alpha$ followed by a copy of $\beta$ . The two routes are independent, and agreeing is the point of the exercise.

In pictures: putting one extra point before a copy of $\omega$ gives a copy of $\omega$ again, since the result still looks like $0, 1, 2, \dots$ ; putting one extra point after it gives something with a greatest element, which $\omega$ does not have.

Facts & Assumptions

Given: The ordinals with the addition of Ordinal addition $\alpha + \beta$ , and $\omega$ the least limit ordinal ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\alpha + 0 = \alpha$ , $\alpha + 1 = \alpha^{+}$ , and $\alpha + \lambda = \bigcup\{\alpha + \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal addition $\alpha + \beta$ ).

[L2]

$\beta \le \alpha + \beta$ (claim (c) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ ).

[L3]

For $m, n \in \omega$ the ordinal $m + n$ lies in $\omega$ (claim (a) of On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product).

[L4]

$\omega$ is a limit ordinal, so $\bigcup \omega = \omega$ and $\omega$ is closed under successor; every nonzero natural number is a successor, and every ordinal in $\omega$ is $0$ or a successor (claims (iii) and (iv) of $\omega$ is the least limit ordinal, Successor and limit ordinals).

[L5]

Every ordinal is transitive, $\mu \subseteq \nu$ iff $\mu \in \nu$ or $\mu = \nu$ , and $\mu \notin \mu$ (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

[L6]

$\alpha + \beta = \mathrm{ot}(\alpha \oplus \beta)$ , where $\alpha \oplus \beta$ is a copy of $\alpha$ with a copy of $\beta$ placed entirely above it ( $\alpha + \beta$ is the order type of $\alpha$ followed by $\beta$ ); every well-order has exactly one order type and order isomorphic well-orders have the same one (Every well-order has a unique order type); a strictly increasing bijection between total orders is an order isomorphism (Order embedding and order isomorphism).

Verification

technique · direct

1.1

From the recursion: for $n \in \omega$ the ordinal $1 + n$ lies in $\omega$ by [L3], hence $1 + n \subseteq \omega$ by [L5]; and $n \le 1 + n$ by [L2], hence $n \subseteq 1 + n$ .

L2L3L5

1.2

From the recursion: $\omega + 1 = \omega^{+} = \omega \cup \{\omega\}$ by [L1], so $\omega \in \omega + 1$ while $\omega \notin \omega$ by [L5], giving $\omega + 1 \ne \omega$ and $\omega < \omega + 1$ .

L1L5

2.1

From the recursion: $1 + \omega = \bigcup\{1 + n : n \in \omega\}$ by [L1], and this equals $\omega$ , since it is contained in $\omega$ by step 1.1 and contains $\bigcup\{n : n \in \omega\} = \bigcup \omega = \omega$ by step 1.1 and [L4].

step 1.1L1L4

2.2

From order types: the map $u : 1 \oplus \omega \to \omega$ with $u(0,0) = 0$ and $u(1,n) = n^{+}$ is a bijection, because every nonzero natural number is a successor by [L4] and $n \mapsto n^{+}$ is injective, and it is strictly increasing, because $(0,0)$ is below every $(1,n)$ and $u(0,0) = 0 \in n^{+}$ , while $(1,m) < (1,n)$ means $m \in n$ and then $m^{+} \in n^{+}$ by [L5]; so $1 + \omega = \mathrm{ot}(1 \oplus \omega) = \mathrm{ot}(\omega) = \omega$ by [L6].

step 1.2L4L5L6

2.3

From order types: $\omega \oplus 1$ has a greatest element, namely the single point of its upper copy, whereas $\omega$ has none, since $\xi \in \omega$ implies $\xi \in \xi^{+} \in \omega$ by [L4]; so the two are not order isomorphic and $\omega + 1 = \mathrm{ot}(\omega \oplus 1) \ne \mathrm{ot}(\omega) = \omega$ by [L6].

step 1.2L4L5L6

3.1

Both routes give $1 + \omega = \omega$ and $\omega + 1 \ne \omega$ , so $1 + \omega \ne \omega + 1$ ; explicitly $1 + \omega = \omega < \omega + 1$ .

step 2.1step 2.2step 2.3step 1.2∎

Remarks

What the two routes cost. The recursive computation needs the limit clause of Ordinal addition $\alpha + \beta$ together with the fact that $\omega$ is closed under adding a natural number, which is On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product. The order-type computation needs only an explicit bijection and $\alpha + \beta$ is the order type of $\alpha$ followed by $\beta$ . Neither is shorter than the other; the second is the one that generalises, since it makes the answer visible before it is computed.

This is the whole of non-commutativity, in miniature. The general statement is FALSE: ordinal addition is commutative, and its proof is exactly the computation above. Everything else about the failure of commutativity is a variation on prepending versus appending.

$\omega + 1$ is not $\omega$ even though the two sets have the same size. $\omega + 1 = \omega \cup \{\omega\}$ is countably infinite, being $\omega$ with one point added, so the difference between $\omega$ and $\omega + 1$ is entirely a difference of order type. That distinction is taken up in $\omega + \omega$ is at most countable although it is not order isomorphic to $\omega$ : order type and cardinality are different invariants.

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$2 \cdot \omega = \omega$ while $\omega \cdot 2 = \omega + \omega$ , pictured as order types

Example

$2 \cdot \omega = \omega, \qquad \omega \cdot 2 = \omega + \omega > \omega.$

Under the convention of Ordinal multiplication $\alpha \cdot \beta$ , $\alpha \cdot \beta$ is $\beta$ copies of $\alpha$ ( $\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ ordered by last differences, that is $\beta$ copies of $\alpha$ ). So $2 \cdot \omega$ is $\omega$ copies of a two element set, laid end to end:

$\underbrace{\bullet\ \bullet}_{}\ \ \underbrace{\bullet\ \bullet}_{}\ \ \underbrace{\bullet\ \bullet}_{}\ \ \cdots$

which is a copy of $\omega$ once the points are counted off $0, 1, 2, \dots$ . And $\omega \cdot 2$ is two copies of $\omega$ , one entirely above the other:

$0, 1, 2, \dots \ \ \text{then} \ \ 0', 1', 2', \dots$

which is $\omega + \omega$ and is strictly larger than $\omega$ .

Both values are computed below from the recursive clauses, and both pictures are justified by the order-type lemmas rather than left as pictures.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal addition $\alpha + \beta$ and Ordinal multiplication $\alpha \cdot \beta$ , and $\omega$ the least limit ordinal ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\alpha \cdot 0 = 0$ , $\alpha \cdot \delta^{+} = \alpha \cdot \delta + \alpha$ , and $\alpha \cdot \lambda = \bigcup\{\alpha \cdot \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal multiplication $\alpha \cdot \beta$ ); $\alpha + 0 = \alpha$ (Ordinal addition $\alpha + \beta$ ).

[L2]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : $1 \cdot \mu = \mu \cdot 1 = \mu$ (claim (a)); $\nu < \theta$ implies $\alpha + \nu < \alpha + \theta$ (claim (b)); $\mu \le \nu$ implies $\mu\gamma \le \nu\gamma$ (claim (e)).

[L3]

For $m, n \in \omega$ the ordinal $m \cdot n$ lies in $\omega$ (claim (a) of On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product).

[L4]

$\omega$ is a limit ordinal, so $\bigcup \omega = \omega$ and $0 \in \omega$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, $\mu \subseteq \nu$ iff $\mu \in \nu$ or $\mu = \nu$ , and $\mu \notin \mu$ (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

[L5]

$\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ under last differences, that is $\beta$ copies of $\alpha$ ( $\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ ordered by last differences, that is $\beta$ copies of $\alpha$ ); $\alpha + \beta$ is the order type of a copy of $\alpha$ followed by a copy of $\beta$ ( $\alpha + \beta$ is the order type of $\alpha$ followed by $\beta$ ); order types are unique (Every well-order has a unique order type) and a strictly increasing bijection between total orders is an order isomorphism (Order embedding and order isomorphism).

Verification

technique · direct

1.1

For $n \in \omega$ the ordinal $2 \cdot n$ lies in $\omega$ by [L3], hence $2 \cdot n \subseteq \omega$ by [L4]; and $n = 1 \cdot n \le 2 \cdot n$ by [L2], since $1 \le 2$ , hence $n \subseteq 2 \cdot n$ .

L2L3L4

1.2

$\omega \cdot 2 = \omega \cdot 1^{+} = \omega \cdot 1 + \omega = \omega + \omega$ by [L1] and [L2]; and $\omega + \omega > \omega + 0 = \omega$ by [L1] and claim (b) of [L2], since $0 \in \omega$ .

L1L2L4

2.1

$2 \cdot \omega = \bigcup\{2 \cdot n : n \in \omega\}$ by [L1], and this equals $\omega$ : it is contained in $\omega$ by step 1.1, and it contains $\bigcup\{n : n \in \omega\} = \bigcup \omega = \omega$ by step 1.1 and [L4].

step 1.1L1L4

3.1

The pictures are the order-type lemmas, not extra assumptions: by [L5], $2 \cdot \omega$ is the order type of $2 \times \omega$ under last differences, which is $\omega$ blocks of two, and step 2.1 evaluates that order type as $\omega$ ; while $\omega \cdot 2$ is the order type of $\omega \times 2$ under last differences, which is two blocks of $\omega$ , and by [L5] again that is the order type of a copy of $\omega$ followed by a copy of $\omega$ , namely $\omega + \omega$ , in agreement with step 1.2.

step 2.1step 1.2L5

4.1

Therefore $2 \cdot \omega = \omega$ and $\omega \cdot 2 = \omega + \omega \ne \omega$ , so the two products differ.

step 2.1step 3.1step 1.2∎

Remarks

Which convention this depends on. Everything above uses the convention fixed in Ordinal multiplication $\alpha \cdot \beta$ , that the successor clause appends a copy of the left factor on the right. Under the opposite convention the two values are exchanged, and $2 \cdot \omega$ would be $\omega + \omega$ . Both conventions appear in the literature; this library uses the one stated, throughout.

The general statement. That ordinal multiplication is not commutative is FALSE: ordinal multiplication is commutative, whose refutation is the computation of step 2.1 and step 1.2. The related failure of right distributivity, $(1+1) \cdot \omega \ne 1 \cdot \omega + 1 \cdot \omega$ , is FALSE: $(\beta + \gamma)\cdot\alpha = \beta\cdot\alpha + \gamma\cdot\alpha$ for all ordinals and uses the same value $2 \cdot \omega = \omega$ .

Why the block picture is a proof and not an illustration. $\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ ordered by last differences, that is $\beta$ copies of $\alpha$ says the product is the order type of the block arrangement, so reading a value off the picture is legitimate once the picture is identified with $\alpha \times \beta$ under last differences. What is not legitimate is reading it off an unlabelled diagram, which is why step 3.1 names the lemma at each use.

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$\omega + \omega$ is at most countable although it is not order isomorphic to $\omega$ : order type and cardinality are different invariants

Example

The ordinal $\omega + \omega$ is at most countable as a set (Finite, countably infinite, countable, uncountable), and it is not order isomorphic to $\omega$ (Order embedding and order isomorphism).

There is no tension. Countability is a statement about bijections and ignores order; order type is a statement about order isomorphisms and is finer. Two well-orders on the same countably infinite set can have different order types: $\omega$ and $\omega + 1$ already show it, as the first example of this page records, and $\omega$ and $\omega + \omega$ show it with the two copies visible.

$\omega + \omega \;=\; \mathrm{ot}\big(\underbrace{0, 1, 2, \dots}_{\text{first copy}} \ \ \underbrace{0', 1', 2', \dots}_{\text{second copy}}\big).$

Facts & Assumptions

Given: The ordinals with the addition of Ordinal addition $\alpha + \beta$ , and $\omega = \mathbb{N}$ the least limit ordinal ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\omega + \omega = \mathrm{ot}(\omega \oplus \omega)$ , where $\omega \oplus \omega$ is the set $(\{0\} \times \omega) \cup (\{1\} \times \omega)$ with the lexicographic order that puts the second copy above the first ( $\alpha + \beta$ is the order type of $\alpha$ followed by $\beta$ ).

[L2]

Every well-order is order isomorphic to exactly one ordinal, its order type, and order isomorphic well-orders have the same order type; in particular an order isomorphism is a bijection (Every well-order has a unique order type, Order embedding and order isomorphism).

[L3]

If $A$ and $B$ are at most countable then so is $A \times B$ (A product of two at most countable sets is at most countable); a set equinumerous with an at most countable set is at most countable, and $\approx$ is symmetric and transitive (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L4]

$\omega$ is at most countable, being equinumerous with $\mathbb{N}$ by the identity, and every natural number is at most countable (Finite, countably infinite, countable, uncountable, The natural numbers $\mathbb{N}$ (von Neumann)).

[L5]

$\nu < \theta$ implies $\alpha + \nu < \alpha + \theta$ , and $\alpha + 0 = \alpha$ (claims (a) and (b) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ , Ordinal addition $\alpha + \beta$ ).

[L6]

$0 \in \omega$ ; every ordinal is transitive and $\mu \notin \mu$ (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals, Successor and limit ordinals).

Verification

technique · direct

1.1

As a set, $\omega \oplus \omega = (\{0\} \times \omega) \cup (\{1\} \times \omega) = 2 \times \omega$ , since $2 = \{0, 1\}$ ; and $2 \times \omega$ is at most countable by [L3] and [L4], both factors being at most countable.

L1L3L4

1.2

$\omega + \omega \ne \omega$ : since $0 \in \omega$ , [L5] gives $\omega = \omega + 0 < \omega + \omega$ , and $\mu \notin \mu$ by [L6].

L5L6

2.1

The order isomorphism of [L1] and [L2] from $\omega \oplus \omega$ onto the ordinal $\omega + \omega$ is in particular a bijection, so $\omega + \omega$ is equinumerous with $2 \times \omega$ and hence at most countable by step 1.1 and [L3].

step 1.1L1L2L3

2.2

$\omega + \omega$ is not order isomorphic to $\omega$ : order isomorphic well-orders have the same order type by [L2], and $\omega$ and $\omega + \omega$ are distinct ordinals by step 1.2, each being its own order type.

step 1.2L2

3.1

So $\omega + \omega$ is an at most countable set carrying a well-order that is not a copy of the well-order $\omega$ : cardinality and order type are different invariants.

step 2.1step 2.2∎

Remarks

How far this goes. Every ordinal strictly below $\omega_1$ is at most countable ( $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF), and there are a great many of them: $\omega + \omega$ and $\omega \cdot \omega = \omega^{2}$ are at most countable by the argument above, and no two distinct ordinals are isomorphic as orders. So a single countably infinite set carries uncountably many mutually non-isomorphic well-orders, one for each infinite ordinal below $\omega_1$ . Nothing here says the same of $\omega^{\omega}$ or of $\varepsilon_0$ ( $\omega^{2}$ , $\omega^{\omega}$ , and $\varepsilon_0 = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, \dots\}$ satisfying $\omega^{\varepsilon_0} = \varepsilon_0$ ): those are ordinals of larger order type, and their cardinality is a question no item on these pages settles, as that item's last remark records.

Why the argument does not need a choice principle. The bijection used at step 2.1 is the collapsing isomorphism of Every well-order has a unique order type, which is unique and therefore never chosen, and A product of two at most countable sets is at most countable is choice free too. Countability of a countable union of countable sets is a different matter and does cost $\mathrm{AC}_\omega$ ; that is Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$ and it is not used here.

The confusion this item exists to prevent. " $\omega + \omega$ is bigger than $\omega$ " is true as a statement about ordinals, where bigger means further along the ordinal order, and false as a statement about size. The two readings of "bigger" are exactly order type and cardinality.

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$\omega^{2}$ , $\omega^{\omega}$ , and $\varepsilon_0 = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, \dots\}$ satisfying $\omega^{\varepsilon_0} = \varepsilon_0$

Example

The first few powers of $\omega$ are

$\omega^{1} = \omega, \qquad \omega^{2} = \omega \cdot \omega, \qquad \omega^{\omega} = \sup\{\, \omega^{n} : 0 < n < \omega \,\},$

and each is strictly larger than the one before ( $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$ ; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ , clause (b)). Iterating the exponential produces the $\omega$ -tower

$t_0 = \omega, \qquad t_{n+1} = \omega^{t_n},$

so $t_1 = \omega^{\omega}$ , $t_2 = \omega^{\omega^{\omega}}$ , and so on. Its supremum

$\varepsilon_0 \;:=\; \sup\{\, t_n : n \in \omega \,\}$

is a limit ordinal satisfying

$\omega^{\varepsilon_0} = \varepsilon_0,$

so it is a fixed point of $\beta \mapsto \omega^{\beta}$ . This is exhibited here by hand: the tower is written down, its supremum is taken, and the fixed point equation is proved from continuity at limits. No fixed-point theorem is used, and none that this library proves applies here: every fixed-point theorem on disk is stated for a set carrying an order or a metric, whereas $\beta \mapsto \omega^{\beta}$ is a class operation on the ordinals, which are not a set.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal multiplication $\alpha \cdot \beta$ and Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ , and $\omega$ the least limit ordinal ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\alpha^{0} = 1$ , $\alpha^{\delta^{+}} = \alpha^{\delta} \cdot \alpha$ , and $\alpha^{\lambda} = \sup\{\alpha^{\beta} : 0 < \beta < \lambda\}$ for limit $\lambda$ (Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ).

[L2]

For $\alpha > 1$ : $\beta < \gamma$ implies $\alpha^{\beta} < \alpha^{\gamma}$ ; and $\alpha^{\lambda} = \sup\{\alpha^{\eta} : \eta \in D\}$ for every nonempty $D \subseteq \lambda$ with $\sup D = \lambda$ (claims (b) and (c) of $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$ ; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ ). Also $\alpha^{1} = \alpha$ (claim (a) of the same).

[L3]

Recursion along the ordinals: a class rule defined on functions with ordinal domain determines exactly one class function on the ordinals (Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal).

[L4]

$\bigcup A$ is an ordinal and the least upper bound of a set $A$ of ordinals; $\mu \subseteq \nu$ iff $\mu \in \nu$ or $\mu = \nu$ ; $\mu \notin \mu$ ; and $\mu < \nu$ iff $\mu^{+} \le \nu$ (Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals, Ordinal (von Neumann)).

[L5]

$\omega$ is a limit ordinal, closed under successor, with $1 \in \omega$ , and every ordinal in $\omega$ is $0$ or a successor ( $\omega$ is the least limit ordinal, Successor and limit ordinals); every ordinal is exactly one of $0$ , a successor, or a limit (Successor and limit ordinals).

[L6]

$\mu \cdot 1 = \mu$ (claim (a) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ ).

[L7]

Induction on $\mathbb{N}$ : a subset of $\mathbb{N}$ containing $0$ and closed under $\sigma$ equals $\mathbb{N}$ , and $\sigma(n) = n^{+}$ (The principle of mathematical induction, The natural numbers $\mathbb{N}$ (von Neumann)).

Verification

technique · direct

1.1

$\omega^{1} = \omega$ by [L2]; $\omega^{2} = \omega^{1^{+}} = \omega^{1} \cdot \omega = \omega \cdot \omega$ by [L1]; and $\omega < \omega^{\omega}$ , since $1 < \omega$ gives $\omega = \omega^{1} < \omega^{\omega}$ by [L2] with base $\omega > 1$ .

L1L2L5L6

1.2

Define a class function $G$ on functions $h$ with ordinal domain $\beta$ by $G(h) = \omega$ if $\beta = 0$ , $G(h) = \omega^{h(\delta)}$ if $\beta = \delta^{+}$ , and $G(h) = 0$ if $\beta$ is a limit; the three cases are exhaustive and exclusive by [L5], so [L3] gives a unique class function $F$ on the ordinals with $F(0) = \omega$ and $F(\delta^{+}) = \omega^{F(\delta)}$ . Write $t_n = F(n)$ for $n \in \omega$ ; then $t_0 = \omega$ , $t_{n^{+}} = \omega^{t_n}$ , and $T = \{t_n : n \in \omega\}$ is a set by Replacement.

L3L5construct

2.1

$t_n < t_{n^{+}}$ for every $n \in \omega$ : let $S = \{n \in \omega : t_n < t_{n^{+}}\}$ ; then $0 \in S$ , because $t_0 = \omega < \omega^{\omega} = t_{0^{+}}$ by step 1.1 and step 1.2; and $n \in S$ implies $\sigma(n) = n^{+} \in S$ , because applying [L2] with base $\omega > 1$ to $t_n < t_{n^{+}}$ gives $\omega^{t_n} < \omega^{t_{n^{+}}}$ , that is $t_{n^{+}} < t_{(n^{+})^{+}}$ ; so $S = \omega$ by [L7].

step 1.1step 1.2L2L7

3.1

$\varepsilon_0 := \bigcup T$ is an ordinal by [L4]; each $t_n$ satisfies $t_n < t_{n^{+}} \subseteq \varepsilon_0$ by step 2.1 and [L4], so $T \subseteq \varepsilon_0$ and $T$ is nonempty with $\sup T = \varepsilon_0$ ; $\varepsilon_0 \ne 0$ because $t_0 = \omega \in \varepsilon_0$ ; and $\varepsilon_0$ is not a successor, since $\varepsilon_0 = \mu^{+}$ would put $\mu \in t_n$ for some $n$ , whence $\mu^{+} \le t_n < t_{n^{+}} \le \varepsilon_0 = \mu^{+}$ by [L4], which [L4] forbids; so $\varepsilon_0$ is a limit ordinal by [L5].

step 2.1step 1.2L4L5

4.1

$\omega^{\varepsilon_0} = \varepsilon_0$ : by [L2] with base $\omega > 1$ applied to the nonempty $D = T \subseteq \varepsilon_0$ with $\sup T = \varepsilon_0$ , one gets $\omega^{\varepsilon_0} = \sup\{\omega^{t_n} : n \in \omega\} = \sup\{t_{n^{+}} : n \in \omega\}$ by step 1.2; and that supremum is $\varepsilon_0$ , because each $t_{n^{+}} \subseteq \varepsilon_0$ while conversely $t_n \subseteq t_{n^{+}}$ for every $n$ by step 2.1 and [L4], so the union over the shifted family contains $\bigcup T = \varepsilon_0$ .

step 3.1step 2.1step 1.2L2L4

5.1

So $\omega^{2} = \omega \cdot \omega$ , the tower $t_0 = \omega$ , $t_{n^{+}} = \omega^{t_n}$ is strictly increasing, and its supremum $\varepsilon_0$ is a limit ordinal with $\omega^{\varepsilon_0} = \varepsilon_0$ .

step 4.1step 3.1step 2.1step 1.1∎

Remarks

Why Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal and not the recursion theorem over $\mathbb{N}$ . The published The recursion theorem builds $g : \mathbb{N} \to A$ from a function $f : A \to A$ on a set $A$ . Here the step is $\beta \mapsto \omega^{\beta}$ , a class operation with no set-sized codomain available at this point, so the recursion theorem does not apply as stated. Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal is exactly the class-valued version, and Replacement then makes the range $\{t_n : n \in \omega\}$ a set.

What is proved and what is not. That $\varepsilon_0$ is a fixed point of $\beta \mapsto \omega^{\beta}$ is proved above. That it is the least such fixed point is true and is not proved here; it would follow from the observation that any fixed point is closed under the tower, and it needs nothing new, but nothing on these pages uses it. No general theory of normal functions or of the Veblen hierarchy is developed, and none is needed for the statement above.

$\varepsilon_0$ and the Cantor normal form. By Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way every nonzero ordinal has a unique base- $\omega$ normal form. For $\varepsilon_0$ that form is $\omega^{\varepsilon_0} \cdot 1$ , whose exponent is $\varepsilon_0$ itself, so the normal form does not reduce $\varepsilon_0$ to strictly smaller data. Below $\varepsilon_0$ it always does, and that is the sense in which $\varepsilon_0$ is where base- $\omega$ notation runs out.

Cardinality is a separate question, and this page does not settle it. Nothing above says how large $\omega^{\omega}$ or $\varepsilon_0$ is as a set. Showing them at most countable would need the countable ordinals to be closed under ordinal exponentiation, which is not proved anywhere in this library; the natural route runs through Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable and a transfinite induction that no item here carries out. What the tower demonstrates is growth in order type, which is the invariant this page is about.

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The Cantor normal form of $(\omega^{2} + \omega\cdot 3 + 5) \cdot \omega^{2}$ , computed by the division algorithm

Example

Put $\alpha = \omega^{2} + \omega \cdot 3 + 5$ , which is already in Cantor normal form (Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way): the exponents $2 > 1 > 0$ strictly decrease and the coefficients $1, 3, 5$ are nonzero natural numbers, since $\omega \cdot 3 = \omega^{1} \cdot 3$ and $5 = \omega^{0} \cdot 5$ . Then

$\alpha \cdot \omega^{2} = \omega^{4},$

whose Cantor normal form is $\omega^{4} \cdot 1$ : the entire tail $\omega \cdot 3 + 5$ is annihilated by multiplying on the right by $\omega^{2}$ . Multiplying on the right by a limit ordinal keeps only the leading behaviour.

Addition behaves quite differently. Also computed below:

$\big(\omega^{2} + \omega \cdot 3 + 5\big) + \big(\omega \cdot 2 + 7\big) = \omega^{2} + \omega \cdot 5 + 7,$

where the coefficients of the matching power add and the lower tail of the left summand is swallowed.

Facts & Assumptions

Given: $\alpha = \omega^{2} + \omega \cdot 3 + 5$ , with the operations of Ordinal addition $\alpha + \beta$ , Ordinal multiplication $\alpha \cdot \beta$ and Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ; the finite ordinals $2, 3, 4, 5, 7$ are elements of $\omega$ (The natural numbers $\mathbb{N}$ (von Neumann), $\omega$ is the least limit ordinal). Products bind tighter than sums, so $\omega \cdot 3 + 5$ is $(\omega \cdot 3) + 5$ .

[L1]

$\alpha \cdot \delta^{+} = \alpha \cdot \delta + \alpha$ and $\alpha \cdot \lambda = \sup\{\alpha \cdot \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal multiplication $\alpha \cdot \beta$ ); $\alpha^{\delta^{+}} = \alpha^{\delta} \cdot \alpha$ and $\alpha^{0} = 1$ (Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ); $\mu + 0 = \mu$ , $\mu + \delta^{+} = (\mu + \delta)^{+}$ and $\mu + \lambda = \sup\{\mu + \xi : \xi \in \lambda\}$ for limit $\lambda$ (Ordinal addition $\alpha + \beta$ ).

[L2]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : $1 \cdot \mu = \mu \cdot 1 = \mu$ and $0 + \mu = \mu$ (claim (a)); $\nu < \theta$ implies $\mu + \nu < \mu + \theta$ (claim (b)); for $\mu > 0$ , $\nu < \theta$ implies $\mu\nu < \mu\theta$ (claim (d)); $\mu \le \nu$ implies $\mu\gamma \le \nu\gamma$ (claim (e)); $\mu + \lambda = \sup\{\mu + \xi : \xi \in \lambda\}$ and $\mu \cdot \lambda = \sup\{\mu\eta : \eta \in D\}$ for $\mu > 0$ , $\lambda$ a limit and $D \subseteq \lambda$ nonempty with $\sup D = \lambda$ (claim (f)).

[L3]

$\cdot$ is associative and $\mu(\nu + \theta) = \mu\nu + \mu\theta$ (Ordinal multiplication is associative, and $\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma$ ); $+$ is associative (Ordinal addition is associative).

[L4]

$\alpha^{\beta + \gamma} = \alpha^{\beta} \cdot \alpha^{\gamma}$ , $\alpha^{1} = \alpha$ , and for $\alpha > 1$ the map $\beta \mapsto \alpha^{\beta}$ is strictly increasing ( $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$ ; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ ).

[L5]

For $\mu > 0$ every $\nu$ is $\mu \cdot \xi + \rho$ with $\rho < \mu$ , uniquely (For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way); every nonzero ordinal has exactly one Cantor normal form (Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way).

[L6]

For $m, n \in \omega$ the ordinals $m + n$ and $m \cdot n$ lie in $\omega$ and agree with the natural-number sum and product (On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product).

[L7]

$\omega$ is a limit ordinal with $\bigcup \omega = \omega$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, $\mu \subseteq \nu$ iff $\mu \in \nu$ or $\mu = \nu$ , and trichotomy holds (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

Verification

technique · direct

1.1

Preliminary identities: $\omega^{1} = \omega$ and $\omega^{2} = \omega^{1^{+}} = \omega^{1} \cdot \omega = \omega \cdot \omega$ by [L1] and [L4]; $\omega \cdot 3 + \omega = \omega \cdot 3^{+} = \omega \cdot 4$ , $\omega + \omega = \omega \cdot 1 + \omega = \omega \cdot 1^{+} = \omega \cdot 2$ , and $\omega^{2} + \omega^{2} = \omega^{2} \cdot 1 + \omega^{2} = \omega^{2} \cdot 2$ by [L1] and [L2].

L1L2L4

1.2

For $n \in \omega$ the ordinals $2 \cdot n$ and $5 + n$ lie in $\omega$ by [L6], hence are subsets of $\omega$ by [L7]; and $n = 1 \cdot n \le 2 \cdot n$ and $n \le 5 + n$ by [L2], so both $\{2 \cdot n : n \in \omega\}$ and $\{5 + n : n \in \omega\}$ are nonempty subsets of $\omega$ with supremum $\bigcup \omega = \omega$ .

L2L6L7

2.1

$\omega^{2} \le \alpha < \omega^{2} \cdot 2$ : the first inequality is $\omega^{2} = \omega^{2} + 0 \le \omega^{2} + (\omega \cdot 3 + 5)$ by [L1] and [L2]; for the second, $5 < \omega$ gives $\omega \cdot 3 + 5 < \omega \cdot 3 + \omega = \omega \cdot 4$ by [L2] and step 1.1, and $4 < \omega$ with $\omega > 0$ gives $\omega \cdot 4 < \omega \cdot \omega = \omega^{2}$ by [L2] and step 1.1, so $\alpha < \omega^{2} + \omega^{2} = \omega^{2} \cdot 2$ by [L2] and step 1.1.

step 1.1L1L2L7

2.2

The addition: $5 + \omega = \sup\{5 + n : n \in \omega\} = \omega$ by [L1] and step 1.2, so $5 + \omega \cdot 2 = 5 + (\omega + \omega) = (5 + \omega) + \omega = \omega + \omega = \omega \cdot 2$ using [L3] and step 1.1; hence $\alpha + (\omega \cdot 2 + 7) = \omega^{2} + \omega \cdot 3 + (5 + \omega \cdot 2) + 7 = \omega^{2} + (\omega \cdot 3 + \omega \cdot 2) + 7 = \omega^{2} + \omega \cdot (3 + 2) + 7 = \omega^{2} + \omega \cdot 5 + 7$ , the regrouping by [L3], the last product by left distributivity in [L3] and $3 + 2 = 5$ by [L6].

step 1.2step 1.1L1L3L6

3.1

$\alpha \cdot \omega = \omega^{3}$ : by [L1], $\alpha \cdot \omega = \sup\{\alpha \cdot n : n \in \omega\}$ , and step 2.1 with claim (e) of [L2] gives $\omega^{2} \cdot n \le \alpha \cdot n \le (\omega^{2} \cdot 2) \cdot n = \omega^{2} \cdot (2 \cdot n)$ for every $n \in \omega$ , using associativity of $\cdot$ from [L3]; the suprema of the outer two families are both $\omega^{2} \cdot \omega$ , the first by [L1] and the second by claim (f) of [L2] applied to the set $\{2 \cdot n : n \in \omega\}$ , which is unbounded in $\omega$ by step 1.2; so $\alpha \cdot \omega = \omega^{2} \cdot \omega = \omega^{2} \cdot \omega^{1} = \omega^{2 + 1} = \omega^{3}$ by [L4].

step 2.1step 1.2step 1.1L1L2L3L4

4.1

$\alpha \cdot \omega^{2} = \omega^{4}$ : by step 1.1, $\omega^{2} = \omega \cdot \omega$ , so associativity in [L3] gives $\alpha \cdot \omega^{2} = \alpha \cdot (\omega \cdot \omega) = (\alpha \cdot \omega) \cdot \omega = \omega^{3} \cdot \omega = \omega^{3} \cdot \omega^{1} = \omega^{3 + 1} = \omega^{4}$ by step 3.1 and [L4].

step 3.1step 1.1L3L4

5.1

The Cantor normal form of $\omega^{4}$ is $\omega^{4} \cdot 1$ : the largest $\beta$ with $\omega^{\beta} \le \omega^{4}$ is $4$ , because $\beta \mapsto \omega^{\beta}$ is strictly increasing by [L4] and $4 < \beta$ would give $\omega^{4} < \omega^{\beta}$ ; dividing by $\omega^{4} > 0$ as in [L5] gives $\omega^{4} = \omega^{4} \cdot 1 + 0$ with remainder $0 < \omega^{4}$ , so the datum of length $1$ with exponent $4$ and coefficient $1$ has value $\omega^{4}$ , and it is the only normal form by the uniqueness in [L5].

step 4.1L1L2L4L5

6.1

So $(\omega^{2} + \omega \cdot 3 + 5) \cdot \omega^{2} = \omega^{4}$ , with Cantor normal form $\omega^{4} \cdot 1$ , while $(\omega^{2} + \omega \cdot 3 + 5) + (\omega \cdot 2 + 7) = \omega^{2} + \omega \cdot 5 + 7$ .

step 5.1step 4.1step 2.2∎

Remarks

Why the tail vanishes under multiplication on the right. The squeeze in step 3.1 is the whole mechanism: $\alpha$ is trapped between $\omega^{2}$ and $\omega^{2} \cdot 2$ , and multiplying either bound on the right by $\omega$ gives $\omega^{3}$ , because $\omega^{2} \cdot 2 \cdot n = \omega^{2} \cdot (2n)$ runs through the same cofinal family as $\omega^{2} \cdot n$ . Anything of the form "leading term plus smaller stuff" therefore behaves, under multiplication by a limit on the right, exactly like its leading term.

Why the tail does not vanish under addition. In step 2.2 the lower part of the left summand, here $5$ , is absorbed by the leading term of the right summand, but the term $\omega^{2}$ of the left summand survives, and the two coefficients of $\omega^{1}$ add. The general rule is the same computation: adding on the left of $\omega^{\beta}$ anything smaller than $\omega^{\beta}$ changes nothing, which is the additive indecomposability used inside the proof of Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way.

A check on the answer. $\omega^{4}$ has Cantor normal form of length $1$ , so $\alpha \cdot \omega^{2}$ is a power of $\omega$ ; that is consistent with the previous remark, since $\alpha$ has leading term $\omega^{2} \cdot 1$ and $\omega^{2} \cdot \omega^{2} = \omega^{4}$ by the sum law in [L4].

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Solving $\omega + \gamma = \omega\cdot 2$ and dividing $\omega^{2} + \omega + 3$ by $\omega$

Example

Two computations with the two "inverse" operations of this page.

Left subtraction. The equation $\omega + \gamma = \omega \cdot 2$ has exactly one solution, and it is $\gamma = \omega$ . Existence and uniqueness are For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ , applicable because $\omega \le \omega \cdot 2$ ; finding the solution is the computation $\omega \cdot 2 = \omega + \omega$ .

Division with remainder. Dividing $\omega^{2} + \omega + 3$ by $\omega$ gives

$\omega^{2} + \omega + 3 \;=\; \omega \cdot (\omega + 1) + 3, \qquad 3 < \omega,$

so the quotient is $\omega + 1$ and the remainder is $3$ , and by the uniqueness in For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way there is no other answer. The step that does the work is $\omega^{2} + \omega = \omega \cdot (\omega + 1)$ , which is left distributivity.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal addition $\alpha + \beta$ , Ordinal multiplication $\alpha \cdot \beta$ and Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ; $\omega$ is the least limit ordinal and $3 \in \omega$ ( $\omega$ is the least limit ordinal, The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

$\mu \cdot \delta^{+} = \mu \cdot \delta + \mu$ and $\mu \cdot 0 = 0$ (Ordinal multiplication $\alpha \cdot \beta$ ); $\mu + 1 = \mu^{+}$ and $\mu + 0 = \mu$ (Ordinal addition $\alpha + \beta$ ); $\mu^{\delta^{+}} = \mu^{\delta} \cdot \mu$ and $\mu^{0} = 1$ (Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ).

[L2]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : $1 \cdot \mu = \mu \cdot 1 = \mu$ (claim (a)); $\nu < \theta$ implies $\mu + \nu < \mu + \theta$ , and $\mu \le \mu + \nu$ (claim (b)).

[L3]

$\mu(\nu + \theta) = \mu\nu + \mu\theta$ (Ordinal multiplication is associative, and $\alpha \cdot (\beta + \gamma) = \alpha\cdot\beta + \alpha\cdot\gamma$ ).

[L4]

$\mu^{1} = \mu$ ( $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$ ; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ ).

[L5]

For $\mu \le \nu$ there is exactly one $\gamma$ with $\mu + \gamma = \nu$ (For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ ).

[L6]

For $\mu > 0$ every $\nu$ is $\mu \cdot \xi + \rho$ with $\rho < \mu$ , in exactly one way (For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way).

[L7]

$\omega$ is a limit ordinal, so $\omega \ne 0$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals); trichotomy and the elementary ordinal facts (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

Verification

technique · direct

1.1

$\omega \cdot 2 = \omega \cdot 1^{+} = \omega \cdot 1 + \omega = \omega + \omega$ by [L1] and [L2], and $\omega \le \omega + \omega$ by [L2].

L1L2

1.2

$\omega^{2} = \omega^{1^{+}} = \omega^{1} \cdot \omega = \omega \cdot \omega$ by [L1] and [L4].

L1L4

2.1

Left subtraction: $\omega \le \omega \cdot 2$ by step 1.1, so [L5] gives exactly one $\gamma$ with $\omega + \gamma = \omega \cdot 2$ ; and $\gamma = \omega$ works, since $\omega + \omega = \omega \cdot 2$ by step 1.1, so $\gamma = \omega$ is the solution.

step 1.1L5

2.2

$\omega^{2} + \omega = \omega \cdot \omega + \omega \cdot 1 = \omega \cdot (\omega + 1)$ by step 1.2, [L2] and left distributivity [L3].

step 1.2L2L3

3.1

Division: $\omega > 0$ by [L7], and $\omega^{2} + \omega + 3 = \omega \cdot (\omega + 1) + 3$ by step 2.2, with $3 < \omega$ ; so by the uniqueness in [L6] the quotient of $\omega^{2} + \omega + 3$ by $\omega$ is $\omega + 1$ and the remainder is $3$ .

step 2.2L6L7

4.1

The unique solution of $\omega + \gamma = \omega \cdot 2$ is $\gamma = \omega$ , and dividing $\omega^{2} + \omega + 3$ by $\omega$ gives quotient $\omega + 1$ and remainder $3$ .

step 3.1step 2.1∎

Remarks

Uniqueness is what makes "the answer" meaningful. Both computations exhibit a solution and then quote a uniqueness theorem. Without For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha + \gamma = \beta$ the equation $\omega + \gamma = \omega \cdot 2$ would only be known to have a solution; without For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way the pair $(\omega + 1, 3)$ would be an answer among possibly many. Both theorems are proved from left cancellation, which is the one cancellation law ordinal addition has.

The equation on the other side has no solution at all. There is no $\gamma$ with $\gamma + \omega = \omega + 1$ , because $\gamma + \omega$ is a limit ordinal for every $\gamma$ (claim (g) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ ) while $\omega + 1 = \omega^{+}$ is a successor. So ordinal subtraction genuinely exists only on the left, and the same asymmetry is what forces the quotient in For $\alpha > 0$ every ordinal $\beta$ is $\alpha \cdot \xi + \rho$ with $\rho < \alpha$ , in exactly one way to be written on the right of $\alpha$ .

Reading the division off the Cantor normal form. $\omega^{2} + \omega + 3$ has normal form with exponents $2 > 1 > 0$ and coefficients $1, 1, 3$ (Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way). In this instance, dividing by $\omega = \omega^{1}$ has put the term of exponent $0$ into the remainder and lowered each of the two remaining exponents by one: $\omega^{2}$ became $\omega$ and $\omega$ became $1$ , which is exactly the quotient $\omega + 1$ , while $3$ is the constant term. That pattern is what the general division algorithm is doing, but no general statement of it is claimed here.

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Assuming countable choice, a strictly increasing $\omega$ -sequence of countable ordinals has a countable supremum, which is a countable limit ordinal below $\omega_1$ ; the instance $\sup_n \omega\cdot(n+1) = \omega^{2}$ needs no choice

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )). Let $(\alpha_n)_{n \in \omega}$ be a strictly increasing sequence of ordinals with every $\alpha_n \in \omega_1$ (The first uncountable ordinal $\omega_1 := \aleph(\omega)$ ). Then

$\sup_{n \in \omega} \alpha_n \;=\; \bigcup\{\alpha_n : n \in \omega\}$

is again an ordinal below $\omega_1$ , hence at most countable (Finite, countably infinite, countable, uncountable), and it is a limit ordinal (Successor and limit ordinals), since a strictly increasing sequence never attains its supremum.

A concrete instance, with everything computed:

$\alpha_n = \omega \cdot (n + 1), \qquad \sup_{n \in \omega} \omega \cdot (n+1) = \omega \cdot \omega = \omega^{2}.$

So $\omega^{2}$ is a countable limit ordinal strictly below $\omega_1$ , reached from below by an $\omega$ -sequence. That is exactly what Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable forbids for $\omega_1$ itself: $\omega_1$ is not the supremum of any such sequence.

Facts & Assumptions

Given: The Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )), a strictly increasing sequence $(\alpha_n)_{n \in \omega}$ of ordinals in $\omega_1$ , and the operations of Ordinal addition $\alpha + \beta$ , Ordinal multiplication $\alpha \cdot \beta$ and Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ ; here $n + 1 = n^{+}$ (Ordinal addition $\alpha + \beta$ ).

[L1]

Assuming $\mathrm{AC}_\omega$ : every at most countable $A \subseteq \omega_1$ has $\sup A = \bigcup A \in \omega_1$ , and $\alpha \le \sup A$ for every $\alpha \in A$ (Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).

[L2]

$\omega_1$ is uncountable, every ordinal in $\omega_1$ is at most countable, and $\omega_1$ is a limit ordinal ( $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF).

[L3]

A nonempty set is at most countable if and only if it is a surjective image of $\mathbb{N}$ (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$ ); a subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable); a product of two at most countable sets is at most countable (A product of two at most countable sets is at most countable); a set equinumerous with an at most countable set is at most countable (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L4]

$\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ under last differences, and an order isomorphism is in particular a bijection ( $\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ ordered by last differences, that is $\beta$ copies of $\alpha$ , Every well-order has a unique order type).

[L5]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ : for $\mu > 0$ , $\nu < \theta$ implies $\mu\nu < \mu\theta$ (claim (d)); $\mu \cdot \theta = \sup\{\mu\eta : \eta \in D\}$ for $\mu > 0$ , $\theta$ a limit and $D \subseteq \theta$ nonempty with $\sup D = \theta$ (claim (f)); $1 \cdot \mu = \mu$ (claim (a)).

[L6]

$\mu^{1^{+}} = \mu^{1} \cdot \mu$ and $\mu^{1} = \mu$ (Ordinal exponentiation $\alpha^{\beta}$ , with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$ , Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$ ); $\omega$ is a limit ordinal, closed under successor, with $\bigcup \omega = \omega$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals, The natural numbers $\mathbb{N}$ (von Neumann)).

[L7]

$\bigcup A$ is an ordinal and the least upper bound of a set $A$ of ordinals; $\mu \subseteq \nu$ iff $\mu \in \nu$ or $\mu = \nu$ ; $\mu \notin \mu$ ; $\mu < \nu$ iff $\mu^{+} \le \nu$ ; and trichotomy holds (Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals, Ordinal (von Neumann)).

Verification

technique · direct

1.1

The set $A = \{\alpha_n : n \in \omega\}$ is a nonempty subset of $\omega_1$ and is at most countable, being the image of $\mathbb{N}$ under $n \mapsto \alpha_n$ , hence a surjective image of $\mathbb{N}$ onto $A$ , so [L3] applies.

L3given

1.2

An ordinal $\mu$ lies in $\omega_1$ if and only if it is at most countable: one direction is [L2]; conversely if $\mu \notin \omega_1$ then $\omega_1 \le \mu$ by [L7], so $\omega_1 \subseteq \mu$ and $\omega_1$ would be at most countable by [L3], contradicting [L2].

L2L3L7

2.1

By [L1] the ordinal $\beta = \sup A = \bigcup A$ lies in $\omega_1$ and is an upper bound of $A$ , so $\beta$ is at most countable by [L2].

step 1.1L1L2

2.2

The instance: each $\omega \cdot (n+1)$ is at most countable, because by [L4] it is order isomorphic, hence equinumerous, to $\omega \times (n+1)$ , which is at most countable by [L3]; so $\omega \cdot (n+1) \in \omega_1$ by step 1.2. The sequence is strictly increasing, since $n + 1 < m + 1$ for $n < m$ gives $\omega \cdot (n+1) < \omega \cdot (m+1)$ by [L5] with $\omega > 0$ .

step 1.2L3L4L5

3.1

$\beta$ is a limit ordinal: it is nonzero because $\alpha_0 \le \alpha_1 \le \beta$ with $\alpha_0 < \alpha_1$ , so $\beta \ne 0$ ; and it is not a successor, since $\beta = \mu^{+}$ would put $\mu \in \alpha_n$ for some $n$ , whence $\mu^{+} \le \alpha_n < \alpha_{n^{+}} \le \beta = \mu^{+}$ by [L7] and strict increase, which [L7] forbids.

step 2.1L7given

3.2

Its supremum is $\omega^{2}$ : the set $\{n + 1 : n \in \omega\}$ is a nonempty subset of $\omega$ with supremum $\omega$ , because $\omega$ is closed under successor and $\bigcup \omega = \omega$ by [L6], so claim (f) of [L5] with $\mu = \omega > 0$ gives $\omega \cdot \omega = \sup\{\omega \cdot (n+1) : n \in \omega\}$ ; and $\omega \cdot \omega = \omega^{1} \cdot \omega = \omega^{1^{+}} = \omega^{2}$ by [L6].

step 2.2L5L6

4.1

So for a strictly increasing $\omega$ -sequence in $\omega_1$ the supremum is a limit ordinal below $\omega_1$ and is at most countable; concretely $\sup_{n} \omega \cdot (n+1) = \omega^{2}$ , a countable limit ordinal below $\omega_1$ .

step 2.1step 3.1step 2.2step 3.2∎

Remarks

Where the choice principle is and is not needed. The general statement uses $\mathrm{AC}_\omega$ , at the single step where Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable is applied. The concrete instance does not: $\omega^{2}$ is shown at most countable directly, from $\alpha \cdot \beta$ is the order type of $\alpha \times \beta$ ordered by last differences, that is $\beta$ copies of $\alpha$ and A product of two at most countable sets is at most countable, both of which are choice free. So the example is available in ZF and only the general statement carries the hypothesis.

The contrast with $\omega_1$ itself. $\omega_1$ is also a limit ordinal, and it is also the supremum of the ordinals below it; what fails there, under $\mathrm{AC}_\omega$ , is that no at most countable family of them suffices. That is Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable again, and it is not a theorem of ZF alone: consistently with ZF the first uncountable ordinal is the supremum of an $\omega$ -sequence of at most countable ordinals, so no choice-free proof of it exists (Choice ledger for this page: $\omega_1$ exists in ZF, and the boundedness theorem does not).

Strict increase is used only for the limit clause. Boundedness needs only that the set of values is at most countable; strictness is what makes the supremum unattained and hence a limit ordinal. A sequence that is eventually constant has its final value as supremum, and that value need not be a limit ordinal at all, which is why step 2.2 quotes the strictness hypothesis.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)‡ rests on unproved material (inherited)Open item page →

‡ Rests on 6 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Cohen 1963: ZF does not prove the Axiom of Choice, ‡ Feferman 1965: ZF does not prove that a free ultrafilter on the naturals exists, ‡ Gödel 1938: ZF does not refute the Axiom of Choice, ‡ Halpern and Lévy 1971: the Boolean prime ideal theorem does not imply the Axiom of Choice and ‡ Schechter 2006: Kelley's cofinite proof yields BPI, not the Axiom of Choice. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Refuted: every limit ordinal has an at most countable cofinal subset — $\omega_1$ has none, assuming countable choice

Statement refuted

False claim: every limit ordinal has an at most countable cofinal subset (Cofinal subset of an ordinal, Finite, countably infinite, countable, uncountable).

The claim is plausible because every limit ordinal a reader meets first does have one. $\omega$ is cofinal in itself and at most countable; and every at most countable limit ordinal $\lambda$ is cofinal in itself and at most countable, so the claim holds for all of them, and $\omega + \omega$ and $\omega^{2}$ are among them, both being shown at most countable earlier on this page. Whether $\omega^{\omega}$ and $\varepsilon_0$ are at most countable is a question no item on these pages settles, so neither is offered here as an instance.

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )). The first uncountable ordinal $\omega_1$ (The first uncountable ordinal $\omega_1 := \aleph(\omega)$ ) refutes the claim: it is a limit ordinal, and no at most countable subset of it is cofinal in it.

The hypothesis is not removable, and the item states it in the title: without a choice principle the refutation itself fails, since consistently with ZF the ordinal $\omega_1$ is the supremum of an $\omega$ -sequence of at most countable ordinals. That is recorded in Choice ledger for this page: $\omega_1$ exists in ZF, and the boundedness theorem does not.

Facts & Assumptions

Given: The Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )) and $\omega_1$ , the first uncountable ordinal (The first uncountable ordinal $\omega_1 := \aleph(\omega)$ ).

[L1]

$C \subseteq \alpha$ is cofinal in $\alpha$ when every $\xi \in \alpha$ satisfies $\xi \le \eta$ for some $\eta \in C$ (Cofinal subset of an ordinal).

[L2]

$\omega_1$ is uncountable, every ordinal in $\omega_1$ is at most countable, and $\omega_1$ is a limit ordinal ( $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF, Successor and limit ordinals).

[L3]

Assuming $\mathrm{AC}_\omega$ : no at most countable subset of $\omega_1$ is cofinal in $\omega_1$ (claim (b) of Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).

[L4]

$\omega$ is a limit ordinal and is at most countable, being $\mathbb{N}$ ( $\omega$ is the least limit ordinal, Finite, countably infinite, countable, uncountable, The natural numbers $\mathbb{N}$ (von Neumann)).

Counterexample

technique · direct

1.1

The claim does hold for every at most countable limit ordinal $\lambda$ : the set $\lambda$ itself is a subset of $\lambda$ , it is at most countable by hypothesis, and it is cofinal in $\lambda$ by [L1], since every $\xi \in \lambda$ satisfies $\xi \le \xi \in \lambda$ . In particular it holds at $\lambda = \omega$ by [L4].

L1L4

1.2

$\omega_1$ is a limit ordinal by [L2], so it is an instance of the claim.

L2

2.1

No at most countable $C \subseteq \omega_1$ is cofinal in $\omega_1$ , by [L3]; so the claim fails at $\omega_1$ .

step 1.2L1L3

3.1

Therefore $\omega_1$ is a limit ordinal with no at most countable cofinal subset, and the claim that every limit ordinal has one is false.

step 2.1step 1.2step 1.1∎

Remarks

What separates $\omega_1$ from the countable limit ordinals. A limit ordinal is always cofinal in itself, so the claim can only fail when the ordinal is itself uncountable. $\omega_1$ is the least uncountable ordinal ( $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF), so it is the first place where the claim can fail at all, and under $\mathrm{AC}_\omega$ it does fail there.

The refutation carries the hypothesis it uses. Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable is stated under $\mathrm{AC}_\omega$ and spends it at exactly one step, so this counterexample inherits the same cost. A page that quotes this item must carry $\mathrm{AC}_\omega$ forward into its own statement; Choice ledger for this page: $\omega_1$ exists in ZF, and the boundedness theorem does not is the ledger, and it names the model in which the conclusion fails outright.

What is deliberately not said at this point in the reading order. In the later vocabulary this item says $\operatorname{cf}(\omega_1)>\omega$ , or that $\omega_1$ is regular. The cofinality and regular/singular vocabulary is introduced later in Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals ↗, so this earlier example stays in the subset language of Cofinal subset of an ordinal. Nothing is lost: the applications, such as the non-normality of the deleted Tychonoff plank, use exactly the subset form.

Ordinal Arithmetic and the First Uncountable Ordinal — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

1+ω=ω1 + \omega = \omega1+ω=ω and ω+1>ω\omega + 1 > \omegaω+1>ω, computed both from the recursion and as order types

Example

Facts & Assumptions

Verification

Remarks

2⋅ω=ω2 \cdot \omega = \omega2⋅ω=ω while ω⋅2=ω+ω\omega \cdot 2 = \omega + \omegaω⋅2=ω+ω, pictured as order types

Example

Facts & Assumptions

Verification

Remarks

ω+ω\omega + \omegaω+ω is at most countable although it is not order isomorphic to ω\omegaω: order type and cardinality are different invariants

Example

Facts & Assumptions

Verification

Remarks

ω2\omega^{2}ω2, ωω\omega^{\omega}ωω, and ε0=sup⁡{ω,ωω,ωωω,… }\varepsilon_0 = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, \dots\}ε0​=sup{ω,ωω,ωωω,…} satisfying ωε0=ε0\omega^{\varepsilon_0} = \varepsilon_0ωε0​=ε0​

Example

Facts & Assumptions

Verification

Remarks

The Cantor normal form of (ω2+ω⋅3+5)⋅ω2(\omega^{2} + \omega\cdot 3 + 5) \cdot \omega^{2}(ω2+ω⋅3+5)⋅ω2, computed by the division algorithm

Example

Facts & Assumptions

Verification

Remarks

Solving ω+γ=ω⋅2\omega + \gamma = \omega\cdot 2ω+γ=ω⋅2 and dividing ω2+ω+3\omega^{2} + \omega + 3ω2+ω+3 by ω\omegaω

Example

Facts & Assumptions

Verification

Remarks

Assuming countable choice, a strictly increasing ω\omegaω-sequence of countable ordinals has a countable supremum, which is a countable limit ordinal below ω1\omega_1ω1​; the instance sup⁡nω⋅(n+1)=ω2\sup_n \omega\cdot(n+1) = \omega^{2}supn​ω⋅(n+1)=ω2 needs no choice

Example

Facts & Assumptions

Verification

Remarks

Refuted: every limit ordinal has an at most countable cofinal subset — ω1\omega_1ω1​ has none, assuming countable choice

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

Standard references

$1 + \omega = \omega$ and $\omega + 1 > \omega$ , computed both from the recursion and as order types

$2 \cdot \omega = \omega$ while $\omega \cdot 2 = \omega + \omega$ , pictured as order types

$\omega + \omega$ is at most countable although it is not order isomorphic to $\omega$ : order type and cardinality are different invariants

$\omega^{2}$ , $\omega^{\omega}$ , and $\varepsilon_0 = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, \dots\}$ satisfying $\omega^{\varepsilon_0} = \varepsilon_0$

The Cantor normal form of $(\omega^{2} + \omega\cdot 3 + 5) \cdot \omega^{2}$ , computed by the division algorithm

Solving $\omega + \gamma = \omega\cdot 2$ and dividing $\omega^{2} + \omega + 3$ by $\omega$

Assuming countable choice, a strictly increasing $\omega$ -sequence of countable ordinals has a countable supremum, which is a countable limit ordinal below $\omega_1$ ; the instance $\sup_n \omega\cdot(n+1) = \omega^{2}$ needs no choice

Refuted: every limit ordinal has an at most countable cofinal subset — $\omega_1$ has none, assuming countable choice