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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-07-26 (claude-opus-5)
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  • AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Basic closure properties of ordinals

Statement

Let α\alpha and β\beta be ordinals (Ordinal (von Neumann)). Then:

(a) every element of α\alpha is an ordinal;

(b) αα\alpha \notin \alpha;

(c) α+=α{α}\alpha^{+} = \alpha \cup \{\alpha\} is an ordinal;

(d) if AA is a nonempty set of ordinals then A\bigcap A is an ordinal;

(e) if AA is any set of ordinals then A\bigcup A is an ordinal;

(f) αβ\alpha \subseteq \beta if and only if αβ\alpha \in \beta or α=β\alpha = \beta;

(g) any two ordinals are comparable under inclusion: αβ\alpha \subseteq \beta or βα\beta \subseteq \alpha.

Everything here is a theorem of ZF and uses no choice principle.

Facts & Assumptions

Given: Ordinals α\alpha, β\beta and, where stated, a set AA all of whose members are ordinals. Claim (g) is not in the usual list of basic facts, but claim (e) needs it, so it is proved here rather than deferred; the trichotomy statement is then read off from it on the next item of this page.

[A1]

An ordinal is a transitive set on which \in is a strict well-order: irreflexive, transitive as a relation, trichotomous, and with a least element in every nonempty subset (Ordinal (von Neumann)).

[L1]

The restriction of a strict well-order to a subset is again a strict well-order, since totality and least elements are inherited by subsets (Well-order and well-ordered set).

Proof

technique · direct
1.1

Claim (a): let xαx \in \alpha; then xαx \subseteq \alpha by transitivity of α\alpha, so \in strictly well-orders xx by [L1], and xx is transitive, because zyxz \in y \in x gives yαy \in \alpha and zαz \in \alpha by transitivity of α\alpha, whence zxz \in x by transitivity of the relation \in on α\alpha; so xx is an ordinal.

A1L1
1.2

Claim (b): if αα\alpha \in \alpha then α\alpha is an element of α\alpha satisfying αα\alpha \in \alpha, contradicting irreflexivity of \in on α\alpha; hence αα\alpha \notin \alpha.

A1
1.3

Claim (f), the easy direction: αβ\alpha \in \beta gives αβ\alpha \subseteq \beta by transitivity of β\beta, and α=β\alpha = \beta gives αβ\alpha \subseteq \beta trivially.

A1
2.1

Claim (f), the substantial direction: assume αβ\alpha \subseteq \beta and αβ\alpha \ne \beta, let γ\gamma be the \in-least element of the nonempty set βαβ\beta \setminus \alpha \subseteq \beta, and check γ=α\gamma = \alpha; indeed δγ\delta \in \gamma forces δβ\delta \in \beta by transitivity of β\beta and then δα\delta \in \alpha, since otherwise δβα\delta \in \beta \setminus \alpha with δγ\delta \in \gamma contradicts minimality, so γα\gamma \subseteq \alpha; conversely δαβ\delta \in \alpha \subseteq \beta compared with γ\gamma by trichotomy in β\beta cannot satisfy δ=γ\delta = \gamma or γδ\gamma \in \delta, since each would put γα\gamma \in \alpha, using transitivity of α\alpha in the second case, so δγ\delta \in \gamma and αγ\alpha \subseteq \gamma; hence α=γβ\alpha = \gamma \in \beta.

step 1.3A1
2.2

Claim (c): α+\alpha^{+} is transitive, because xα+x \in \alpha^{+} means xαx \in \alpha, whence xαα+x \subseteq \alpha \subseteq \alpha^{+}, or x=αα+x = \alpha \subseteq \alpha^{+}; the relation \in is irreflexive on α+\alpha^{+} by [A1] and step 1.2, transitive there because xyzx \in y \in z with z=αz = \alpha gives xyα=zx \in y \subseteq \alpha = z and with zαz \in \alpha reduces to transitivity in α\alpha, and trichotomous there because two elements of α\alpha are comparable in α\alpha while xαx \in \alpha satisfies xαx \in \alpha and neither αx\alpha \in x nor x=αx = \alpha, both of which would give αα\alpha \in \alpha; finally a nonempty Sα+S \subseteq \alpha^{+} has an \in-least element, namely the \in-least element of SαS \cap \alpha when that is nonempty and α\alpha otherwise.

step 1.2step 1.1A1
2.3

Claim (d): A\bigcap A is transitive, since xAx \in \bigcap A gives xδx \in \delta and hence xδx \subseteq \delta for every δA\delta \in A, so xAx \subseteq \bigcap A; and A\bigcap A is a subset of any fixed member of the nonempty set AA, so \in strictly well-orders it by [L1].

step 1.1A1L1
3.1

Claim (g): γ=αβ\gamma = \alpha \cap \beta is an ordinal by claim (d) applied to {α,β}\{\alpha, \beta\}, and γα\gamma \subseteq \alpha and γβ\gamma \subseteq \beta, so claim (f) gives γα\gamma \in \alpha or γ=α\gamma = \alpha, and likewise for β\beta; both memberships at once would give γαβ=γ\gamma \in \alpha \cap \beta = \gamma, contradicting claim (b), so γ=α\gamma = \alpha or γ=β\gamma = \beta, that is αβ\alpha \subseteq \beta or βα\beta \subseteq \alpha.

step 2.3step 2.1step 1.3step 1.2
4.1

Claim (e): A\bigcup A is transitive, since xδAx \in \delta \in A gives xδAx \subseteq \delta \subseteq \bigcup A; its elements are ordinals by claim (a), so \in is irreflexive on it by claim (b) and transitive on it because xyzx \in y \in z with zδAz \in \delta \in A puts x,y,zx, y, z all in the ordinal δ\delta; any two of its elements lie in a common member of AA by claim (g) and are therefore comparable, which gives trichotomy by claim (b) and claim (f); and a nonempty SAS \subseteq \bigcup A has an \in-least element, namely the \in-least element of SδS \cap \delta for any δA\delta \in A meeting SS, since an element of SS lying \in-below it would lie in δ\delta by transitivity and contradict minimality.

step 3.1step 1.1step 1.2step 2.1A1
5.1

Claims (a) to (g) are established.

step 1.1step 1.2step 1.3step 2.1step 2.2step 2.3step 3.1step 4.1

Remarks

The successor is the immediate successor. Claim (c) makes α+\alpha^{+} an ordinal, and it is the least ordinal strictly above α\alpha: any γ\gamma with αγ\alpha \in \gamma satisfies α+γ\alpha^{+} \subseteq \gamma, hence α+γ\alpha^{+} \le \gamma by claim (f). So the ordinals have no gaps immediately above a given point, which is what makes the successor and limit dichotomy of Successor and limit ordinals exhaustive.

Suprema come for free. Claim (e) says a set of ordinals always has a least upper bound, namely A\bigcup A: it contains every member of AA as a subset, hence lies weakly above each by claim (f), and any ordinal weakly above all of them contains A\bigcup A. Claim (d) gives the dual statement for a nonempty set. Neither needs any completeness assumption, in sharp contrast with the situation for R\mathbb{R}.

Nonemptiness in claim (d) is essential. The intersection of the empty family is not a set, so the hypothesis cannot be dropped. The union of the empty family, by contrast, is =0\emptyset = 0, which is why claim (e) needs no such hypothesis.

Claim (g) is a departure from the usual bookkeeping. It is normally derived alongside trichotomy. It is proved here because claim (e) cannot be proved without it, and separating them would either duplicate the argument or create a circular dependency between this lemma and Trichotomy and well-ordering of the ordinals.

The naturals are the model case. Every natural number is a transitive set and satisfies nnn \notin n (Every natural number is a transitive set and is not a member of itself), which is exactly claims (a) and (b) in the situation this definition abstracts, and σ(n)=n{n}\sigma(n) = n \cup \{n\} is the successor operation of claim (c). That every natural number, and ω\omega itself, really is an ordinal is proved in ω\omega is the least limit ordinal.

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