The closed long ray $\omega_1 \times [0,1)$ under the lexicographic order, and the long line, with the order topology

Let $\omega_1$ be the first uncountable ordinal (The first uncountable ordinal $\omega_1 := \aleph(\omega)$, $\omega_1$ 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), whose elements are the at most countable ordinals (Ordinal (von Neumann)) ordered by membership (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals), and let

$[0,1) \;=\; \{\, t \in \mathbb{R} : 0 \le t < 1 \,\}$

be the half-open unit interval of the complete ordered field $\mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Order on the reals, Ordered field, Complete ordered field (least-upper-bound property)).

The closed long ray is the set

$R \;:=\; \omega_1 \times [0,1)$

with the lexicographic order

$(\alpha, s) \;<\; (\beta, t) \quad :\Longleftrightarrow \quad \alpha < \beta, \ \text{ or } \ \bigl(\alpha = \beta \text{ and } s < t\bigr),$

and $(\alpha,s) \le (\beta,t)$ meaning $(\alpha,s) < (\beta,t)$ or $(\alpha,s) = (\beta,t)$; $R$ carries the order topology of this order (The order topology of a linearly ordered set, with the open rays as a subbasis; order-convex sets, order-density, the least upper bound property, and linear continua, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

The lexicographic order is a linear order, and this is discharged here. Antisymmetry and irreflexivity of $<$: if $(\alpha,s) < (\beta,t)$ then either $\alpha < \beta$, which by trichotomy of the ordinals (Trichotomy and well-ordering of the ordinals) forbids $\beta \le \alpha$ and hence forbids $(\beta,t) < (\alpha,s)$, or $\alpha = \beta$ and $s < t$, which by trichotomy in $\mathbb{R}$ (Ordered field) forbids $t < s$; in particular no element is $<$ itself. Transitivity: if $(\alpha,s) < (\beta,t) < (\gamma,u)$ then $\alpha \le \beta \le \gamma$, so $\alpha \le \gamma$; if $\alpha < \gamma$ we are done, and if $\alpha = \gamma$ then $\alpha = \beta = \gamma$ and $s < t < u$ gives $s < u$. Comparability: given two elements, compare the first coordinates by Trichotomy and well-ordering of the ordinals and, if they are equal, the second by trichotomy in $\mathbb{R}$. So $(R, \le)$ is a totally ordered set (Partial order and partially ordered set).

Least element, and the open long ray. $R$ has the least element $0_R := (0, 0)$, since $0 = \varnothing$ is the least ordinal and $0$ the least element of $[0,1)$. The open long ray is $R \setminus \{0_R\}$ with the restricted order and its order topology. $R$ has no greatest element: given $(\alpha, s)$, the element $(\alpha^{+}, 0)$ is strictly above it, and $\alpha^{+} \in \omega_1$ because $\alpha^{+}$ is again at most countable (Basic closure properties of ordinals, $\omega_1$ 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).

The long line. Let $R' := R \setminus \{0_R\}$ be the open long ray. The long line is the set

$\mathbb{L} \;:=\; (\{0\} \times R') \;\cup\; (\{1\} \times R)$

with the order

$(0,x) < (0,y) :\Longleftrightarrow y < x, \qquad (0,x) < (1,y) \ \text{ always}, \qquad (1,x) < (1,y) :\Longleftrightarrow x < y,$

that is: a reversed copy of the open long ray laid before a copy of the closed long ray, the two halves meeting at the single centre point $(1, 0_R)$. This is again a total order, by the same three checks applied within each copy and by the third clause across them, and $\mathbb{L}$ carries its order topology. One copy is open and one is closed deliberately: were both copies open, each half would be a union of open rays of $\mathbb{L}$, so the two halves would form a pair of disjoint nonempty open sets covering $\mathbb{L}$ — a separation — and the order would have a gap at the seam instead of the centre point $(1, 0_R)$ that closes it.

Blocks. The set $\{0\} \times [0,1)$ is an initial segment of $R$ order-isomorphic to $[0,1)$, and for each $\alpha \in \omega_1$ the block $\{\alpha\} \times [0,1)$ is order-isomorphic to $[0,1)$; the blocks are laid end to end in the order type of $\omega_1$. A block $\{\alpha\} \times [0,1)$ has a least element $(\alpha,0)$ and no greatest element. No element of $R$ has an immediate predecessor or an immediate successor. Within a block this is the corresponding fact for $[0,1)$. At a block boundary $(\alpha, 0)$ with $\alpha \ne 0$: if $\alpha = \beta^{+}$ then the elements below it are the $(\beta,s)$, $s < 1$, among which there is no greatest, so it has no immediate predecessor; and if $\alpha$ is a limit ordinal (Successor and limit ordinals) the elements below it include $(\xi, 0)$ for every $\xi < \alpha$, again with no greatest, since $\alpha$ is a limit. Immediate successors fail because no block has a greatest element.

• Why $[0,1)$ and not $[0,1]$. With $[0,1]$ the element $(\alpha, 1)$ would be the greatest element of its block and $(\alpha^{+}, 0)$ would be its immediate successor, producing a jump; the order would then fail to be order-dense and the long ray would be disconnected. Half-open blocks glue without a seam, which is the whole point of the construction.

• Why $\omega_1$ and not a larger ordinal. The construction makes sense for any ordinal, and for $\omega$ it produces an order isomorphic to $[0,\infty)$. What is special about $\omega_1$ is that it is uncountable while each of its elements 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), which is what makes every proper initial segment of $R$ look like an ordinary half-line while $R$ itself does not.

• Naming. The obligation that the long ray deserves to be called a continuum — that it is order-dense and has the least upper bound property — is discharged by The long ray is a linear continuum, hence connected; every one of its at most countable subsets is bounded above, assuming countable choice ↗, recorded in this item's justified_by, and is not assumed anywhere above.

• What is not defined here. Nothing above asserts that $R$ or $\mathbb{L}$ is path-connected, or metrizable, or that either is homeomorphic to any space built earlier. Those questions need machinery this page does not develop, and no statement on this page depends on their answers.