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 ${\omega }_1:=\mathrm{\aleph }(\omega )$, ${\omega }_1$ 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

$(\alpha ,s)<(\beta ,t):⟺\alpha <\beta ,\text{ or }(\alpha =\beta \text{ and }s<t),$

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 $0=\mathrm{\varnothing }$ 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 $R^{\mathrm{\prime }}:=R\setminus \{0_R\}$ be the open long ray. The long line is the set

$\mathbb{L}:=(\{0\}\times R^{\mathrm{\prime }})\cup (\{1\}\times R)$

with the order

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

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 $\alpha \mathrm{\ne }0$: 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 $[0,\mathrm{\infty })$. 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.