# Is there a “deep line” topological space in analogue to the “long line” $\omega_1\times[0,1)$?

I was reading about the "long line" $L=\omega_1\times[0,1)$ in the lexicographic order topology, which is locally like $\Bbb R$ except that it is "long" on one end, so there is no countable sequence that runs off to infinity unlike $\Bbb R$. My question is whether one could construct a "deep line" that is homogeneous and totally ordered (i.e. one dimensional), and is only as "long" as $\Bbb R$ as you run off to infinity to either end, but is much deeper in the sense that no point is the intersection of countably many neighborhoods; you need to intersect at least $\omega_1$ open sets to get a singleton.

I'm using some informal language above because I'm a bit new to topology; does my idea correlate to any known or studied spaces? My intuition is telling me that a model of this space is the surreal numbers up to generation $\omega_1$, except that you cut off all the infinite elements: which is to say, I want to define $D=\{x\in{\sf No}\mid{\rm bday}(x)<\omega_1\wedge\exists y\in\Bbb N\,|x|<y\}$ in the order topology. Does this set behave the way I think it should?

I find it curious that this space is not metrizable, even though there is an obvious "metric" $d(x,y)=|x-y|$.

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$\def\RR{\mathbb{R}}$Let $X = \RR^{\omega_1}$; the set of all maps $\omega_1 \to \RR$. Make $X$ into an abelian group by pointwise addition. Let $f\in X$ be nonzero. Let $a$ be the least element of $\omega_1$ for which $f(a) \neq 0$. (Since $\omega_1$ is well ordered, there is a least such $a$.) Define $f$ to be positive if $f(a) >0$ and negative if $f(a) < 0$. This makes $X$ into an ordered abelian group. Equip $X$ with the order topology. Since I built $X$ as the order topology on an ordered abelian group, $X$ is homogenous and one-dimensional in your sense.
Let $f_n$ map the minimal element of $\omega_1$ to $n$ and map everything else to $0$. Then the sequence $f_n$ marches all the way to the end of $X$, so this line is not "long".
I claim that $X$ is deep. Let $U_i$ be a counteable collection of open sets in $X$ containing $0$; we must show that $\bigcap U_i$ is not $\{ 0 \}$. Replacing each $U_i$ by a smaller open set, we may assume that $U_i$ is of the form $(f_i, g_i)$ with $f_i < 0 < g_i$. Let $a_i$ be the least element of $\omega_1$ for which $f_i(a_i) \neq 0$, so $f_i(a_i) < 0$. Similarly, let $b_i$ be minimal with $g_i(b_i) > 0$. Choose $c \in \omega_1$ with $a_i < c$ and $b_i < c$ for all $i$. Let $f(c) = -1$ and $f(d) = 0$ for all other $d \in \omega_1$. Then $(-f,f) \subset (a_i, b_i)$ for all $i$, so the intersection contains the whole interval $(-f,f)$.
Very clever. I'm trying to imagine what a homomorphism from your space to mine would look like, and I think it would take $f\mapsto f(0)+f(1)\varepsilon + f(2)\varepsilon^2+\dots$, but I'm not totally convinced that the behavior is the same, especially at limit ordinals. Note that the surreals are sometimes represented as functions $f:\alpha\to\{-1,1\}$ for some ordinal $\alpha$ (lexicographically ordered the same way as your space), but since $\{-1,1\}$ has a maximum element, you get long increasing sequences, unlike your example. – Mario Carneiro Jun 12 '13 at 23:42
@Mario: It isn’t quite lexicographic order, since it has to compare functions $\sigma$ and $\tau$ when $\sigma\subsetneqq\tau$. In this case the non-value undefined is treated as lying between $-1$ and $1$, so that $\sigma\prec\tau$ iff $\tau(\mathrm{dom}\;\sigma)=1$. – Brian M. Scott Jun 13 '13 at 12:09
@BrianM.Scott Correct; it is perhaps more appropriate to view them as functions on $\omega_1$ (or ${\sf On}$ for the whole construction) that take on the value $0$ outside their domain. I thought for a while about it, but I'm not sure if the surreals up to $\omega_1$ can be recovered (with their order topology) by taking the subspace of ${\Bbb R}^{\omega_1}$ given by functions which have $f(x)\in\{-1,1\}$ for $x<\alpha$ and $f(x)=0$ for $\alpha\le x<\omega_1$ (for some $\alpha<\omega_1$). – Mario Carneiro Jun 13 '13 at 14:37