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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 countable 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 $g_i(b_i) > 0$. Countable subsets of $\omega_1$ have upper bounds so we may 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 $(-f,f) \subset (f_i, g_i)$ for all $i$, so the intersection contains the whole interval $(-f,f)$.

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  • $\begingroup$ 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. $\endgroup$ – Mario Carneiro Jun 12 '13 at 23:42
  • $\begingroup$ @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 $\tau(\mathrm{dom}\;\sigma)=1$. $\endgroup$ – Brian M. Scott Jun 13 '13 at 12:09
  • $\begingroup$ @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 $\alpha<\omega_1$). $\endgroup$ – Mario Carneiro Jun 13 '13 at 14:37
  • $\begingroup$ @BrianM.Scott [con't] (It is definitely the same in the topology from the inherited order, because the inherited order is the surreal order, but the inherited order topology is not always the same as the subspace topology.) $\endgroup$ – Mario Carneiro Jun 13 '13 at 14:40
  • $\begingroup$ Hi @David. I've been thinking about your example and may have found an unfortunate property. I believe this set is not a linear continuum because it does not satisfy the upper bound property. Consider the subset of functions $\{f \in X : f(a) = 0, a \neq 1\}$. This is upper bounded by the function I'll briefly write as $(1,0,\ldots{})$ but has no least upper bound. It is therefore not connected and so does not seem like a good analogue to the long line. (Note $\mathbb{R}^2$ isn't connected with the dictionary order either since all vertical lines are open and the same thing happens here.) $\endgroup$ – Geoffrey Sangston Jun 10 '17 at 3:28
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I just thought of this the other night and came up with something similar to the other answer.

Take {0,1}^(your favorite ordinal) to create "long" sequences of binary numbers. Putting them in order is simple. Find where two different sequences first differ (by well-orderedness of ordinals) and the one assigned a 1 is the larger element.

Not quite done yet though. We want a line, not a closed interval type space with a first and last element, so go ahead and remove the "all-zeros" and "all-ones" elements from the space.

Still not quite done. This space is not connected. So identify all elements such that all positions past a certain point are 1s (meaning there is a last element of the domain of the "sequence" assigned 0) with the element flipping that last 0 and the following 1s. This is the analogue of 0.9999... = 1.

Okay, so now we have a line that is both deep and long (assuming your favorite ordinal is one that is not a successor, has no countable increasing sequences converging to it etc.). Interestingly you'll find elements in this deep line having countable sequences converging from behind (like omega 1s followed by all 0s), but never from ahead so that there is no countable basis at any point -- thus the line is indeed "deep".

That's the line I was interested in, but you asked for one which isn't long. That's an easy fix, somewhat inspired by the other answer. Just glue countably many of these together, and you're done!

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    $\begingroup$ Now, R has the property that all of its open intervals are homeomorphic to R. If we call my space X, X doesn't quite have the analogous property, but it has one similar: Each open interval in X contains a copy of X. Can anyone think of a deep line with the propterty that all of its open intervals are copies of itself? $\endgroup$ – Bryce Blackham May 27 '18 at 22:16

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