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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" $d(x,y)=|x-y|$.

Mario Carneiro
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2 Answers2

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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 $(-f,f)$.

Geoffrey Sangston
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David E Speyer
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  • 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
  • @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.) – Mario Carneiro Jun 13 '13 at 14:40
  • 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.) – Geoffrey Sangston Jun 10 '17 at 03:28
  • @David. I'm hope that it can be remedied however. Consider the set $[0,1]^{\omega_1}$ with the dictionary order topology. Your argument works for this set and additionally it should have upper bounds (like the ordered square). I don't think it is homogenous though and my idea for fixing this breaks the requirement that it be as short at $\mathbb{R}$. – Geoffrey Sangston Jun 10 '17 at 03:35
  • @GeoffreySangston I was looking at this question again, and I couldn't figure out what I meant by "one dimensional in your sense". Once I understand what the definition of one dimensional is in this question, I'll be able to respond better to the details of your suggestion. – David E Speyer Jun 11 '17 at 12:21
  • @David. Another potentially unfortunate observation is that all nontrivial closed intervals of this space (and of $[0,1]^{\omega_1}$) have cardinality equal to the power set of $\mathbb{R}$, which is unlike the long line. I'm not sure if the "deep line" property implies this must happen though. – Geoffrey Sangston Jun 11 '17 at 13:53
  • @GeoffreySangston Why would you view it as a bad thing that "all nontrivial closed intervals have cardinality $2^\frak{c}$"? This is a property shared with $\Bbb R$, and also with the long line (because every closed interval in the long line is bounded and hence is homeomorphic to a closed interval in $\Bbb R$). – Mario Carneiro Nov 19 '17 at 01:12
  • @GeoffreySangston I think it is inevitable that the upper bound property will fail when making this kind of modification to the real line. Specifically, if you have a linearly ordered abelian group $R$ with an archimedean element (call it $1$), such that the order is complete, then it is divisible ($x/n$ is the supremum of $y$ such that $ny<x$), and so we can build $\Bbb Q\subseteq R$, and then to each $x\in R$ we can define $st(x)=\sup{q\in\Bbb Q\mid q<x}$, and - here's the kicker - $st(x)$ is always less than $x$! If $st(x)=x$, then $x$ is accessible with countably many open sets. – Mario Carneiro Nov 19 '17 at 01:49
  • (...) But then that means ${q\in\Bbb Q\mid q<x}\ne{q\in\Bbb Q\mid q<st(x)}$, so there exists $q\in \Bbb Q$ between $st(x)$ and $x$, for every $x\in R$. This is a contradiction since $\Bbb Q$ is countable but there are uncountably many disjoint open intervals between $0$ and $1$ (because there is an $\omega_1$ descending sequence from $1$ approaching $0$). – Mario Carneiro Nov 19 '17 at 01:56
  • @MarioCarneiro It's not really a bad thing but I was hoping to find an analog which shares as many properties with the reals as possible. The space David gave and the space I gave don't locally have the cardinality of the reals but again maybe this isn't that bad. My apologies for not understanding what you mean when you say the property is shared by the reals since the cardinalities of the closed intervals are different in either case. I think the reals are the unique Dedekind complete linearly ordered abelian group so the property has to fail. – Geoffrey Sangston Nov 22 '17 at 01:19
  • You may be aware that the property you give in the original post is the statement that no point is a $G_\delta$ set. Reading the wikipedia page, in a metrizable space every closed set is a $G_\delta$ set. Since we're interested in normal spaces (order topologies are normal), our space must not be separable (Urysohn's metrization theorem). So now I don't really expect the property I called foul on to be shared. – Geoffrey Sangston Nov 22 '17 at 01:36
  • I made an error two comments ago. The reals should be the unique Dedekind complete, dense, linearly ordered abelian group with a nontrivial element. – Geoffrey Sangston Nov 22 '17 at 04:11
  • @GeoffreySangston "the cardinalities of the closed intervals are different in either case" My mistake, I miscalculated the size of a closed interval in $\Bbb R$ as $2^\frak{c}$ instead of $\frak{c}$. What can be said is that the cardinality of a nontrivial closed interval is the same as the cardinality of the whole space, although I don't think this property necessarily holds in all "deep lines", just $\Bbb R^{\omega_1}$ with this construction. – Mario Carneiro Nov 22 '17 at 04:19
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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!

Bryce Blackham
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    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? – Bryce Blackham May 27 '18 at 22:16