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It is a well known result that the long line (namely, the topological space $S_\omega \times [0, 1)$ in the order topology, where $S_\omega$ is the minimal uncountable well-ordered set) cannot be embedded in Euclidean space of any dimension. Is it possible, however, to embed the long line in $\Bbb{R}^2$ in the lexicographic order topology? What about the ordered square?

My intuition about this space is that it is composed of an "uncountable number of real lines" which makes me conjecture it could be possible to embed it in such a space; as it stands, I'm not yet able to give a rigorous proof of the matter - or the opposite.

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  • $\begingroup$ I believe $\mathbb{R}^2$ with the dictionary topology is first-countable while the long line is not, so certainly the long line cannot be embedded in the ordered plane. That doesn't rule out the converse. $\endgroup$ – Sofie Verbeek Sep 15 '18 at 13:37
  • $\begingroup$ en.m.wikipedia.org/wiki/Long_line_%28topology%29?wprov=sfla1 This article on Wikipedia lists the (non-extended) long line or ray (actually, my example is on the non extended closed long ray) as first countable. As much as I trust SE more than most other sites, I would like to clarify this mismatch as well. $\endgroup$ – Niki Di Giano Sep 15 '18 at 13:58
  • $\begingroup$ Ah, my bad. I was looking at the extended version. That said, the extended version is obtained by adding a single point, so maybe you can still extract something useful from the observation. (For instance given an embedding of the non-extended long line into R2, maybe we can we extend the embedding to an embedding of the extended long line into R2 and thus find a contradiction.) $\endgroup$ – Sofie Verbeek Sep 15 '18 at 14:15
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For $\mathbb{R}^2$ with the lexicographic order, this is very simple, Note that for each $x\in\mathbb{R}$, $\{x\}\times\mathbb{R}$ is clopen in $\mathbb{R}^2$ and homeomorphic to $\mathbb{R}$ with the usual topology. Since the long line is connected, any continuous map from the long line to $\mathbb{R}^2$ must have its image contained in $\{x\}\times\mathbb{R}$ for some $x$, and thus cannot be an embedding since the long line does not embed in $\mathbb{R}$.

For the unit square the argument is a bit more complicated. Let $L$ denote the long line and let $S$ denote $[0,1]^2$ with the lexicographic order. Let $p:S\to [0,1]$ be given by $p(x,y)=x$; note that $p$ is continuous (for the usual topology on $[0,1]$). So, if $f:L\to S$ is any continuous map, we have a continuous composition $p\circ f:L\to [0,1]$. Any continuous map $L\to[0,1]$ is eventually constant, so $p\circ f$ is eventually constant. However, this means that there exists $a\in L$ and $x\in [0,1]$ such that $f(b)\in \{x\}\times[0,1]$ for all $b>a$. Since $\{x\}\times[0,1]$ just has the usual topology of $[0,1]$, this again implies that $f$ is eventually constant, and so cannot be an embedding.

(This second argument applies equally well to the space $\omega_1$ instead of $L$, to show that it does not embed in either $\mathbb{R}^2$ or $[0,1]^2$ with the lexicographic order topology.)


Just for completeness, here is a proof that any continuous map $f:L\to\mathbb{R}$ is eventually constant (the same argument also applies with $\omega_1$ in place of $L$). First, fix $\epsilon>0$. Suppose that for all $a\in L$ there exist $b,c>a$ such that $|f(b)-f(c)|>\epsilon$. We can then choose an increasing sequence $b_1<c_1<b_2<c_2<\dots$ such that $|f(b_n)-f(c_n)|>\epsilon$ for each $n$. Every sequence in $L$ is bounded above, so this sequence has a supremum $x$ which it converges to. But this contradicts the continuity of $f$, since the sequence $(f(b_1),f(c_1),f(b_2),f(c_2),\dots)$ is not Cauchy and so cannot converge to $f(x)$.

So for every $\epsilon>0$, there exists $a\in L$ such that $|f(b)-f(c)|<\epsilon$ for all $b,c>a$. For each $n$, choose $a_n$ that works as such an $a$ for $\epsilon=1/n$. Letting $a$ be an upper bound for all these $a_n$, we have $|f(b)-f(c)|<1/n$ for all $n$ if $b,c>a$. That is, $f(b)=f(c)$ for all $b,c>a$, so $f$ is constant above $a$.

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  • $\begingroup$ Yes! Originally I thought of the argument from continuity of the long line, but couldn't tell if the lines of the ordered plane were connected or not. Thank you for both proofs. $\endgroup$ – Niki Di Giano Sep 15 '18 at 18:57
  • $\begingroup$ I was trying far too hard. +1! $\endgroup$ – Noah Schweber Sep 15 '18 at 18:59
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Great question! Here is partial work towards a negative answer.

(Incidentally, note that it's easy to show that the answer is no if we consider the long line and the ordered plane as linear orders, and look at order-preserving embeddings. That doesn't answer your question, but lends some plausibility to a negative answer.)


Below, I'm going to write "$\omega_1$" for what you call "$S_\omega$" - this is in fact the standard notation.

Suppose $f$ is an embedding (in the sense of topology) of the long line into the ordered plane. Then we get an embedding $g$ from $\omega_1$ into the ordered plane, by composing $f$ with the obvious embedding $e$ of $\omega_1$ into the long line (sending $\alpha$ to the "$0$" of the "$\alpha$th copy of $[0,1)$").

Let $h:\omega_1\rightarrow\mathbb{R}$ be the map sending $\alpha$ to the index for the "copy" of $\mathbb{R}$ which $g(\alpha)$ lives in. Then $h$ is a continuous map from $\omega_1$ into $\mathbb{R}$. It is not guaranteed to be an embedding, since there is no reason it should be injective, so this isn't immediately a contradiction; we're going to have to do some work.

The key claim - which I can't prove yet - is:

Claim: For each $r\in\mathbb{R}$, $h^{-1}(\{r\})$ is countable.

That is, $h$ only "comes back" to a given real countably many times.

Taking the claim above on faith for the moment, I'm going to define a set $A\subseteq\omega_1$ with the following two properties:

  • $h$ restricted to $A$ is injective.

  • $A$ with the subspace topology is homeomorphic to $\omega_1$. Note that for this, it is not enough to have $A$ be uncountable: the set of successor elements of $\omega_1$ with the subspace topology is discrete, after all.

This will give a contradiction, since - composing the restriction of $h$ to $A$ with a homeomorphism between $A$ and $\omega_1$ - it gives us an embedding of $\omega_1$ into $\mathbb{R}$.

(Well, we'll still have to prove the claim! But we'll be close to done.)

So here's how we build $A$. By transfinite recursion, we can define a sequence $(\theta_\eta)_{\eta\in\omega_1}$ of elements of $\omega_1$ satisfying $$\theta_\eta=\min(\{\alpha\in\omega_1: \forall\beta<\eta(\mbox{$h(\alpha)\not=h(\theta_\beta)$})\}).$$ The fact that $\theta_\eta$ is well-defined for each $\eta\in\omega_1$ follows from the claim above. Now let $$A=\{\theta_\eta:\eta\in\omega_1\}.$$ Clearly $h$ restricted to $A$ is injective, and the fact that $A$ (with the subspace topology) is homeomorphic to $\omega_1$ follows from the easily-checked fact that for every limit ordinal $\lambda\in\omega_1$, we have $$\theta_\lambda=\sup\{\theta_\beta:\beta<\lambda\}.$$


So all that's left to do is prove the claim above. I don't immediately see how to do this, although there are a couple tempting approaches (in particular, Fodor's lemma seems potentially quite useful).

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  • $\begingroup$ Thank you for your beautifully attempted proof! $\endgroup$ – Niki Di Giano Sep 15 '18 at 18:58

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