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.
Best Answer
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 $a$.