While doing some exercises about different metrics, I had a doubt about my comprehension of the matter.
Recall that the lexicographic order on $\mathbb{R}^2$ is defined as follows :
$$
(a,b) < (c,d) \iff (a<c) \quad \mathrm{or} \quad (a=c \ \mathrm{and} \ b<d)
$$
We put the order topology on the real plane, so the open intervals of the form $](a,b), (c,d)[$ look like a vertical open interval if $a=c$ or the union of a vertical line over $(a,b)$ with the band of all the points $(x,y)$ such as $a<x<c$ and the vertical line at $x=c$ up to $y=d$. (I don't know if it makes sense, it was just to get the picture in a sentence… but drawing an example on a sheet of paper should also work.)
If I did not make any mistakes, the real plane with such a topology is metrizable (is it true ?).
For the record, I defined $d$ a metric as :
$$
d( (x,y), (x',y') ) =
\begin{cases}
\infty \ \mathrm{if} x \neq x' \\
|y'-y| \ \mathrm{otherwise}
\end{cases}
$$
I'll skip the proof (but if $d$ is not a good candidate, please feel free to correct me on that point) that $d$ is a metric and defines the same topology that we had from the lexicographic order.
Then, the next question was about the existence of a metric on $[0,1]^2$ that gives the same topology as the one from the order defined above. I think I did the usual proof that consists of saying that the square is compact and therefore, if it was metrizable, it would separable which is not the case with the lexicographic order topology (the line passing through the middle $\{(s,1/2)|s\in[0,1]\}$ is discrete and uncountable).
First, is that statement correct ?
Second, what prevent us to use to induced metric on the square ? I couldn't think of a counter example when someone told me that restricting $d$ to the square was enough to get a metric space. But, as it would contradict the "proof" above, one of them is wrong (or something in broken in the math world, which I highly doubt and fear).
Best Answer
Let $\tau$ be the lexicographic order topology on $\Bbb R^2$ and $\tau_s$ the lexicographic order topology on the unit square $S=[0,1]\times[0,1]$. The main thing that you’re missing, I think, is that $\tau_s$ is not the topology that $S$ inherits as a subspace of $\langle\Bbb R^2,\tau\rangle$.
In $\langle\Bbb R^2,\tau\rangle$ each vertical $\{x\}\times\Bbb R$ is a clopen set, and in fact the lexicographically ordered plane is homeomorphic to the product $\Bbb R_d\times\Bbb R$, where $\Bbb R_d$ denotes $\Bbb R$ with the discrete topology, and the second factor $\Bbb R$ has the usual topology. Since $\Bbb R_d$ is metrizable with the discrete metric, this is a product of two metrizable spaces and as such is metrizable. One compatible metric is given by
$$d(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle)=\begin{cases} \min\{|y_0-y_1|,1\},&\text{if }x_0=x_1\\ 1,&\text{otherwise}\;. \end{cases}$$
In the relative topology induced on $S$ by $\tau$ the sets $\{x\}\times[0,1]$ for $x\in[0,1]$ would be clopen, and the restriction of $d$ to $S\times S$ would be a metric inducing the relative topology.
In $\tau_s$, however, those sets are not clopen: they’re closed, but they’re not open. If $0<a\le 1$, every open $\tau_s$-nbhd of $\langle a,0\rangle$ contains points $\langle x,0\rangle$ with $x<a$, and if $0\le a<1$, every $\tau_s$-open nbhd of $\langle x,1\rangle$ contains points $\langle x,1\rangle$ with $x>a$. Thus, none of the sets $\{x\}\times[0,1]$ with $x\in[0,1]$ is open in $\tau_s$. It’s this difference between $\tau_s$ and the relative topology inherited from $\tau$ that makes $S$ compact in the topology $\tau_s$, as it clearly is not in the relative topology, in which
$$\big\{\{x\}\times[0,1]:x\in[0,1]\big\}$$
is an open cover of $S$ with no finite subcover — indeed, no proper subcover.
In short, you can start with a compatible metric on the $\langle\Bbb R^2,\tau\rangle$ and restrict it to a metric on $S$, and it will generate the subspace topology on $S$, but this isn’t $\tau_s$, the lexicographic order topology on $S$. Once you realize that you’re looking at two different topologies on $S$, the apparent problem disappears.
As an aside, it’s worth noting that the two topologies actually do agree at all points of $S$ except those of the form $\langle x,0\rangle$ with $0<x\le 1$ and those of the form $\langle x,1\rangle$ with $0\le x<1$.