Trouble constructing metric for $\mathbb R \times \mathbb R$ in the dictionary order.

Question

I'm having trouble constructing, and understanding the construction, of a metric that induces the dictionary order topology on $\mathbb R \times \mathbb R$. There are example metrics posted here on Math.SE, but that is not what I'm looking for. I'd like to understand how to approach constructing the proper metric. My first attempt was to define $$d((x_1,x_2),(y_1,y_2)) = \begin{cases} |y_1 - x_1|, & \mbox{if } 0 < |y_1 - x_1| \\ |y_2 - x_2|, & \mbox{if } x_1 = y_1, \mbox{and } 0 < |y_2 - x_2| \\ 0, & \mbox{if } x = y \end{cases}$$ However, even if I didn't make a mistake in my calculations, and this is in fact a metric, I don't know how to approach proving that $d$ induces the dictionary order topology. To summarize, I'd like to know how to approach constructing the metric and if my first attempt even works.

Answer 1

Intuitively, the plane $\Bbb R^2$ in the lexicographic order, is just a set of vertical lines, all of which are topological copies of $\Bbb R$ and such that different vertical lines do not really interact: the horizonal line $\mathbb{R} \times \{0\}$ has the discrete topology as $(x,0)$ has the neighbourhood $O(x):= \{x\} \times (-1,1)$ which equals the open interval $((x,-1),(x,1))$ in the order, such that $O(x) \cap \left(\mathbb{R} \times \{0\}\right) = \{(x,0)\}$. Some more thought will tell you that in fact $\Bbb R^2$ in the order topology is just homeomorphic to the product $(\Bbb R, \mathcal{T}_d) \times (\Bbb R, \mathcal{T}_e)$, where the first factor has the (metrisable) discrete topology and the second the normal Euclidean one. This suggests a metric: we can use the sum (or $\max$-metric) of the component metrics of the factors to get the right product topology, so define

$$d((x_1, y_1), (x_2,y_2)) = \begin{cases} |x_2-y_2| & \text{ if } x_1 = y_1 \\ 1+|x_2-y_2| & \text{ if } x_1 \neq y_1 \\ \end{cases}$$

which is then (by standard facts, I hope for you) a metric that induces the product topology of the discrete metric and the Euclidean metric, and so the right topology for the lexicographically ordered plane.

Answer 2

I assume that by "dictionary" order you mean lexicographical order. Note that there is more than one way to define ordering on a product of posets. In particular $(a,b)\leq (a',b')$ if and only if $a<a'$ or ($a=a'$ and $b<b'$).

So the order topology is generated by open intervals. To avoid confusion I will write $I[x,y]$ to denote an open interval in a given poset and $(x,y)$ to denote a pair of points. What exactly are these intervals in the case of lexicographical ordering? If $a<a'$ then $(a,b)<(a',b')$ for any $b,b'$ and so

$$I[(a,b), (a',b')]=\{a,a'\}\times I[b,b']\cup \{x\in\mathbb{R}\ |\ a<x<a'\}\times\mathbb{R}$$

These can be further reduced by noticing that each such subset is a union of $\{x\}\times I[b,b']$ which are elements of the basis as well. So this is the basis of the order topology. Intuitively vertical, one-dimensional lines are open.

Now given any metric we construct a topology out of it by taking open balls as a basis. So what are open balls in your metric? Assume that $r>0$ and let $(a,b)\in\mathbb{R}^2$. If $d((a,b), (a',b'))<r$ then we have two cases: (1) if $a\neq a'$ then $|a-a'|<r$ and $b,b'$ can be arbitrary and (2) if $a=a'$ then $|b-b'|<r$. In particular the open ball $B((a,b), r)$ can be written as

$$(I[a-r, a+r]\backslash\{a\})\times\mathbb{R}\cup\{a\}\times I[b-r, b+r]$$

Intuitively this is sort of opposite of the order topology basis. These are "squares" infinitely wide everywhere except for the middle.

But do these collections generate the same topology? Unfortunately, they don't. First of all note that given two basis $\mathcal{B},\mathcal{B'}$ and corresponding topologies $\mathcal{T},\mathcal{T'}$ we have $\mathcal{T}\subseteq\mathcal{T'}$ if and only if for any $B\in\mathcal{B}$ and any $x\in B$ there is $B'\in\mathcal{B'}$ such that $x\in B'\subseteq B$. See here.

With that you can check that your metric topology is a subset of the order topology. But not the other way around. Balls in your metric topology are too big, they don't fit in one-dimensional lines.

So how to fix that? Well, you were on the right track. You just gave too much freedom on the first coordinate, in particular you allowed it to be arbitrarily small. And so open balls always "catch" infinitely many elements from the first coordinate. You can fix that by slightly modifying your metric. Put

$$d((x_1,x_2),(y_1,y_2)) = \begin{cases} 1+|x_1-y_1|, & \mbox{if } 0 < |y_1 - x_1| \\ |y_2 - x_2|, & \mbox{if } x_1 = y_1, \mbox{and } 0 < |y_2 - x_2| \\ 0, & \mbox{if } x = y \end{cases}$$

Now note that for small enough $r$, i.e. $0<r<1$ the open ball $B((a,b), r)$ is indeed a vertical line $\{a\}\times I[b-r,b+r]$.