Prob. 2, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: Does this metric on $\mathbb{R} \times \mathbb{R}$ induce the dictionary order topology?

Question

Here's Prob. 2, Sec. 20 in the book Topology by James R. Munkres, 2nd edition:

Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable.

This question has already been asked here.

Now my question is, can the following metric induce the dictionary order topology on $\mathbb{R} \times \mathbb{R}$?

Let the $d_0 \colon \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ be defined by $$d_0 (\alpha, \beta) \colon= \begin{cases} 1 \ \mbox{ if } \ \alpha \neq \beta, \\ 0 \ \mbox{ otherwise;} \end{cases} $$ for all $\alpha, \beta \in \mathbb{R}$. Then $d_0$ is of course a metric on $\mathbb{R}$.

Now let $d \colon \left( \mathbb{R} \times \mathbb{R} \right) \times \left( \mathbb{R} \times \mathbb{R} \right) \to \mathbb{R}$ be defined by $$d \left( x_1 \times y_1, \ x_2 \times y_2 \right) \colon= \max \left\{ d_0 \left( x_1, x_2 \right), \ \left\vert y_1 - y_2 \right\vert \right\} \ \mbox{ for all } \ x_1 \times y_1, \ x_2 \times y_2 \in \mathbb{R} \times \mathbb{R}.$$ I know that this $d$ is a metric on $\mathbb{R} \times \mathbb{R}$.

Does this metric induce the dictionary order topology on $\mathbb{R} \times \mathbb{R}$?

Answer 1

Yes, it does. It induces the product topology on $\mathbb{R}_d \times \mathbb{R}$, where the former set is the set of reals in the discrete topology. And the linked answer shows that $\mathbb{R} \times \mathbb{R}$ in the lexicographic order is equal to this product topology (which is itself again homeomorphic to the disjoint sum of continuum many copies of the reals, although Munkres does not treat sum spaces as a general construction).

Added later

Suppose that $B((x,y), r)$, where $(x,y)$ is in the plane and $r>0$, is an open ball in your metric. We want to show it is a neighbourhood of $(x,y)$ in the lexicographical order topology.

Now take $0 < s \le r$ such that $s < 1$, or e.g. define $s = \min(\frac{1}{2},r)$, say. Then $d((x,y),(x',y')) < s$ iff $x = x'$ and $|y - y'| < s$, by the definition of the metric (as $s<1$ the first coordinates must be equal, or we'd already have a metric value of $1$), so $\{x\} \times B_e(y,s) = B((x,y),s)$, where $B_e$ denotes the ball in Euclidean metric. Also, this equals the lexicographical open interval $( (x,y-s), (x,y+s) )_{lex}$ in which $(x,y)$ also lies. As $B((x,y),s ) \subseteq B((x,y), r)$.

The previous implies that any metric open set $O$ is lexicographically open: for all $(p,q)$ in $O$ we find an open ball $B$ around $(p,q)$ that sits inside $O$, and as $B$ is a lexicographical neighbourhood of $(p,q)$, and so it is one for all points of $O$.

On the other hand, if $O = (\leftarrow, (p,q))_{lex}$ is a segment in the order topology, and $(x,y) \in O$, we know that $x < p$ or $x = p, y < q$.

In the former case, $B((x,y),1) = \{x\} \times (y-1,y+1) \subseteq O$ (as all points in it have first coordinate $<p$) and in the latter case, pick $0 < r < 1$ such that $y + r < q$ and note that $B((x,y), r) = \{x\} \times (y-r,y+r) \subseteq O$ (as all points have first coordinate $x$ and second one $< q$). In both cases, $O$ is open in the metric topology (every point around it has some ball around it that sits inside it). One can do exactly the same thing for the other type of subbasic open set, the open segments $((p,q),\rightarrow)_{lex}$, as you can easily check. So all lexicographically open sets are metric open as well.

This shows that both topologies coincide, and are equal to the product topology as I mentioned. "Small neighbourhoods" all have the same form: singletons times open intervals, which equal small open intervals lexicographically, or small balls in the $d$-metric.