Dictionary Order Topology on $\mathbb{R}^2$ Metrizable?

1. Feb 1, 2017

Bashyboy

1. The problem statement, all variables and given/known data
I am trying to show that there exists a metric on $\mathbb{R}^2$ that induces the dictionary order topology on the plane.

2. Relevant equations

3. The attempt at a solution
If I recall correctly, vertical intervals in the plane form basis elements for the dictionary order topology. With this in mind, I am trying to define a metric such that the corresponding $\epsilon$-balls are vertical strips. The only function that I could come up with is

$$d((x,y),(w,z)) = \begin{cases} \infty ,& x \neq w \\ |y-z| ,& x=w \\ \end{cases}$$

Based on my work, it seems that this is a well-defined metric. However, somehow I feel that I am cheating by using $\infty$. Is this in fact a valid metric?

2. Feb 3, 2017

Bashyboy

Okay, I think I may have a way to "fix" this, though I would still be interested in knowing if the above defines a valid metric. When dealing with a metric induced topology whose basis elements are $\epsilon$-balls, it suffices to look at those $\epsilon$-balls with radius less than $1$. Now, if we replace $\infty$ in the above function and confine ourselves to dealing with $\epsilon$-balls of radius less than $1$ , then the metric will restrict the balls from containing points "off" the vertical line.

More precisely, if $(a,b) \in B((x,y), \epsilon)$ and $a = x$, then $d((x,y),(a,b)) = 1 > \epsilon$, a contradiction. So, if $(a,b)$ is a point in the ball, then necessarily $a \neq x$ and therefore $d((x,y),(a,b)) = |y-a| < \epsilon$ or $a \in (y-\epsilon,y+\epsilon)$. This, of course, suggests that $B((x,y),\epsilon) = \{x\} \times (y-\epsilon,y+\epsilon)$, which is a basis element of the dictionary order topology. In this case, $1$ effectively acts as $\infty$.

As I mentioned above, I would still be interested in knowing whether the function in my first post defines a valid metric.