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

  • Thread starter Bashyboy
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  • #1
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Homework Statement


I am trying to show that there exists a metric on ##\mathbb{R}^2## that induces the dictionary order topology on the plane.

Homework Equations




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?
 

Answers and Replies

  • #2
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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.
 

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