[topology] The metric topology is the coarsest that makes the metric continuous

[nonequilibrium]

[topology] "The metric topology is the coarsest that makes the metric continuous"

Homework Statement

Let (X,d) be a metric space. Show that the topology on X induced by the metric d is the coarsest topology on X such that $d: X \times X \to \mathbb R$ is continuous (for the product topology on $X \times X$).

Homework Equations

N.A.

The Attempt at a Solution

I can prove that d is continuous, but I'm having trouble proving that the topology is the coarsest. Let V be a subset of R, then denote $U := d^{-1}(V)$ and suppose U is open. I want to prove that V is open, but I'm not sure how.

 

[micromass]

Try to prove that if d is continuous, then the open balls are open sets in that topology.

 

[nonequilibrium]

Thanks.

Hm, I seem to be confused.

The best thing I can do, is to say: take an open ball $B(x, \varepsilon)$, then we know that $d^{-1}(]-\epsilon,\epsilon[)$ is open, and hence $d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)$ is open in $\{x\} \times X$, which proves your statement if we identify
$B(x, \varepsilon) \leftrightarrow d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)$
$X \leftrightarrow \{x\} \times X$
but I would doubt this round-about method is necessary...

Part of my confusion seems to be about the following: I think I'm wrong about my initial assumption that the statement is equivalent to saying it's (also) the coarsest on $X \times X$ such that d is continuous? Because I'm confused by the fact that (take $X = \mathbb R$ for the moment) $d^{-1}(]a,b[)$ is always an unbounded region, consequently the initial topology on $\mathbb R \times \mathbb R$ induced by d cannot coincide with the product topology on $X \times X$ inherited from the metric topology on X (which would have to be the case if the above statement were true).

I hope my confusion makes sense...

 

[micromass]

Thanks.

Hm, I seem to be confused.

The best thing I can do, is to say: take an open ball $B(x, \varepsilon)$, then we know that $d^{-1}(]-\epsilon,\epsilon[)$ is open, and hence $d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)$ is open in $\{x\} \times X$, which proves your statement if we identify
$B(x, \varepsilon) \leftrightarrow d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)$
$X \leftrightarrow \{x\} \times X$
but I would doubt this round-about method is necessary...

That is correct.

Part of my confusion seems to be about the following: I think I'm wrong about my initial assumption that the statement is equivalent to saying it's (also) the coarsest on $X \times X$ such that d is continuous? Because I'm confused by the fact that (take $X = \mathbb R$ for the moment) $d^{-1}(]a,b[)$ is always an unbounded region, consequently the initial topology on $\mathbb R \times \mathbb R$ induced by d cannot coincide with the product topology on $X \times X$ inherited from the metric topology on X (which would have to be the case if the above statement were true).

Yes, I indeed do not think that is true. Note that the actual initial topology of d (on the set XxX) is not even Hausdorff. Indeed, the point (x,x) and (y,y) can never be separated by open sets. If (x,x) is in $d^{-1}(]a,b[)$ then $0\in ]a,b[$. But then we also have $(y,y)\in ]a,b[$.

Some extra information:
As you can see, the set $\{d^{-1}(]a,b[~\vert~a,b\in \mathbb{R}\}$ does not induce a nice topological space. But it does introduce a nice structure called a uniformity. Uniform spaces (spaces equipped with a uniformity) are a generalization of metric spaces which can be used to define notions such as uniform continuity, uniform convergence, totally boundedness and Cauchy sequences. Every uniform space induces a topological space. So we actually have "metric space --> uniform space --> topological space".

 

[nonequilibrium]

Ah, thank you :)

And is the way I proved it the most efficient way/the way you had in mind? With the identifications? Or is it a round-about way for something shorter that I'm overlooking?

 

[micromass]

Ah, thank you :)

And is the way I proved it the most efficient way/the way you had in mind? With the identifications? Or is it a round-about way for something shorter that I'm overlooking?

It was the way I had in mind. Perhaps there is a shorter way, but I doubt it.

 

[nonequilibrium]

Ah perfect :) You have settled my qualms.