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I'm trying to prove that a map between a metric and topological space is continuous at x iff for every converging sequence in the materic space, the map of the sequence converges in the topological space.

I was able to prove the forward case without too much trouble, but I'm having trouble proving that converging sequences imply continuity.

I've assumed for contradiction that the map is not continuous and then invoked the definition of continuity to say there is an open neighborhood of x whose preimage is closed, but I'm having trouble piecing together a contradiction on convergence from this.

user201756
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1 Answers1

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Your idea (proof by contradiction) is a good one, but there's a problem. I will denote the metric space and its metric by $M,d;$ the topological space, by $T.$ Assuming $f:M\to T$ is not continuous means that there is an open subset $U$ of $T$ whose pre-image is not open. This is not the same as saying that the pre-image is closed--sets (unlike doors) may be closed, open, both, or neither.

Since $f^{-1}[U]$ is not open, then there is some $x\in f^{-1}[U]$ such that for all $r>0,$ there is some $y\in M$ with $d(x,y)<r$ and $y\notin f^{-1}[U].$

Can you use this fact to show that there is a sequence of points of $M$ converging to $x$, each of which lies outside of $f^{-1}[U]$? Once you've done that, can you take it from there?

Cameron Buie
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  • Is the proposition true if the map is the other way around (from topological space to metric one)? –  Dec 23 '21 at 12:18
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    @Cedric Only in part. If $f:T\to M$ is continuous, then for any sequence of points converging in $T$, the sequence of their images under $f$ converges in $M.$ However, the converse need not hold unless $T$ is a sequential space. See for example here and here, as well as the discussion in the comments here. – Cameron Buie Dec 23 '21 at 13:30