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There exists a function $f: X\to Y$ between metric spaces $X$ and $Y$ that is not continuous but has the property that, for each closed ball $B$ of $Y$, $f^{-1}(B)$ is closed in $X$.

What is an example in support of the above statement?

Is this result true for all metric spaces $X$ and $Y$?

If I replace "Closed" with "Open" in the above problem, then the function is becoming continuous.

Saikat
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    Certainly not: if $X$ and $Y$ are singleton sets, then any map between them is continuous! – Theo Bendit Feb 22 '21 at 03:32
  • Actually, all we really need is $X$ to be the discrete metric space, and any map from $X$ to any metric space $Y$ will be continuous. – Theo Bendit Feb 22 '21 at 03:37

1 Answers1

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HINT: Try $X=[0,2]$, $Y=[0,1)\cup[2,3]$, and

$$f:X\to Y:x\mapsto\begin{cases} x,&\text{if }0\le x<1\\ x+1,&\text{if }1\le x\le 2\,. \end{cases}$$

Here $X$ and $Y$ have their usual metrics as subspaces of $\Bbb R$ with its usual metric.

Brian M. Scott
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  • Teacher can you explain a bit more – James Feb 22 '21 at 04:13
  • @James: Show that $f$ is not continuous. (HINT: It’s not continuous at $x=1$.) You also need to show that if $B$ is a closed ball in $Y$, then $f^{-1}[B]$ is always a closed ball in $X$. You’ll have to deal with several cases here, depending on where the centre of $B$ is and what it’s radius is, but each of the cases is pretty straightforward. – Brian M. Scott Feb 22 '21 at 04:48