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Give a function $f : X \to Y$ such that $(X,d)$ and $(Y,p)$ are metric spaces, and $f$ is continues and one-to-one function and onto $Y$ but $f^{-1}$ is not continues ?

I know that compactness of $X$ must play a role here, because if $X$ is compact then there is no such counter-example, so $X$ must be no compact, but other than this i don't have a direction for the solution, please help.

Ahmad
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2 Answers2

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If $X:=(\mathbb{R},d)$ where $d(x,y)=1$ for all $x,\ y$ and $Y:=(\mathbb{R},|\ |)$ where $|\ |$ is a canonical metric, then $f=id: X\rightarrow Y$ is continuous. But $f^{-1}(1/n) = 1/n$ does not converge to $0$

HK Lee
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Take the identity function from $\mathbb R$ into itself, with the discrete metric on the lect and the usual one on the right.

José Carlos Santos
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