Suppose $f$ is a continous bijection from compact metric space $X$ onto a metric space $Y$ . Then $f^{-1}$ is continuous mapping of $Y$ onto $X$

Asked Oct 06 '18 at 07:39

Active Oct 06 '18 at 07:39

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I read this theorem in rudin's book:

Suppose $f$ is a continous bijection from compact metric space $X$ onto a metric space $Y$ . Then $f^{-1}$ is continuous mapping of $Y$ onto $X$

I have some questions about hypothesis of the theorem.. Clearly bijection is required for the existence of inverse. But we need compact? Do compact a necessary condition?

I want to see few counterexample,that if $X$ is not a compact metric space,then $f^{-1}$ is not continuous.

Thanks

asked Oct 06 '18 at 07:39

Cloud JR K

2,466

I see an example in rudin mapping [0,$2\pi)$ onto unit circle.please give me some more examples other than this – Cloud JR K Oct 06 '18 at 07:41
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$X$ discrete and $Y$ non-discrete so $X= \mathbb{N}$, $0\to 0$ and $n \to \frac1n$. – Henno Brandsma Oct 06 '18 at 07:41
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Here's many examples – Giuseppe Negro Oct 06 '18 at 07:42
@Henno Brandsma then $Y={\frac{1}{n}:n\in\mathbb N}$ – Cloud JR K Oct 06 '18 at 07:47
@Giuseppe Negro that helps a lot thanks – Cloud JR K Oct 06 '18 at 07:50
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And ${0}$ too is in $Y$ – Henno Brandsma Oct 06 '18 at 08:42
Thanks @Henno Brandsma – Cloud JR K Oct 06 '18 at 12:31

Suppose $f$ is a continous bijection from compact metric space $X$ onto a metric space $Y$ . Then $f^{-1}$ is continuous mapping of $Y$ onto $X$

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