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If there is none, why?

And for the other side, what about open set $(0, 1)$ to closed set $[0, 1]$ with a continuous function?

Thanks

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    Related: https://math.stackexchange.com/questions/42308/2018-11-29

2 Answers 2

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HINT: For the first one use the fact that, Continuous image of a compact set is compact.

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    @haohaolee: Myke did my job!2010-11-09
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For the other side consider $f: (0,1) \to [0,1]$ defined as $f(x)= |\cos(2\pi x)|^{2}$

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    Perhaps mapping $[1/3,2/3]$ linearly to $[0,1]$ and mapping $(0,1/3)$ to $f(1/3)$ and $(2/3,1)$ to $f(2/3)$ would be a simpler way to see there is an example.2011-12-05