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$\begingroup$

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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    @learner: I disagree with this edit (and would have rejected it myself), the readability of the post did not improve, and all the effect was bumping up a question from two and a half years ago, instead of letting the newer questions (and lord knows, we don't have a shortage of those) to be seen on the front page. Every edit bumps the question, so bump with care!2013-04-08
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    @AsafKaragila: The system already discriminates so strongly in favour of the youngest among questions, which aren't necessarily the most interesting ones, that I don't mind if an older question returns to the front page now and then. Let's say, if I find an annoying typo in a question or answer, even if it is one of my own, I won't hesitate to edit it away. I have no particular opinion about the current question though. But replacing non-math by a math formula is almost always an improvement.2013-04-08
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    @Marc: Typos are one thing; adding `$` symbols to math which wasn't any less readable than it is now (say, compared to things like cos x=sin x) is really just pigging out on the main page resources.2013-04-08
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    May I know where you came across this problem?2015-03-31
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    Related: https://math.stackexchange.com/questions/1677642018-11-29
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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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    Thanks. But for the other side, can a non-compact set be mapped to compact? Actually I know a little about topology, but I cannot find info about mapping open set to closed.2010-11-09
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    @haohaolee: Thinking!2010-11-09
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    Yes. I can even figure out some examples, e.g., sinx, but it would be better if I can get the serious proof2010-11-09
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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