So I'm supposed to come up with a bijection that fulfills the criteria above but I have no idea how to approach this problem. I know that $f(x)=\tan(\frac{\pi}{2}x)$ would work if the domain was restricted to $(-1,1)$ but I don't know how to expand this to include the endpoints of the interval? Any hints in the right direction appreciated!! EDIT: I also know that no such continuous map exists but just don't know an alternate approach or how to split the problem up.
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3Find a bijection which maps $[-1, 1]$ to $(-1, 1)$, and you can use what you have already. – Alex Wertheim Oct 14 '15 at 05:11
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2Iterate this, twice. – Did Oct 14 '15 at 05:27
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@AlfredYerger Sorry but what is your point? – Did Oct 14 '15 at 05:31
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Okay, so what I will do is give you a bijection from a closed interval to an open interval that I just happen to know off the top of my head. Then by composing with translations/dilations/homeomorphisms to the real line, you can make the composition yourself.
We will biject $[0,1]$ to $(0,1)$. Define the function by mapping $0$ to $1/2$, and then writing $f(x) = \frac{1}{2^{n+2}}$ if $x=1/2^n$, and otherwise writing $f(x) = x$. That is to say, $f$ shrinks powers of two, sends 0 into the middle (which needs to be hit since there are no larger powers of two), and then keeps everything else right where it's at.
Now you can compose and do the rest.
A. Thomas Yerger
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