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I'm trying to prove that all continuous maps of pairs $f:([-1,1], \{-1,1\})\to (\{-1,1\},\{-1,1\})$ are constant, and I've almost got a working argument, but it reduces down to the following situation:

Since $\{-1\}$ and $\{1\}$ are both open in $\{-1,1\}$, so too must be $f^{-1}(\{-1\})$ and $f^{-1}(\{1\})$. By definition of $f$, their union must equal all of $[-1,1]$.

So now I have a pair of disjoint open subsets of $[-1,1]$ whose union is all of $[-1,1]$. This is impossible, and easy to prove if they are both basic open sets (but I can't assume that).

Any suggestions?

  • 0
    Is there a typo in the first line? Should it be $f:([0,1],\{0,1\}) \to (\{-1, 1\}, \{-1,1\})$?2012-01-18
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    Fixed. Thanks for pointing it out. :)2012-01-18
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    Do you know the definition of a connected space? If so, can you prove that the interval $[-1,1]$ is connected?2012-01-18
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    Got it!!! This contradicts the fact that $[0,1]$ is connected! :)2012-01-18
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    Ahh you beat me to it :P thanks very much :)2012-01-18
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    er. $[-1,1]$. I keep messing that up...2012-01-18
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    If you've got it, you should write it up as an answer and, assuming no one shoots it down over the next couple of days, accept it. It may seem odd to answer your own question and accept the answer, but it's actually the way the site works.2012-01-18

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