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I am trying to prove the following result:

"Let $f:[a,b] \to [c,d]$ be a homeomorphism on the indicated intervals. Prove that $f$ maps endpoints to endpoints."

This problem follows the section in which it is proved that the connected subsets of $\mathbb{R}$ are exactly the intervals. I have really not made much progress. Can someone give me an idea of how to prove this?

Additionally, can you give me a sense of what your initial thoughts were while starting to think about the problem? I'm very interested not only in how to prove the statement, but in the thought process involved. And an attempt at generalization to any problem is most welcome (I may try to make this sort of idea into its own question.)

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up vote 5 down vote accepted

My first thought is to see what happens if $f(a)$, say, is not $c$ or $d$: suppose that $f(a)=x\in(c,d)$. Now look at what $f$ does to the interval $(a,b]$: it must map it onto $[c,x)\cup(x,d]$. But $[c,x)\cup(x,d]$ is not an interval, so it’s not connected, contradicting the fact that the continuous map $f$ must map the connected set $(a,b]$ to a connected set. And that does it: the same argument clearly shows that $b$ must also go to one of the endpoints $c$ and $d$.

Can you use a similar idea to show that the space $X=\{\langle x,y\rangle\in\Bbb R^2:x=0\text{ or }y=0\}$ cannot be homeomorphic to $\Bbb R$? If you manage that, try showing that neither $X$ nor $\Bbb R$ is homeomorphic to $Y=\{\langle x,y\rangle\in X:y\ge 0\}$; the same basic trick works there, too.

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Sorry for the delay, I've had a very busy weekend. Thanks for the good answer. – Alex Petzke Sep 16 '12 at 22:51

If $f(a) \in (c,d)$ (open interval), then $f(a,b] = [c,f(a)) \cup (f(a),d]$, since $f$ is a bijection. However $(a,b]$ is connected, but $f(a,b]$ is not, which contradicts the continuity of $f$. Hence $f(a) \notin (c,d)$, which means $f(a) \in \{c,d\}$. The same argument, mutatis mutandis, applies for the other endpoint.

An alternative approach:

Suppose $f(a) < f(b)$ (they can't be equal since $f$ is a bijection, if $>$ then consider $-f$). Then if $[\alpha, \beta] \subset [a,b]$ is a non-empty interval, we must have $f(\alpha) < f(\beta)$. If $f(\alpha) \geq f(\beta)$, then the intermediate value theorem would yield a point $x \in [a,\alpha]$ such that $f(x)=f(\beta)$, which contradicts $f$ being a bijection. It follows then that $f$ is strictly monotonically increasing. It follows by continuity that $f[a,b] = [f(a),f(b)]$, and since $f$ is a bijection with $[c,d]$, it follows that $[f(a),f(b)]=[c,d]$, which is the desired result.

These approaches are, of course, equivalent in some sense.

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Sorry for the delay, I've had a very busy weekend. I really like the second approach. Thanks for the answer. – Alex Petzke Sep 16 '12 at 22:50

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