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I need a hint: prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic without referring to compactness. This is an exercise in a topology textbook, and it comes far earlier than compactness is discussed.

So far my only idea is to show that a homeomorphism would be monotonic, so it would define a poset isomorphism. But the can be no such isomorphism, because there are a minimal and a maximal elements in $[0, 1]$, but neither in $(0, 1)$. However, this doesn't seem like an elemenary proof the book must be asking for.

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    Closely related: http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-12011-06-05

4 Answers 4

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There is no continuous and bijective function $f:(0,1) \rightarrow [0,1]$. In fact, if $f:(0,1) \rightarrow [0,1]$ is continuous and surjective, then $f$ is not injective, as proved in my answer in Continuous bijection from $(0,1)$ to $[0,1]$. This is a consequence of the intermediate value theorem, which is a theorem about connectedness. Are you allowed to use that?

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    I am allowed to use anything, since nobody actually standing behind my back :) I was just curious about ways to prove this fact with just elementary tools. Theo Buehler's answer is great, BTW.2011-06-05
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Let $f:[0,1]\to (0,1)$ be a homeomorphism. Let us suppose, for a contradiction, that $f$ is not monotonic. In particular, either there exists:

(1) $a with $f(a) and $f(b)>f(c)$

                    or 

(2) $a with $f(a)>f(b)$ and $f(b).

Let us consider case (1) without loss of generality.

Exercise 1: Prove that the image of $(a,c)$ under $f$ is not open in $(0,1)$. In particular, we have a contradiction and $f$ must be monotonic.

Let us assume without loss of generality that $f$ is monotonically increasing.

Exercise 2: Prove that $f$ is not surjective. In particular, we have a final contradiction.

Therefore, $f$ is not a homeomorphism.

The following exercises are relevant:

Exercise 3: Prove that there is a continuous surjection $f:(0,1)\to [0,1]$. Do you think that there is a continuous surjection $f:[0,1]\to (0,1)$?

Exercise 4: Does there exist a surjective open map (i.e., open sets are mapped to open sets) $f:[0,1]\to (0,1)$. Do you think that there is a surjective open map $f:(0,1)\to [0,1]$?

Exercise 5: Find an example of a continuous bijection between topological spaces that is not a homeomorphism.

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Here's a hint. If $X$ and $Y$ are homeomorphic (via $f$) then $X \setminus \{x\}$ and $Y \setminus \{f(x)\}$ (subspace topologies) are homeomorphic via the restriction of $f$, for any $x \in X$.

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    You are right of course. It is just more likely to have been treated in a preliminary course on real analysis.2011-06-05
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$\text{id}_{(0,1)}$ has no maximum (easy)

Every continuous function $f$ on $[0,1]$ has a maximum, otherwise pick $x_n$ such that $f(x_n)>\text{sup}f([0,1])-\frac{1}{n}$ for all $n$. Then Bolzano-Weierstraß does it.