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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.

Srivatsan
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Aleksei Averchenko
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    Closely related: http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1 – lhf Jun 05 '11 at 12:15

4 Answers4

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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?

Community
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lhf
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  • Why the down vote? – lhf Jun 05 '11 at 12:18
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    I have no idea, but the voting on this question seems relatively strange to follow "live", to say the least. By the way, the OP excluded the use of connectedness in his comment to mt_'s answer. – t.b. Jun 05 '11 at 12:29
  • 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. – Aleksei Averchenko Jun 05 '11 at 12:49
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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) $a<b<c$ with $f(a)<f(b)$ and $f(b)>f(c)$ 

                    or

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

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.

Amitesh Datta
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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 $x \in X$.

Matthew Towers
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    So $(0, 1] \subset [0, 1]$ will have just one connected component, but $(0, 1) \setminus {f(0)}$ will necessarily have two? That's very clever, but connectedness hasn't been mentioned yet either :) – Aleksei Averchenko Jun 05 '11 at 11:56
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    you can easily prove what you need by elementary real analysis, e.g. it's clear there's a continuous function from one of the spaces onto the discrete space ${0,1}$, but for the other no such function exists. – Matthew Towers Jun 05 '11 at 11:59
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    You don't really need to use the notion of connectedness. It might be more elementary to observe that if f restricted to X{x} maps to Y{f(x)}, then by the intermediate value theorem, f(x) must be in the image, which is a contradiction. – gfes Jun 05 '11 at 12:23
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    @user11751: it might seem more elementary, but the IVT is of course based on connectedness. – wildildildlife Jun 05 '11 at 12:29
  • You are right of course. It is just more likely to have been treated in a preliminary course on real analysis. – gfes Jun 05 '11 at 12:33
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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.

Srivatsan
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Blah
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