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How to prove that $$(a,b)\not\cong[a,b]$$ (not homeomorphic) as subsets of real line?

Is it true that in some topology $(a,b)$ is closed?

Thanks a lot!

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What do you mean by $\ncong$? – Asaf Karagila Oct 13 '12 at 12:03
To answer the last question: In the discrete topology, every subset is closed. – Harald Hanche-Olsen Oct 13 '12 at 12:04
@Asaf: it means not homeomorphic – Aspirin Oct 13 '12 at 12:06
@Aspirin: Then my answer works. You should add this to the question. – Asaf Karagila Oct 13 '12 at 12:07
Given any interval $(a,b)$ for fixed $a,b$ you can always make it a closed set in the topology $\tau$ on $\Bbb{R}$ consisting of $$\tau = \{\emptyset, \Bbb{R}, (a,b)\}.$$ – user38268 Oct 13 '12 at 12:08
up vote 7 down vote accepted

To show that two spaces are not homeomorphic you can find a property that holds for one and fails for the other, and is invariant under homeomorphism.

For example $[a,b]$ is compact, whereas $(a,b)$ is not. The continuous image of a compact set is compact, so if $f\colon[a,b]\to(a,b)$ were a homeomorphism you would have that $(a,b)$ is the continuous image of a compact set and therefore compact. Contradiction.

As for the second question, you can always declare a set is closed. Namely, $A\subseteq\mathbb R$ in the topology generated by $\mathbb R\setminus A$ you have that $A$ is closed.

Here is a slightly more hands-on approach:

  1. Show that if $f\colon[a,b]\to(a,b)$ is continuous and injective then it is either strictly increasing or strictly decreasing. Assuming without loss of generality that $f$ is increasing.

  2. Denote by $b'=f(b)$, then we have to have that $f(x)\leq b'$ for all $x\in[a,b]$ by the above argument.

  3. Since $b'<b$ we have to have some point $x\in(b',b)$ which cannot be in the range of $f$.

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its right but it problem gives immediately after the definition of homeomorphism (before topological invariants), so I interested in "head-on" solution – Aspirin Oct 13 '12 at 12:13
@Aspirin: I see. You can always prove that compactness is preserved by continuous functions. It's quite a simple proof actually (assuming that you define continuity as "preimage of an open set is open"). – Asaf Karagila Oct 13 '12 at 12:16
i.e. there is no any continuous map from $[a,b]$ to $(a,b)$? – Aspirin Oct 13 '12 at 12:24
@Aspirin: There is no continuous map from $[a,b]$ onto $(a,b)$. There is certainly a map into $(a,b)$, but its range can never be the entire interval. – Asaf Karagila Oct 13 '12 at 12:25
@Aspirin: No. For half-open intervals connectedness works better, it is disguised in my "hands-on" suggestion as well (the connectedness argument). – Asaf Karagila Oct 13 '12 at 12:36

This is more of a continuation of Asaf's answer, but there is an even stronger sense that $(0,1)$ can be closed in a topology on $\mathbb{R}$. One can actually construct a metric $d$ on $\mathbb{R}$ satisfying the following properties:

  1. every open set in the usual topology on $\mathbb{R}$ is open in the $d$-metric topology;
  2. the metric $d$ is complete (all Cauchy sequences with respect to this new metric converge);
  3. $\mathbb{R}$ is separable (has a countable dense subset) with respect to the new topology; and
  4. $(0,1)$ is a clopen (closed and open) set in the new topology.
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It should be remarked that $\mathbb R$ is no longer connected in this topology. – Asaf Karagila Oct 13 '12 at 12:29
Here's an explicit construction: Take the function $$f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{R} \setminus (0,1) \\ \frac{1}{x(1-x)} & \text{if } x \in (0,1)\end{cases}$$ and put $d(x,y) = \lvert(x,f(x)) - (y,f(y))\rvert$ (the distance of two points $x,y \in \mathbb{R}$ is the Euclidean distance on the graph of $f$). Then $(\mathbb{R},d)$ has all the claimed properties. – commenter Oct 13 '12 at 12:53

Another way similar to the answer already given:

Let $f\colon[a,b]\to(a,b)$ is a continuous map then $[a,b]\setminus\{b\}=[a,b)$ is connected set but $f([a,b))$ is not connected as we have deleted $f(b)$ from $(a,b)$.

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Assume that $f: (a,b) \rightarrow [a,b]$ is a homeomorphism, i.e. in particular it's continuous and bijective (one-to-one and onto). Then there are $u,v \in (a,b)$ with $f(u)=a$, $f(v)=b$. From the intermediate value theorem, it follows that $f([u,v]) \supset [a,b]$. Since $\text{rng } f = [a,b]$, it cannot be an actual superset, so it further follows that $f([u,v]) = [a,b]$. Now, since $f$ is bijective, it must be that $(a,b) \setminus [u,v] = \emptyset$, since there's no place to map any remaining points to without destroying bijectivity. You obviously also have $[u,v] \subset (a,b)$, and together this shows $[u,v] = (a,b)$, which is impossible.

There is thus no homeomorphism from $(a,b)$ to $[a,b]$, hence the two sets are not homeomorphic.

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To say that $f$ is a homeomorphism is more than a continuous bijection. You require it is open as well. Of course we don't need that in your answer, but that is irrelevant to how you wrote it in the first sentence. – Asaf Karagila Oct 13 '12 at 12:33
@AsafKaragila Yes, but I didn't need that. Should have worded it in a way that doesn't imply every continuous bijection is a homeomorphism, though. Will fix. – fgp Oct 13 '12 at 12:35

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