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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 $\ncong$?2012-10-13
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    To answer the last question: In the discrete topology, *every* subset is closed.2012-10-13
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    @Asaf: it means not homeomorphic2012-10-13
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    @Aspirin: Then my answer works. You should add this to the question.2012-10-13
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    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 $$\tau = \{\emptyset, \Bbb{R}, (a,b)\}.$$2012-10-13
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    @BenjaLim: You mean $\mathbb R\setminus(a,b)$ perhaps. Topology is the collection of *open* sets.2012-10-13
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    @AsafKaragila Sorry yes.2012-10-13

4 Answers 4

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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' 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" solution2012-10-13
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    @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").2012-10-13
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    i.e. there is no any continuous map from $[a,b]$ to $(a,b)$?2012-10-13
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    @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.2012-10-13
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    I understand. But is non-compactness preserved under continuous map?2012-10-13
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    @Aspirin: Of course not. Consider the constant function which sends everything to a single point. See http://math.stackexchange.com/questions/156859/discontinuous-function-sending-compacts-to-compacts2012-10-13
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    but proof with compactness does not work for $[a,b)\not\cong(a,b)$?2012-10-13
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    @Aspirin: No. For half-open intervals connectedness works better, it is disguised in my "hands-on" suggestion as well (the connectedness argument).2012-10-13
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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.2012-10-13
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    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.2012-10-13
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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.2012-10-13
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    @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.2012-10-13