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Prove that an open interval $(a,b)$ and a closed interval $[c,d]$ are not homeomorphic.

I'm trying to prove this statement but have only vague ideas on how to start. How may I use the property of connectedness to show this?

nomadicmathematician
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4 Answers4

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If we remove either $c$ or $d$ from $[c,d]$ we get a connected subspace.

If we remove any point $\gamma$ from $(a,b)$, the resulting subspace $J$ can be written as

$\tag 1 J = (a, \gamma) \sqcup (\gamma, b)$

which is a disconnected space.

If $(a,b)$ was homeomorphic to $[c,d]$, removing some point from $(a,b)$ would give us a connected space. But this contradicts (1).

CopyPasteIt
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Here's a proof not using connectedness properties.

Suppose $f: [a,b] \to (c, d)$ is a homeomorphism. Observe that $f(a)\not=f(b)$ as $f$ is injective and consider the point $f(a)+f(b)\over 2$ at half the distance between $f(a)$ and $f(b)$. As $f$ is surjective, $f(x) = {f(a)+f(b)\over 2 }$ for some $x \in [a,b]$.

Now let $\delta$ be enough for both the distances between $f(x)$ and $f(a)$, and between $f(x)$ and $f(b)$ to be less than $\delta$ and yet the open interval centered at $f(x)$ of radius $\delta$ not to cover the whole $(c,d)$.

As $f$ is continuous, there must be an open interval centered at $x$ large enough to contain both $a$ and $b$ and whose image under $f$ is within a distance of $\delta$ from $f(x)$. As such an interval must ecompass the whole $[a,b]$, the function $f$ cannot be surjective for $\delta$ is chosen in such a way that there're points in $(c,d)$ with a distance from $f(x)$ greater than $\delta$.

Tarc
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Or one can do the following. By scaling arguments one can show that $(0,1)$ is homeomoprhic to $(a,b)$ and $[0,1]$ to $[c,d]$. Now, $(0,1)$ is not compact but $[0,1]$ is. I learned compactness before connectedness. I am using $(0,1)$ and $[0,1]$, because those I think are the most famous examples of non-compactness and compactness.

The very fluffy Panda
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  • Why is $(0,1)$ not compact? Its compactness needs to be evaluated relative to its induced topology, not some unrelated superset like $\mathbb{R}$. – VF1 May 07 '14 at 00:48
  • I don't understand what you mean, to me it's clear that it should be evaluated with respect to the induced usual topology from $\mathbb{R}$. $\mathbb{R}$ or any space with more than one point is totally disconnected with respect to the discrete topology. So the same is true for connectivity. – The very fluffy Panda May 07 '14 at 05:51
  • @VF1 Forgot to write @ – The very fluffy Panda May 07 '14 at 06:16
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    Right, it should be the induced topology. What I'm saying is that it isn't clear that on the induced topology that the whole unit open interval is not a compact set. On the induced topology it is closed after all. – VF1 May 07 '14 at 13:47
  • @VF1 it is clear, since there are Cauchy sequences without convergent subsequence (e.g. $a_n = 1/n$). You are right that the interval is closed on the induced topology, but it is not compact. – lasik43 Jul 19 '22 at 11:29
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Hint

Suppose this was true, and $\exists f:(a,b)\longrightarrow [c,d]$, where $f$ is a homeomorphism.

Now, consider the inverse image of $f$, $g=f^{-1}$, which must be a homeomorphism.

As Daniel Fischer suggests, look at the image under $g$ of a set in $[c,d]$ less a particular point. Another useful property you may wish to consider is the fact that the homeomorphism $g$ is (1) an open mapping and (2) preserves connectivity by continuity.

VF1
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  • I've considered f: [c,d] -> (a,b) and the restriction of f from (c,d], which then produces a disconnected image, contradicting that continuity preserves connectivity. But where do I need to use the fact that a homeomorphism is an open mapping? – nomadicmathematician May 06 '14 at 21:21
  • Didn't you make assumptions about how the image looks like? – VF1 May 06 '14 at 21:26