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In my Real Analysis class I got a bit frisky and broke out a homeomorphism in a problem to show that a set was closed (that is, I had a closed set, and I made a homeomorphism between it and the set in question to show that the set in question was closed). My reason for doing this was simple: A homeomorphism maps closed sets to closed sets.

My instructor made a note saying that this is not true in general. He says homeomorphisms do not in general map closed sets to closed sets.

Everything I have ever read about homeomorphisms contradicts this. But maybe I'm wrong, so if someone on here could provide a counterexample (that is, a homeomorphism mapping a closed set to an open set), I would certainly appreciate it.

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Just to clarify, you have a set $X$ and two subsets $A,B$. You tried to show $A$ is closed by constructing a homeomorphism between $A$ and $B$ and showing that $B$ is closed in $X$. Correct? –  Alex Becker Sep 6 '12 at 0:00
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It would help if you could be more specific. One thing to be careful about is the following sort of situation: $\arctan\colon \mathbb R \to \mathbb R$ induces a homeomorphism $\mathbb R \to (-\pi/2, \pi/2)$. This does not mean that $(-\pi/2, \pi/2)$ is closed in $\mathbb R$. This is the difference between being relatively closed (in a subspace of $X$) and closed in $X$. –  Dylan Moreland Sep 6 '12 at 0:02
    
That is correct Alex. –  user25326 Sep 6 '12 at 0:04
    
@user25326 then your instructor is correct. See my answer. –  Alex Becker Sep 6 '12 at 0:04

1 Answer 1

up vote 4 down vote accepted

It is possible to have a space $X$ and subsets $A,B$ such that $A$ is closed, $A$ is homeomorphic to $B$ (when both are endowed with the subspace topology), and yet $B$ is not closed. If this is what you did, your instructor is correct. Consider the Sierpiński space, that is the set $\{0,1\}$ with open sets $\emptyset,\{0\}$ and $\{0,1\}$. Then $\{0\}$ is not closed yet $\{1\}$ is and $\{0\},\{1\}$ are homeomorphic in the subspace topology.

What you actually need is a homeomorphism $f:X\to X$ such that $f(A)=B$. Then $A$ is closed iff $B$ is.

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Alright, I see what he was trying to say. Though in my construction, f(A) = B, so I was right, I just didn't word my reasoning as well as he would have liked I suppose. Though what if f maps between two different spaces? Would that still hold? That is: f:X->Y with A a subset of X and B a subset of Y. –  user25326 Sep 6 '12 at 0:06
    
Why do you call this the "Klein space"? I always knew it as the Sierpiński space. –  t.b. Sep 6 '12 at 0:07
    
@t.b. I recall it vaguely being called that somewhere. I'll look into it. Edit: It seems I was imagining things. Thanks for correcting me. –  Alex Becker Sep 6 '12 at 0:08
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@user25326: The important point in Alex's last paragraph is not $f(A)=B$, but $f: X\to X$. That is, the domain/range of $f$ must be the entire space that $A$/$B$ are to be closed in. (Being closed is not a property of a set in itself, but of how it is a subset of the toplogical space it is a subset of). –  Henning Makholm Sep 6 '12 at 0:17
    
@HenningMakholm +1, thanks for clarifying my point. –  Alex Becker Sep 6 '12 at 0:23

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