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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.

user25326
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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 06 '12 at 00:00
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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 06 '12 at 00:02
  • That is correct Alex. – user25326 Sep 06 '12 at 00:04
  • @user25326 then your instructor is correct. See my answer. – Alex Becker Sep 06 '12 at 00:04

1 Answers1

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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.

BCLC
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Alex Becker
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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 06 '12 at 00:06
  • Why do you call this the "Klein space"? I always knew it as the Sierpiński space. – t.b. Sep 06 '12 at 00:07
  • @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 06 '12 at 00:08
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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). – hmakholm left over Monica Sep 06 '12 at 00:17
  • I'd completely forgotten about the Sierpenski space. I gotta review my general topology, yikes. I used to be great at it. : ( – Mathemagician1234 Aug 16 '23 at 04:33
  • This caught my eye due to the recent edit: an observation: there are very simple examples in euclidean space: e.g., take $A \subseteq (\Bbb{R} \setminus {0}) \times \Bbb{R}$ to be the graph of $x \mapsto 1/x$ and $B$ to be $(\Bbb{R} \setminus {0}) \times {0}$. Projection onto the $x$-axis gives a homeomorphism between $A$ and $B$, but $A$ is closed in $\Bbb{R}^2$ while $B$ is not. – Rob Arthan Jun 11 '24 at 20:10
  • ... or even in the real line: take $A = [0, \infty)$ and $B = [0, 1)$. – Rob Arthan Jun 11 '24 at 20:12