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Let be $X$ a topological space and we suppose that $E$ and $F$ are homeomorphic though a map $f$ from $F$ to $E$. So I ask to me if $E$ is open/closed when $F$ is open/closed but unfortunatley I was not able to prove or to disprove this so that I thought to put a specific question where I ask some clarification: in particular if the result is generally false I'd like to know if it can be true we additional hypotesis, e.g. Hausdorff separability, First Countability, Metric Topology, etc....

So could someone help me, please?

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  • $\begingroup$ @SassatelliGiulio Is the disjoint union topology the quotient topology of the canonical injections? $\endgroup$ Apr 21, 2022 at 13:38
  • $\begingroup$ The disjoint union topology is this. See also here. $\endgroup$ Apr 21, 2022 at 13:41

3 Answers 3

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Consider $E = (0,1)$ and $F = \mathbb{R}$.

Obviously, $E,F \subset \mathbb{R}$. It is quite easy to show that $F$ and $E$ are homeomorphic.

$F$ is closed and open. But $E$ is not closed.

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  • $\begingroup$ Oh, yeah: this is a very simple counterexample. Anyway, perhaps I found another counterexample: what do you think about? I would like to know if it is correct. $\endgroup$ Apr 21, 2022 at 16:39
  • $\begingroup$ @andrécaldas Incidentally you have presented a case of a space $X$ where if $E\subseteq X$ is open and $F\subseteq X$ is homeomorphic to $E$, then $F$ is open. See the "Added" question to my answer. $\endgroup$ Apr 21, 2022 at 16:58
  • $\begingroup$ @SassatelliGiulio: your answer is much better... more complete... and you offer lots of thoughts... I am upvoting yours. :-) $\endgroup$ Apr 22, 2022 at 12:55
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    $\begingroup$ @AndréCaldas Thanks, but truth be told I've expanded it a few times. $\endgroup$ Apr 22, 2022 at 13:55
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The only condition I can think of is the usual property that in compact T2 spaces the closed sets coincide with the compact sets, and therefore if $X$ is compact T2, then $F$ will be closed as soon as $E$ is.

Added: Thinking about it, there is a famous special case given by the invariance of domain theorem. If $M$ is a (Hausdorff, second countable) topological manifold without boundary, $E\subseteq M$ is open and $F\subseteq M$ is homeomorphic to $E$, then $F$ is open as well.

If you put these two facts together, you obtain that for a compact manifold $X$ without boundary (such as $S^n$, $P^n\Bbb R$, $P^n\Bbb C$, $U(n)$ et cetera) you indeed have that open subsets are homeomorphic only to open subsets and closed subsets are homeomorphic only to closed subsets.


There are, however, all sorts of counterexamples with seemingly good properties, for instance:

Counterexample 0: (Which kind of defeats the purpose of the other two) Let $X=[0,\infty)$. It's a contractible manifold with (non-empty) boundary and a separable complete metric space. It's homeomorphic to its subspaces $[2,\infty)$, $[0,1)$ and $[1,2)$ which are respectively closed and not open, open and not closed, and neither.


Counterexample 1: Consider $X=\{(x,y)\in\Bbb R^2\,:\, x=0\lor x^2+y^2=1\lor x=5\}$ with the usual distance. This is a locally compact, second-countable complete metric space. The four subspaces $A=\{0\}\times\Bbb R$, $B=\{5\}\times \Bbb R$, $C=\{0\}\times(10,\infty)$ and $D=\{(x,y)\,:\, x^2+y^2=1\land y\ne 1\}$ are all homeomorphic to $\Bbb R$. However:

  • $A$ is closed and not open;
  • $B$ is open and closed;
  • $C$ is open and not closed;
  • $D$ is neither open nor closed.

Counterexample 2: Consider $X=\Bbb R\times [0,1]^{\Bbb N}$. This is a product of countably many completely metrizable spaces, and therefore it's completely metrizable. It's also locally compact, second countable and path-connected. Notice that:

  • $X$ is open and closed;
  • $A=(0,\infty)\times[0,1/2]^{\Bbb N}$ is neither open (it does not contain a neighbourhood of $\langle 1,f\rangle$, where $f$ is the constant $1/2$ function) nor closed (the sequence $\langle 1/n, f\rangle$ converges to $\langle 0,f\rangle\notin A$);
  • $B=\{\langle x,f\rangle\in X\,:\, f_2=1\}$ is closed and, therefore, not open (in fact, its interior is empty);
  • $C=(0,\infty)\times[0,1]^{\Bbb N}$ is open and, therefore, not closed.

However, $A$, $B$ and $C$ are all homeomorphic to $X$.

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So it is a well know result that the graph $\mathcal G$ of a continuous function is homeomorphic to its domain: see here for details. Now, let be $f:\Bbb R^+\rightarrow\Bbb R$ the function $$ f(x):=\frac 1 x $$ for any $x\in\Bbb R^+$ so that we know that the sets $\Bbb R^+$ and $\mathcal G(f)$ are homeomorphic. Now $\Bbb R^+$ is homeomorphic via the natural immersion to the set $\Bbb R^+\times\{0\}$ so that the sets $\Bbb R^+\times\{0\}$ and $\mathcal G(f)$ are homeomorphic but the first is not closed whereas the second is closed because the origin is an accumulation point for $\Bbb R^+\times\{0\}$ and it is not there contained so that this set cannot be closed, whereas $\mathcal G(f)$ is the intersection between $[0,+\infty)\times\Bbb R$ and the zero set of the continuous function $g:\Bbb R^2\rightarrow\Bbb R$ defined though the equation $$ g(x,y):=xy-1 $$ for any $x,y\in\Bbb R^+$ and thus by continuity $\mathcal G(f)$ is closed.

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    $\begingroup$ It's true that $\mathcal G(f):=\{(x,y)\,:\,x>0\land xy=1\}$ is closed in $\Bbb R^2$ and $\Bbb R^+\times\{0\}$ isn't closed in $\Bbb R^2$. However, your proof doesn't work: the closed graph theorem will only carry you up to proving that $\mathcal G(f)$ is closed in $\Bbb R^+\times\Bbb R$, which doesn't tell much about whether or not it's closed in $\Bbb R^2$. By comparison, notice that $\Bbb R^+\times\{0\}$ is closed in $\Bbb R^+\times\Bbb R$, but not in $\Bbb R^2$. The characterization of $\mathcal G(f)$ as $\{(x,y)\,:\,xy=1\}$ doesn't work because $\{(x,y)\,:\, xy=1\}$ has points with $x<0$ $\endgroup$ Apr 21, 2022 at 16:49
  • $\begingroup$ So for I want characterize $\mathcal G(f)$ as the graph of the function $xy-1$ defined in the positive space $\Bbb R^+\times\Bbb R^+$. $\endgroup$ Apr 21, 2022 at 17:10
  • $\begingroup$ Anyway if $f:A\rightarrow\Bbb R$ is a continuous function defined in an open set then you are claim that the map $$\Gamma:=A\owns x\rightarrow\big(x,f(x)\big)\in\mathcal G(f)$$ is not a homeomorphism? Sorry but I do not understand this: $\endgroup$ Apr 21, 2022 at 17:10
  • $\begingroup$ indeed, the compositions $\pi_i\circ\Gamma$ for $i=1,\dots,n$ and $\pi_{n+1}\circ\Gamma$ are continuous (is this false?) so that $\Gamma$ is continuous and moreover the map $$h:\mathcal G(f)\owns\big(x,f(x)\big)\rightarrow x\in A$$ is by analogous arguments continuous with respect the subspace topology (is this false?) so that observing that $\Gamma$ and $h$ are the inverse of each other then we conclude that $\Gamma$ is an homeomorphism: where's now the wrong? Forgive my confusion. $\endgroup$ Apr 21, 2022 at 17:11

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