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Let $f:A \subset \mathbb{R}^n \to \mathbb{R}^n$ be a homeomorphism onto its image, where $A$ is an arbitrary subset of $\mathbb{R}^n.$ I want to show that for every open set $U \subset A$ (where $U$ is open in $\mathbb{R}^n$) the set $f(U)$ is open.

I saw related questions where other users mention about the Invariance of domain theorem. I know that such theorem has a hard proof. But note that the hypotesis of that theorem is "$f$ is a continuous bijection" and I am asumming something a little stronger: "$f$ is an homeomorphism".

I'm trying to prove this theorem by using the following proposition:

Proposition 1. $f:A \subset \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if for every open set $U \subset \mathbb{R}^m$, there exist an open subset $V\subset \mathbb{R}^n$ such that $f^{-1}(U)=V\cap A.$

It is possible doing it using the proposition 1?

This is my attempt:

Since $f$ is a homeomorphism, then $f^{-1}:f(A) \to \mathbb{R}^n$ is continuous. If $U\subset \mathbb{R}^n$ is open, then by Proposition 1 we have some open set $V \subset \mathbb{R}^n$ such that $(f^{-1})^{-1}(U)=f(U)=V\cap f(A). $ then... I was trying to proof that $V\cap f(A)$ is open using the continuity of $f,$ but I get stuck in this step.

Can someone please help me with this? Thanks in advance.

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    $\begingroup$ If you mean that $f[U]$ is open in $f[A]$, this is automatic from the definition of homeomorphism. If you mean that $f[U]$ is open in $\Bbb R^n$, it’s not necessarily true. $\endgroup$ – Brian M. Scott Jul 30 '20 at 0:21
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    $\begingroup$ @BrianM.Scott Huh? You are given $f[A]=\mathbb{R}^n$. $\endgroup$ – user10354138 Jul 30 '20 at 0:24
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    $\begingroup$ @user10354138 $f(A)$ is not necessarily equal to $\mathbb{R}^n$ $\endgroup$ – rowcol Jul 30 '20 at 0:30
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    $\begingroup$ @user10354138: No, we’re simply told that $\Bbb R^n$ is the codomain of $f$. And indeed if $f$ is a homeomorphism, and $A$ is compact or not connected, for instance, then $f[A]$ cannot possibly be all of $\Bbb R^n$. $\endgroup$ – Brian M. Scott Jul 30 '20 at 0:38
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    $\begingroup$ @BrianM.Scott "homeomorphism" != "homeomorphism onto its image". We are given $f\colon A\to\mathbb{R}^n$ is a homeomorphism, not $f\colon A\to\mathbb{R}^n$ is a homeomorphism onto its image. $\endgroup$ – user10354138 Jul 30 '20 at 0:42
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A homeomorphism is a bicontinuous bijection (onto its image -- redundant, since that is always what "$f:X \rightarrow Y$ is a [particular map class]" means). As it is a bijection, it has an inverse. Bicontinuous means that both the map and its inverse are continuous; equivalently, the map is both an open map and a continuous map. (... and both properties hold for the inverse map.)

Proposition 1 is too general to prove the theorem. Proposition 1 must write "$V \cap A$" since all that can be promised is that the preimage of an open set is relatively open in $A$. As a very easy example, take $A$ closed in $\Bbb{R}^n$ and for $U$ take any open set containing $f(A)$. Then $f^{-1}(U) = A$, so is closed in $\Bbb{R}^n$ and relatively open in $A$.

A concrete example when $n < m$: Let $A = [0,1]$, a closed set in $\Bbb{R}^1$, and let $f:A \subset \Bbb{R}^1 \rightarrow \Bbb{R}^2 : (x) \mapsto (x,0)$. Then take for $U$ the open ball centered at $(0,0)$ of radius $2$. Since $f(A) \subset U$, we can take any open set in $\Bbb{R}$ that contains $A$ as $V$, for instance, $V = (-1,3)$. Then $f^{-1}(U) = V \cap A = A$, but that intersection is not open (in $\Bbb{R}$); that intersection is relatively open in $A$.

A concrete example when $n > m$: Let $A = [0,1] \times [0,1]$, a closed set in $\Bbb{R}^2$, and let $f:A \subset \Bbb{R}^2 \rightarrow \Bbb{R}^1: (x,y) \mapsto (x)$. Then take for $U$ the open ball centered at $(0)$ of radius $2$. Since $f(A) \subset U$, we can take any open set in $\Bbb{R}^2$ that contains $A$ for $V$, for instance $(-1,3) \times (-1,3)$. Then $f^{-1}(U) = V \cap A = A$, but that intersection is not open in $\Bbb{R}^2$; that intersection is relatively open in $A$.

From these two examples, we see that proposition 1 must have "$V \cap A$" in its conclusion when $n \neq m$.

Something special happens when $n = m$ and that something special is not captured by proposition 1. In particular, the homeomorphism in the proposition for $n \neq m$ "crushes" subsets of the larger space to subsets of the smaller space (along either $f$ or $f^{-1}$ depending, respectively, on whether $n$ or $m$ is larger). (Think about this in the context of invariance of domain: the image of a homeomorphism looks like a possibly folded, twisted, and distorted embedding of the domain. When $n<m$, the image cannot be open because the image cannot contain an open ball in $\Bbb{R}^m$. When $n > m$, the same argument applies to the inverse.) When $n = m$, there are no "directions" along which a homeomorphism is permitted such crushing, but proposition 1 does not have a separate conclusion for this case, so does not capture this additional constraint when $n = m$.

Of course, anything provable by proof system $P$ can be proven in $P \cup \text{Prop. 1}$, by ignoring proposition 1. This can't really be said to meet your criterion "possible doing it using the proposition 1".

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  • $\begingroup$ $f(U)$ is of course open in $f(A)$, but I think the OP wants to show that $f(U)$ is open in $\mathbb{R}^n$. That's not trivial. $\endgroup$ – Daniel Fischer Jul 30 '20 at 19:35
  • $\begingroup$ @DanielFischer : OP does not ask for a proof of the theorem. In spite of this, I have added excruciating detail to the argument that $f(U)$ is open in $\Bbb{R}^n$. The only potential nontriviality is $f(\mathrm{int}\,A) = \mathrm{int}\,f(A)$, which seems at most epsilon far from a triviality to my eye. OP asks for a proof using proposition 1, which proposition isn't present in my argument. (It does appear in the linked proof.) $\endgroup$ – Eric Towers Jul 30 '20 at 21:58
  • $\begingroup$ I think we have a terminological problem. By "a homeomorphism onto its image" the OP means an embedding, as I understand it. And that doesn't a priori guarantee that $f(A)$ has nonempty interior. In this setting, invariance of domain gives us that, but that's the thing OP wants to avoid. $\endgroup$ – Daniel Fischer Jul 30 '20 at 22:06
  • $\begingroup$ @DanielFischer : OP gives $U \subset A$ and $U$ open in $\Bbb{R}^n$. This forces $U \subset \mathrm{int}\,A$ so $\mathrm{int}\,A \neq \varnothing$ (both "interior"s taken relative to $\Bbb{R}^n$). The linked proof then shows that $\mathrm{int}\,f(A) = f(\mathrm{int}\,A)$, so is nonempty. $\endgroup$ – Eric Towers Jul 30 '20 at 22:13
  • $\begingroup$ For a homeomorphism. But if I'm not misunderstanding, the OP means an embedding. $\endgroup$ – Daniel Fischer Jul 30 '20 at 22:15
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There is no need to carry $f(A)$ around since you know it is $\mathbb{R}^n$.

$f\colon A\to\mathbb{R}^n$ is a homeomorphism, so $f^{-1}\colon\mathbb{R}^n\to A$ is also a homeomorphism. In particular, $f^{-1}$ is continuous and so $f(U)=(f^{-1})^{-1}(U)$ is open.

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    $\begingroup$ But in this case $f(A)$ is not necessarily $\mathhbb{R}^n$ $\endgroup$ – rowcol Jul 30 '20 at 0:38
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    $\begingroup$ This is wrong. Nothing in the problem implies that $f[A]=\Bbb R^n$. In fact, since we are told that $A$ is an arbitrary subset of $\Bbb R^n$, there are definitely cases in which $f[A]\ne\Bbb R^n$. To take an obvious example, what if $A$ is finite? Or what if $A$ is compact? Continuous maps preserve compactness, and $\Bbb R^n$ is not compact. $\endgroup$ – Brian M. Scott Jul 30 '20 at 0:41
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    $\begingroup$ @BrianM.Scott Huh? "$g\colon X\to Y$ is a homeomorphism" necessarily means $g(X)=Y$. You seems to be confusing it with "$g\colon X\to Y$ is a homeomorphism onto its image" in which case we only have $g\colon X\to g(X)$ is a homeomorphism. $\endgroup$ – user10354138 Jul 30 '20 at 0:44
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    $\begingroup$ @user10354138: No, it does not necessarily mean anything of the kind. It can mean that; it does not always mean that. You may not have encountered any other usage; that just means that my experience is a bit wider than yours. $\endgroup$ – Brian M. Scott Jul 30 '20 at 0:51
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    $\begingroup$ $f$ is a homeomorphism onto its image. This means that $f : A \to f(A)$ is a homeomorphism, but not that $f(A) = \mathbb R^n$. $\endgroup$ – Paul Frost Jul 30 '20 at 10:47
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I would say this more or less follows from the definition of homeomorphism, or at least that's how one should think about it after they've seen a proof at least once. But in general, homeomorphism means topological isomorphism, so "of course" $f$ will send open sets to open sets. I'd argue that a "proof" of this fact makes the claim a little less believable, but never mind, we can prove it.

To "prove" this, it helps to look at things a little more generally. Let $B$ be the image of $f$. Now forget where $A$ and $B$ came from: they are both topological spaces and $f\colon A\to B$ is a homeomorphism. The inverse $g=f^{-1}$ is continuous, so if $U\subset A$ is open, then $g^{-1}(U) = (f^{-1})^{-1}(U) = f(U)\subset B$ is open, by the definition of continuity.

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Brouwer theorem for invariance of domains says that if n is a natural number and U is open in $R^n$ and $f:U->R^n$ is a continuous injection, then f is an open map.

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