# Is there an elementary proof for the fact that homeomorphism preserves open sets in Euclidean Spaces?

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.

• 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. – Brian M. Scott Jul 30 '20 at 0:21
• @BrianM.Scott Huh? You are given $f[A]=\mathbb{R}^n$. – user10354138 Jul 30 '20 at 0:24
• @user10354138 $f(A)$ is not necessarily equal to $\mathbb{R}^n$ – rowcol Jul 30 '20 at 0:30
• @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 $\Bbb R^n$. – Brian M. Scott Jul 30 '20 at 0:38
• @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. – user10354138 Jul 30 '20 at 0:42

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 $$n, the image cannot be open because the image cannot contain an open ball in $$\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".

• $f(U)$ 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. – Daniel Fischer Jul 30 '20 at 19:35
• @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.) – Eric Towers Jul 30 '20 at 21:58
• 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. – Daniel Fischer Jul 30 '20 at 22:06
• @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. – Eric Towers Jul 30 '20 at 22:13
• For a homeomorphism. But if I'm not misunderstanding, the OP means an embedding. – Daniel Fischer Jul 30 '20 at 22:15

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.

• But in this case $f(A)$ is not necessarily $\mathhbb{R}^n$ – rowcol Jul 30 '20 at 0:38
• 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. – Brian M. Scott Jul 30 '20 at 0:41
• @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. – user10354138 Jul 30 '20 at 0:44
• @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. – Brian M. Scott Jul 30 '20 at 0:51
• $f$ is a homeomorphism onto its image. This means that $f : A \to f(A)$ is a homeomorphism, but not that $f(A) = \mathbb R^n$. – Paul Frost Jul 30 '20 at 10:47

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.

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.