Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U\subset\Bbb R^n$ be an open set ($n > 1$), $f : U\to\Bbb R^n$ a continuous function with the following property: There exists a closed discrete subset $X\subset U$ such that $ f\left| {_{U - X} } \right. $ is locally a homeomorphism. Prove that $f$ is an open map

I have no idea what can I do here :S

share|cite|improve this question
Any ball $B$ intersects $X$ in finitely many points, since $X$ is closed and discrete. Maybe this implies that $f(B)$ is open... – Siminore May 1 '12 at 8:15
Hint: have a look at the case $n=1$, for which the statement is not true (try to construct a counterexample). What's the difference to $n>1$? – user20266 May 1 '12 at 9:13
Actually, when $n>2$ you can get more from that: $f$ is a locally homeomorphism from $U$ to $\mathbb{R}^n$ by simply-connected $S^{n-1}$. – Yuchen Liu May 2 '12 at 5:06
up vote 0 down vote accepted

Since $f$ is a homeomorphism near any point of $U-X$, hence we need only to prove $f$ is open near every point $p$ of $X$. Because $X$ is discrete, we can choose a small open ball $B(p,r)$ around $p$ such that $B(p,r)\cap X=\{p\}$. Denote $q=f(p)$, so we need only to prove that $f(B(p,r))$ is a neighborhood of $q$.

Since $f$ is locally homeomorphism in $U-X$, $f^{-1}(q)-X$ must be a discrete set, hence $f^{-1}(q)$ is also discrete. Therefore, we can choose $r>0$ properly such that $q\notin f(\partial B(p,r))$.

By continuousness of $f$, we have $f(\partial B(p,r))$ compact, hence there exists $\delta>0$ such that $B(q,2\delta)\cap f(\partial B(p,r))=\emptyset$.

Denote $B_0(p,r)=B(p,r)-\{p\}$, $A_{\epsilon}=\overline{B(q,\delta)}-B(q,\epsilon)$, where $0<\epsilon<\delta$. Let's look at the set $$C_\epsilon:=A_\epsilon\cap f(B(p,r)).$$

On the one hand, $C_\epsilon=A_\epsilon\cap f(\overline{B(p,r)})$ is compact hence closed in $A_\epsilon$; on the otherhand $C_\epsilon=A_\epsilon\cap f(B_0(p,r))$ is open in $A_\epsilon$ (since $f(B_0(p,r))$ is open by $f$ locally homeomorphism).

Conclusion: $C_\epsilon$ is both open and closed in $A_\epsilon$.

Note that $A_\epsilon$ is connected if and only if $n>1$, so if $n>1$ we have $C_\epsilon=A_\epsilon$ or $\emptyset$. For sufficiently small $\epsilon$, we always have $C_\epsilon \neq \emptyset$, hence $C_\epsilon =A_\epsilon$ for every $\epsilon>0$ sufficiently small. That is to say, $A_{\epsilon}\subset f(B(p,r))$ for $\epsilon>0$ sufficiently small. Therefore, put $\epsilon\rightarrow 0$ we have $B(q,\delta)\subset f(B(p,r))$, hence finish the proof.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.