Let $X$ be a topological space, for any $x\in X$, there exists a set $U_x$ containing an open neighborhood of $x$ such that $U_x$ is homeomorphic to an open subset $V_x$ of $\Bbb R^n(n\in \Bbb N)$. Please prove $U_x$ is open for any $x\in X$.
Is a (not a priori open) neighborhood of every point which is homeomorphic to an open subset of $\mathbb{R}^n$ in fact open?
2
$\begingroup$
general-topology
algebraic-topology
-
0I added the tag "algebraic-topology" because I know if you haven't studied algebraic topology, you may not solve the problem. – 2017-01-25
-
0This looks like a homework. What did you try to solve this problem? – 2017-01-25
-
0Not a homework, just posted by myself. – 2017-01-25
-
1Still, share you own thought. – 2017-01-25
-
0For instance: Why do you think algebraic topology is needed here? Are you familiar with the proof of the invariance of domain theorem? Did you try to repeat the arguments and where did you get stuck?... – 2017-01-25
-
0You know the answer, O(∩_∩)O~. – 2017-01-25
-
0http://www.math.org.cn/forum.php?mod=viewthread&tid=30132&extra=page%3D12&page=2 – 2017-01-25