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Let $W$ be a subset of an $n$-dimensional complex topology vector space $Y$ such that $0\notin W$.We have known that $Y$ is homeomorphic to $C^n$ and let $S$ be the unit sphere of $C^n$.Can anyone show me that $W$ is homeomorphic to $S$?

mathon
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1 Answers1

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This is certainly not true. For a counterexample, let $W$ be the set $\{|z|=1\}\cup\{|z|=2\}$, where $Y=\mathbb C$. The unit sphere $S$ is connected, but $W$ is not, so they cannot be homeomorphic.

Potato
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  • Let me repeat your statement,there are two connect components of $W$ in your counterexample,but $S$ is connected,so they cannot be homeomorphic. – mathon Jun 20 '13 at 08:26
  • @mathon Precisely. Yes. – Potato Jun 20 '13 at 08:27
  • @Potato.If I replace $W$ by $Y-V$ where $V$ is a balanced neighborhood of $0$ in $Y$,how about the conclusion? – mathon Jun 21 '13 at 01:53
  • @mathon What is a balanced neighborhood? – Potato Jun 21 '13 at 02:08
  • For simplicity,consider neighborhood.In fact,every neighborhood of $0$ contains a balanced neighborhood of $0$ in a topological vector space.And a set $B$ in a topological vector space is said to be balanced if $tB\subset B$ for every $t\in C$ with $|t|\leq 1$. – mathon Jun 21 '13 at 02:17
  • @mathon Will $W={.9\le |z| \le 1}\cup {1.9\le|z|\le 2}$ work, using the same setup as above? From your definition, it seems that $Y-W=V$ is a balanced neighborhood. – Potato Jun 21 '13 at 02:21