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I was reading the proof of the following theorem in Rudin 2/e:

Theorem 1.21 If $n$ is a positive integer and $Y$ is an $n$-dimensional subspace of a complex topological vector space $X$, then every isomorphism of $\mathbb{C}^n$ onto $Y$ is a homeomorphism.

The initial part of the proof goes:

Let $S$ be the sphere which bounds the open unit ball $B$ of $\mathbb{C}^n$. Suppose $f:\mathbb{C}^n\to Y$ is an isomorphism. Put $K=f(S)$. Since $f$ is continuous, $K$ is compact. Since $f(0)=0$ and $f$ is one-to-one, $0 \not\in K$, and therefore there is a balanced neighborhood $V$ of $0$ in $Y$ which does not intersect $K$. The set $$E=f^{-1}(V)=f^{-1}(V\cap Y) $$ is therefore disjoint from $S$. Since $f$ is linear, $E$ is balanced, and hence connected. Thus $E\subset B$, because $0\in E$, and this implies that the linear map $f^{-1}$ takes $V\cap Y$ into $B$...

I understood everything in the proof except the last two sentences in the quoted part. Particularly,

  1. How do we go from $E$ being balanced to its being connected?
  2. How does this imply that $E\subset B$?

Please pardon me for my lack of formal training in topology (and particularly the concept of connectedness); but at least from the section on Connected Spaces in Munkres 2/e that I have, I couldn't figure out the above two points. Any help appreciated!

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What is $\mathcal{C}^n$? – Sanath K. Devalapurkar May 8 '14 at 0:47
Same as ${\mathbb C}^n$. – studiosus May 8 '14 at 5:32
@SanathDevalapurkar I changed it to $\mathbb{C}^n$ which is probably more common. – Fang Jing May 8 '14 at 13:24

1 Answer 1

up vote 3 down vote accepted
  1. To show that $E$ is connected, we could show that $E$ is path-connected. By definition, this means that there is a continuous path joining any two points $x$ and $y$ in $E$. Such a path can be constructed using the balancedness of $E$ by taking a path from $x$ to $0$, and then from $0$ to $y$. (I could be more precise if you like.) (Note that path-connected implies connected; in fact, in nice spaces like $\mathbb{R}^n$ and $\mathbb{C}^n$, the two notions are equivalent.)

  2. The unit sphere $S$ divides $\mathbb{C}^n$ into two connected components (i.e., two disjoint open sets): the unit ball $B$ (points $x$ where $|x|<1$), and $A$, which I'll use to denote the "exterior" of $S$ (points $x$ where $|x|>1$). Since $E$ is connected, it must lie entirely in either $A$ or $B$. (This is because it has been shown that $E\cap S=\emptyset$, so $E \subset A\cup B$. Suppose $E$ intersects both $A$ and $B$, then the intersections are both open, hence there exists a separation of $E$, contradicting that $E$ is connected.) Since $0 \in E$ and $B$ contains $0$, $E$ must be a subset of $B$.

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From your explanation, I think it can also be said that "any balanced set in a topological vector space is (path-) connected." – Fang Jing May 8 '14 at 15:14
Yes, that's correct. – Phillip Andreae May 8 '14 at 16:18
Nice edit. You filled in the details as to why $E$ lies entirely in $A$ or $B$. I made one small improvement to your edit: it should be $E \subset A \cup B$, not $E \in A \cup B$. – Phillip Andreae May 8 '14 at 16:20

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