Functional Analysis (Big Rudin)

[Edwin]

I had a quick question on a part of a proof in chapter 1 of Functional Analysis, by Professor Rudin.

Theorem 1.10 states

"Suppose K and C are subsets of a topological vector space X. K is compact, and C is closed, and the intersection of K and C is the empty set. Then 0 has a neighborhood V such that

$$(K+V) \cap (C+V) = \emptyset$$"

In the proof of this theorem, Professor Rudin starts out by proving the following proposition

"If W is a neighborhood of 0 in X, then there is neighborhood U of 0 which is symmetric (in the sense that U = -U) and which satisfies

$$U + U \subset W$$."

The question I have is about the next part of Rudin's proof

"Suppose K is not empty, and consider x in K, since C is closed, and since x is not in C, and since the topology of X is invariant under translations, the preceding proposition shows that 0 has a symmetric neighborhood

$$V_{x}$$

such that

$$x + V_{x} + V_{x} + V_{x}$$

does not intersect C..."

Is Professor Rudin's reasoning as follows:

Since C is closed, the the complement C* of C is open in X by definition. Since x is not contained in C, then x is contained in the complement of C, C*. Since C* is open, and contains x, then C* is a neighborhood of x. Since C* is a neighborhood of x, then the set

$$-x+C^{*}$$ is a neighborhood of 0 in X. Thus by the preceding proposition, there exists a symmetric neighborhood

$$V_{x}$$

of 0 in X such that

$$V_{x} + V_{x} + V_{x} \subset -x + C^{*}$$.

Since the topology of X is translation invariant, then

$$V_{x} + V_{x} + V_{x} \subset -x + C^{*}$$ iff

$$x+V_{x} + V_{x} + V_{x} \subset x+(-x) + C^{*} = C^{*}$$,

so that $$x+V_{x} + V_{x} + V_{x}$$ does not intersect C...?

Is this line of reasoning correct?

 

[MathematicalPhysicist]

It looks correct, but I'm not Rudin. (-: