Consequences of theorem 1.10 Rudin functional analysis.

Question

Theorem 1.10 states

Let $X$ be a topological vector space, $K$ compact, $C$ closed and $K \cap C = \emptyset$ then there's a neighborhood of $0$ such that $$ (K + V) \cap (C+V) = \emptyset $$

As consequence of this we have later

1. Since $C+V$ is open then $\overline{K+V}\cap (C+V) = \emptyset$
2. In particular $\overline{K+V}\cap C = \emptyset$
3. The case where $K = \left\{ 0 \right\}$ is of particular interest.

I'm trying to work out by myself the details of 3, but I think the key is understanding both 1. and 2.

I think 1. is true because... if $x$ is a limit point for $K+V$ then any neighborhood $V_x$ is such that $V_x \cap (K+V) \neq \emptyset$ however since we have $(K+V) \cap (C+V) = \emptyset$ we also have $$V_x \cap (K+V) \cap (C + V) = \emptyset$$ I guess this is the argument for 1. About 2. I'm not sure how to prove it. And I then don't get how 3. follows from these two.

Can you clarify?

There's actually an answer given to such question, but that's expressed in terms that Rudin doesn't use and I'd like to understand how to use his terminology for that.

Answer 1

For 1. we have the following, assume for the sake of contradiction that $\exists x\in \overline{K+V}\cap C+V$ but that also $K+V\cap C+V =\emptyset$. Then, we have that $x\not\in K+V$ so $x$ is a boundary point of $K+V$. Since $x\in C+V$, there is a neighborhood $U_x\subset C+V$ with $x\in U_x$. Since $x$ is a boundary point of $K+V$, $U_x\cap K+V\neq\emptyset$. This contradicts that $K+V\cap C+V =\emptyset$.

For 2. we know that $0\in V$ since it is a neighborhood of $0$. Thus $C\subset C+V$ (just take $c+0$ for all $c\in C$), and so $C\cap\overline{K+V}=\emptyset$.

For 3. if $K=\{0\}$ then $0\not\in C$. The theorem then says that we can find a neighborhood of $0$ ($V$) which does not intersect $C$ (Note that $V+\{0\}=V$). This doesn't seem that significant to me because we know that if $C$ is closed and doesn't contain $0$ then the compliment $C^c$ must contain $0$ and is an open set, therefore it contains a neighborhood about $0$ contained entirely in $C^c$.