Statement on a neighborhood of $0$ in topological vector spaces

Question

Consider a topological vector space $E$ and let $U\subseteq E$ be a neighborhood of $0$. I want to understand the proof of the following statement:

If $K\subseteq E$ is compact and $U$ is open with $K\subseteq U$, then there exists an open $W\subseteq E$ with $0\in W$, such that $K+W\subseteq U$.

The author of the book starts the proof as follows:

For every $x\in K$, one can pick $V$ a neighborhood of $0$ , such that $x+V\subseteq U$.

Unfortunately, I do not understand this step. Of course, since $U$ is a neighborhood of $0$, I can find an open set, which contains $0$ and translating this by $x$ would be a neighborhood of $x$, since the translation is an open map. However, I do not understand, why this should still be contained in $U$.

I have basically the following question: How can I guarantee, that the set $V$ is 'small' enough, such that $x+V$ is still contained in $U$?

Besides that, I want to mention that the statement above is only one bullet point of a Lemma. Only for this statement, the author requires, that $U$ should be open. Maybe it has to do with the openness of $U$?

Any help is appreciated! Thank you in advance!

Answer 1

This follows by the fact that $U$ is open and continuity of translation. Consider $x\in K\subset U$. Because $U$ is open there exists an open $x\in V'\subset U$. Now $V:=V'-x$ is an open containing $0$ by continuity of translation and has the property that $V+x\subset U$.

Answer 2

In regards to your question

I have basically the following question: How can I guarantee, that the set V is 'small' enough, such that x+V is still contained in U?

Every topological vector space is regular. See: Every Topological Vector Space is Regular

This means that you can always separate points and closed sets. In particular consider the closed set $C=E\setminus U$ and the point $x$, by this property you can find $W_1$ an open neighbourhood of $x$ and $W_2$ an open neighbourhood of $C$ that are disjoint. The fact that they are disjoint implies that $W_1\subset U$.

This choice also implies that considering $$V=(W_1 - x)\cap U$$ we find that $V$ is a neighborhood of 0 and $V \subset U$. Furthermore

$$x+V \subset W_1 \subset U$$