If $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric and which satisfies $U+U\subset W$

Question

What is shown below is a reference from the text Functional Analysis by Walter Rudin.

If $X$ s a vector space over the field $\Bbb F$ then the following notation well be used $$ x+A=\{x+a:a\in A\}\\ x-A:=\{x-a:a\in A\}\\ A+B:=\{a+b:a\in A\wedge b\in B\}\\ \lambda A:=\{\lambda a:a\in A\} $$ for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.

Definition

If $\tau$ is a topology on a vector space $X$ such that

1. every point of $X$ is a closed set
2. the vector space operation are continuous with respect to $\tau$

we say that $X$ is a topological vector space equipped with the vector topology $\tau$.

So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.

Proposition

Let be $X$ a topological vector space over the field $\Bbb F$ and for some $v\in X$ and $\lambda\in\Bbb F$ we define the $v$-traslation and $\lambda$-multiplication functions through the condition $$ T_v(x):=x+v\,\,\,\text{and}\,\,\,M_\lambda(x):=\lambda x $$ for any $x\in X$. So the above two defined functions are homeomorphisms.

Proof. Omitted.

One consequence of the above proposition is that every vector topology $\tau$ is translation-invariant, that is a set $E\subseteq X$ is open if and only if each of its translates $v+E$ is open and thus $\tau$ is completely determined by any local base so that the open sets of X are then precisely those that are unions of translates of memebrs of any local base.

Lemma

Let be $X$ a topological vector space over the field $F$. So if $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric (that is $U=-U$) and which satisfies $U+U\subseteq W$.

Proof. To see this, note that $0+0=0$ and that addition is continuous and that $0$ therefore has neighborhoods $V_1$ and $V_2$ such that $V_1+V_2\subseteq W$. So if we put $$ U=V_1\cap V_2\cap (-V_1)\cap(-V_2) $$ the $U$ has the required properties.

So I ask to explain why effectively $0$ has neighborhoods $V_1$ and $V_2$ such that $V_1+V_2\subseteq W$ and why the neighborhood $U$ above defined has the required properties. So could someone help me, please?

Answer 1

Let $P: X \times X \rightarrow X$ denote the addition. It’s continuous at $(0,0)$ so $P^{-1}(W)$ is a neighborhood of $(0,0)$. In particular, it contains a “basic” open subset (for the product topology) which itself contains $(0,0)$. Such basic open subsets are the $V_1 \times V_2$ where $V_1,V_2$ are open subsets of $X$ with $0 \in V_1 \cap V_2$. Then, $U=V_1 \cap (-V_1) \cap (V_2) \cap (-V_2)$ is an open subset of $X$ containing $0$, clearly it’s symmetric, and $U+U=P(U \times U) \subset P(V_1 \times V_2) \subset W$, since $U \subset V_1$ and $U \subset V_2$.