Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1

Theorem
Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$. Let $W$ be an open neighborhood of ${\mathbf 0}_X$.

Then there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that:


 * $U + U \subseteq W$

Proof
Since ${\mathbf 0}_X + {\mathbf 0}_X = {\mathbf 0}_X$, we can apply Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods to obtain an open neighborhood $V_1$ of ${\mathbf 0}_X$ and an open neighborhood $V_2$ of ${\mathbf 0}_X$ such that:


 * $V_1 + V_2 \subseteq W$

Note at this point we do not necessarily have $V_1 = V_2$.

From Dilation of Open Set in Topological Vector Space is Open, $-V_1$ and $-V_2$ are open neighborhoods of ${\mathbf 0}_X$.

Let:


 * $U = V_1 \cap V_2 \cap \paren {-V_1} \cap \paren {-V_2}$

Since $U$ is the finite intersection of open sets, it is itself open.

Since ${\mathbf 0}_X$ is contained in each of the four sets of the intersection, we have ${\mathbf 0}_X \in U$.

So $U$ is an open neighborhood of ${\mathbf 0}_X$.

We show that $U$ is symmetric and that:


 * $U + U \subseteq W$

Let $x \in U$.

Then $x \in V_1$, $x \in V_2$, $x \in -V_1$ and $x \in -V_2$.

So we have that $-x \in -V_1$, $-x \in -V_2$, $-x \in V_1$ and $-x \in -V_2$.

So:


 * $-x \in \paren {-V_1} \cap \paren {-V_2} \cap V_1 \cap V_2 = U$

So we have that $U$ is symmetric.

Now let $x, y \in U$.

We want to show that $x + y \in W$.

Since $x, y \in U$, we have $x \in V_1$ and $y \in V_2$.

So, we have $x + y \in V_1 + V_2$.

We have $V_1 + V_2 \subseteq W$, so $x + y \in W$.

So, we have $U + U \subseteq W$.

Hence $U$ is our desired open neighborhood.