Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods

Theorem
Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $a, b \in X$.

Let $W$ be an open neighborhood of $a + b$.

Then there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:


 * $W_1 + W_2 \subseteq W$

where $W_1 + W_2$ denotes the sum of $W_1$ and $W_2$.

Proof
Equip $X \times X$ with the product topology.

From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.

Define $S : X \times X \to X$ by:


 * $\map S {x, y} = x + y$

for each $\tuple {x, y} \in X \times X$.

From the definition of a topological vector space, $S$ is continuous.

In particular, $S$ is continuous at $\tuple {a, b}$.

Hence, there exists an open neighborhood $V$ of $\tuple {a, b}$ such that:


 * $S \sqbrk V \subseteq W$

From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:


 * $W_1 \times W_2 \subseteq V$

Then we have, from Image of Subset under Mapping is Subset of Image:


 * $S \sqbrk {W_1 \times W_2} \subseteq S \sqbrk V$

Since:


 * $S \sqbrk {W_1 \times W_2} = W_1 + W_2$

this gives:


 * $W_1 + W_2 \subseteq W$

as required.