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 $W_2$.

Corollary 1

Let $W$ be an open neighborhood of ${\mathbf 0}_X$.

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

$U + U \subseteq W$

Corollary 2

Let $K \in \set {\R, \C}$.

Let $W$ be an open neighborhood of ${\mathbf 0}_X$.

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

$U + U \subseteq W$

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.

$\blacksquare$