Sum of Closures is Subset of Closure of Sum in Topological Vector Space

Theorem
Let $K$ be a topological field.

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

Let $A, B \subseteq X$.

Then:


 * $A^- + B^- \subseteq \paren {A + B}^-$

where $A^-$, $B^-$ and $\paren {A + B}^-$ denote the closures of $A$, $B$ and $A + B$.

Proof
Let $a \in A^-$ and $b \in B^-$.

We aim to show that $a + b \in \paren {A + B}^-$.

It suffices to show that for each open neighborhood $W$ of $a + b$ we have:


 * $W \cap \paren {A + B} \ne \O$

Fix an open neighborhood $W$ of $a + b$.

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:


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

Since:


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

this gives:


 * $W_1 + W_2 \subseteq W$

Since $W_1$ is an open neighborhood of $a$ and $a \in A^-$ we have:


 * $W_1 \cap A \ne \O$

Since $W_2$ is an open neighborhood of $b$ and $b \in B^-$ we have:


 * $W_2 \cap B \ne \O$

Let:


 * $x \in W_1 \cap A$

and:


 * $y \in W_2 \cap B$

Then we have $x \in W_1$ and $y \in W_2$ so that:


 * $x + y \in W_1 + W_2$

Since $W_1 + W_2 \subseteq W$, we then obtain $x + y \in W$.

Since we also have $x \in W_1$ and $y \in W_2$, we have $x + y \in A + B$.

So:


 * $x + y \in W \cap \paren {A + B}$.

We can therefore conclude that:


 * $W \cap \paren {A + B} \ne \O$

Since $W$ was arbitrary, we have that $a + b \in \paren {A + B}^-$ as required.