Sum of Closures is Subset of Closure of Sum in Topological Vector Space/Proof 1

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 $a + b$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods, there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:

$W_1 + W_2 \subseteq W$

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.

$\blacksquare$