Sum of Set and Open Set in Topological Vector Space is Open

Theorem
Let $K$ be a topological field.

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

Let $A$ and $B$ be subsets of $X$, with $A$ an open set.

Then the linear combination $A + B$ is open.

Proof
It can be shown that:


 * $\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$

If:


 * $\ds v \in \bigcup_{b \mathop \in B} \paren {A + b}$

then $v = a + b$ for some $a \in A$ and $b \in B$.

So $v \in A + B$, giving:


 * $\ds \bigcup_{b \mathop \in B} \paren {A + b} \subseteq A + B$

Conversely, suppose that $v \in A + B$.

Then there exists $a \in A$ and $b \in B$ such that $v = a + b$.

Then, we have $v \in A + b$, and so $v \in A + B$.

We therefore obtain:


 * $\ds A + B \subseteq \bigcup_{b \mathop \in B} \paren {A + b}$

So, by the definition of set equality, we have:


 * $\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$

From Translation of Open Set in Topological Vector Space is Open:


 * $A + b$ is open for each $b \in B$.

So $A + B$ is the union of open sets.

Since topologies are closed under union, we therefore have that $A + B$ is open.