Sum of Set and Open Set in Topological Vector Space is Open
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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.
$\blacksquare$