Interior of Translation of Set in Topological Vector Space is Translation of Interior

Theorem

Let $K$ be a topological field.

Let $\struct {X, \tau}$ be a topological vector space over $K$.

Let $E \subseteq X$.

Let $x \in X$.

Then:

$\paren {E + x}^\circ = E^\circ + x$

Proof

Let $y \in \paren {E + x}^\circ$.

Then there exists an open neighborhood $U$ of $y$ such that $U \subseteq E + x$.

Then from Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of $y - x$ such that $U - x \subseteq E$.

Hence $y - x \in E^\circ$.

So $y \in E^\circ + x$.

We therefore obtain $\paren {E + x}^\circ \subseteq E^\circ + x$.

Now let $y \in E^\circ + x$.

Then $y - x \in E^\circ$.

So there exists an open neighborhood $U$ of $y - x$ such that $U \subseteq E$.

Then from Translation of Open Set in Topological Vector Space is Open, $U + x$ is an open neighborhood of $y$ such that $U + x \subseteq E + x$.

Hence $y \in \paren {E + x}^\circ$.

We therefore obtain $E^\circ + x \subseteq \paren {E + x}^\circ$.

So $\paren {E + x}^\circ = E^\circ + x$.

$\blacksquare$