Proposition 8.1.8. Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
If $A$ is open, then $A + B$ is open.
If $A$ is closed and $B$ is compact, then $A + B$ is closed.
Proof. $(1)$: For every $x \in B$, $A + x$ is open by Definition 8.1.2, so
is open.
$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
By Proposition 8.1.7, $\overline{(-B) + U}\subset (-B) + U + U$, so
Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
so $x \in y + B \subset A + B$.$\square$