Sum of closed and compact set in a TVS

I am trying to prove: $A$ compact, $B$ closed $\Rightarrow A+B = \{a+b | a\in A, b\in B\}$ closed (exercise in Rudin's Functional Analysis), where $A$ and $B$ are subsets of a topological vector space $X$. In case $X=\mathbb{R}$ this is easy, using sequences. However, since I was told that using sequences in topology is "dangerous" (don't know why though), I am trying to prove this without using sequences (or nets, which I am not familiar with). Is this possible?

My attempt was to show that if $x\notin A+B$, then $x \notin \overline{A+B}$. In some way, assuming $x\in\overline{A+B}$ should then contradict $A$ being compact. I'm not sure how to fill in the details here though. Any suggestions on this, or am I thinking in the wrong direction here?

Solution 1:

Suppose $x\notin A+B$, then for each $a\in A$, $x \notin a+B$, which is a closed set (since $v \mapsto a+v$ is a homeomorphism). Since every TVS is regular, there are open sets $U_a$ and $V_a$ such that $$ x\in U_a, \quad a+B \subset V_a, \quad \text{ and } U_a\cap V_a = \emptyset $$ Now, $$ V_a - B = \cup_{b\in B} (V_a - b) $$ is open and contains $a$. Hence, $A\subset \cup_{a\in A}(V_a - B)$. Since $A$ is compact, there is a finite set $\{a_1, a_2, \ldots a_n\}$ such that $$ A \subset \cup_{i=1}^n (V_{a_i} - B) $$ Let $U = \cap_{i=1}^n U_{a_i}$, then $U$ is a neighbourhood of $x$.

We claim that $U\cap (A+B) = \emptyset$. If not, then $y = a+b \in U\cap(A+B)$, then $$ y \in V_{a_i} \quad\text{ for some } i $$ and $y \in U_{a_i}$, which is a contradiction.

Solution 2:

If $x\notin (A+B)$, then $A\cap(x-B)=\varnothing$. Since $(x-B)$ is closed, it follows from Theorem 1.10 in Rudin's book that there exists a neighborhood $V$ of $0$ such that $(A+V)\cap(x-B+V)=\varnothing$. Therefore $(A+B+V)\cap(x+V)=\varnothing$ and, in particular, $(A+B)\cap(x+V)=\varnothing.$ As $(x+V)$ is a neighborhood of $x$, this shows that $x\notin \overline{(A+B)}$. (Proof taken from Berge's book.)

Solution 3:

Do not hesitate to use nets and subnets because there is no loss of generality (i.e. they are sufficiently powerful to match the needs of general topology). In general, filters are suited when subsets or domains (as, for example, for germs of functions or asymptotic analysis) are under consideration, and nets are suited for computation of limits (which can be made the case here).

Proof Let $z$ be a boundary point of $A+B$, there exists a net $(z_\alpha)_{\alpha\in X}$ in $A+B$ which converges to $z$ (not yet known as belonging to $A+B$ and to be proved to be so) i.e. putting $z\in \lim_{\alpha \in X}z_\alpha$ in the non-Hausdorff case.

One writes $z_\alpha=x_\alpha+y_\alpha$ with $x_\alpha\in A,\ y_\alpha\in B$. As $A$ is compact, there exists a subnet $x_{\phi(\beta)}$ with acceptable $\phi:Y\rightarrow X$ which converges in $A$. Set $x=\lim_{\beta}x_{\phi(\beta)}$, from what precedes $y_{\phi(\beta)}=z_{\phi(\beta)}-x_{\phi(\beta)}$ converges to $z-x$ and this point lies in $B$ (as $B$ is closed). So $z=x+y\in A+B$. QED