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?
 A: 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
A: 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.
A: 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.)
A: Here is another proof inspired by the sequential proof in the $\mathbb{R}^n$ case.
Let $x\in \overline{A + B}$. Let us show that it is in fact in $A+B$.
I assume the topological vector space $X$ is Hausdorff so that $\displaystyle \bigcap_{i\in I} V_i = \{x\}$ where $V_i$ runs over all neighborhoods of $x$.
By definition of being in the adherence, $\forall\ i\in I,\enspace V_i \cap (A+B) \neq \emptyset$. Let us introduce the compact subset
$$K_i := \overline{\left\lbrace a \in A,\ \exists\, b\in B,\ a+b \in V_i  \right\rbrace } \subset A$$
(compact as a closed subset of the compact subset $A$).
If $\ \displaystyle \bigcap_{i\in I} K_i = \emptyset $ then by the finite intersection property of compact sets, $\exists\ J \subset I,\ J$ finite such that $\ \displaystyle\bigcap_{j\in J} K_j = \emptyset $ but this means that
$$ \left(\bigcap_{j\in J} V_j \right)\cap (A+B)= \emptyset  $$
but since a finite intersection of neighborhoods of $x$ is also a neighborhood of it, this contradicts the fact that $x\in \overline{A + B}$.
Hence $\ \displaystyle \bigcap_{i\in I} K_i \neq \emptyset ,\ \exists\ a_0\in A$ s.t. $\forall\ i\in I,\ \exists\ b_i \in B,\ a_0 + b_i \in V_i$.
One may say that the function $i \mapsto a_0 + b_i $ has limit $x$ "along the filter $I$" and thus that
$$\lim_{\overset{\longrightarrow}{I}} b_i = x - a_0=: b_0 \in B$$
since the latter is closed. Finally, one does have $x=a_0 + b_0 $ with $a_0 \in A,\ b_0\in B$.
Comment:

*

*I was interested in a (general if possible) proof of this statement while reading the proof of Thm 1.7 p.7 of "Functional Analysis, Sobolev Spaces and Partial Differential Equations" (UTX, 2011) by H. Brezis, or while reading (ii) of the proof of Proposition 1.3 p.69 of "Eléments de distributions et d'équations aux dérivées partielles" by C. Zuily

*In fact there is a sequential proof for a slightly more general case in "Eléments de distributions et d'équations aux dérivées partielles" by C. Zuily, Proposition 6.3 p.79

