Is sum of infinite many compact sets closed? Consider a sequence of compact sets $\{S_i\}_{i=1}^{\infty}$, where $S_i \subset \mathbb{R}^n$ and $0\in S_i$ for all $i$. Define
$$\sum_{k = 1}^\infty S_k := \Big\{\sum_{k = 1}^\infty s_k \;\;  \Big| \;\; \text{ this sum exists and } s_k \in S_k \;\; \forall k\Big\}$$
How to show that $\sum_{k = 1}^\infty S_k$ is closed in $\mathbb R^n$?

Here is what I tried.
Define $S:= \sum_{k = 1}^\infty S_k$.
I want to prove it by contradiction. So suppose there exists a vector $s \notin S$ but there exists a sequence of $\{s_i\}_{i=1}^{\infty}, s_i \in S$ such that $\lim_{n \to \infty} s_n = s$.
Since $s_i \in S$, then there exists a sequence $z_{ij} \in S_j, j=1, 2, \cdots$, such that $\lim_{m \to \infty} \sum\limits_{j=1}^m z_{ij} = s_i$. Then $s = \lim\limits_{n \to \infty} \lim\limits_{m \to \infty} \sum\limits_{j=1}^m z_{nj}$.
Then I want to use the interchange of these two limits but I do not know which theorem I should refer to.
If I use the interchange directly, I can get $s = \lim\limits_{m \to \infty} \sum\limits_{j=1}^m \lim\limits_{m \to \infty}  z_{nj} = \lim\limits_{m \to \infty} \sum\limits_{j=1}^m  z_j$, where $z_j \in S_j$ since $S_j$ is compact. Then $\lim\limits_{m \to \infty} \sum\limits_{j=1}^m  z_j \in S$, which forms a contradiction.

But the problem is I do not know if I can use the interchange of two limits.
 A: This is false. Counterexample: $n=1$,
$$S_i = \{0, 1+ 1/i\}$$
for each $i=1, 2,\cdots$.
Then $1$ is a limit point of $\sum S_i$ since $1+1/i\in \sum S_i$, but $1$ is not in it: every points in $\sum S_i$ is either
$$ 0 = 0+ 0 + \cdots +$$
or is $>1$ (when one of $s_i \neq 0$).
A: Assume $n = 1$. Let $\mathbf S = \sum_{k = 1}^\infty S_k$. Take a sequence $\mathbf x_1, ..., \mathbf x_n$ in $\mathbf S$. We must show that the limit point of that sequence is in $\mathbf S$. Wlog assume $\mathbf x_1 \le \mathbf x_2 \le \cdots$ (this is easy to justify). For each $\mathbf x_i$, let $(s_{i,1}, s_{i,2}, ...)$ be such that $s_{i,k} \in S_k$ for all $k$, and $\sum_{k = 1}^\infty s_{i,k} = \mathbf x_i$.
Claim. We can construct the $\mathbf x_i$ from underlying summands that are monotonically increasing across the sequence $\mathbf x_1, \mathbf x_2, ...$ -- formally, there are elements $(s'_{1,1}, s'_{1,2}, ...)$, $(s'_{2,1}, s'_{2,2}, ...)$ such that $\sum_{k = 1}^\infty s_{i,k} = \sum_{k = 1}^\infty s'_{i,k} = \mathbf x_i$ for all $i$, and $s'_{1,i} \le s'_{2,i} \le s'_{3,i} \le \cdots $ for all $i$. (I'm not going to prove this here, but I'm pretty confident it's true -- think about it.)
Given this, we can construct the summands $s^*_1, s^*_2, ...$ with $s^*_k \in S_k$ of the limit point $\mathbf x^*$ in the obvious way. Namely, $s_j^* := \lim_{k = 1}^\infty s'_{k,j}$. We know this limit point exists because the $s'_{k,j}$ are monotonically increasing and $S_j$ is compact.
By construction, $\mathbf x^* \in \mathbf S$. All that's left to prove is that $\mathbf x_1, \mathbf x_2, \cdots$ does, in fact, converge to $\mathbf x^*$. I'm also not doing this, but I'm about 90% confident that it's true. I think you need to assume something like, there is a sequence $\epsilon_1, \epsilon_2, ... $ with $\epsilon_i \in \mathbb R$ and $\lim_{i = 1}^\infty \epsilon_i = 0$ and $S_k \subseteq [-\epsilon_k, \epsilon_k]$. You can assume this without loss of gneerality because if it isn't true, then $\mathbf S = \mathbb R$ so the problem is trivial.
If all that is correct, you still have to extend it to $n > 1$, but I think the core ideas are here.
