Prove that the sum of two compact sets in $\mathbb R^n$ is compact. 
Given two sets $S_1$ and $S_2$ in $\mathbb R^n$ define their sum by 
  $$S_1+S_2=\{x\in\mathbb R^n; x=x_1+x_2, x_1\in S_1, x_2\in S_2\}.$$
  Prove that if $S_1$ and $S_2$ are compact, $S_1+S_2$ is also compact.

Prove that the sum of two compact sets in $\mathbb R^n$ is compact. 
Compact set is the one which is both bounded and closed. The finite union of closed sets is closed. But union is not the same as defined in the task. I so not know how to proceed. I do understand that I need to show that the resulting set is both bounded and closed, but I do know how to do that. 
 A: As $S_1$ and $S_2$ are bounded, every element in these sets are bounded as well. That is there exists $M_1 , M_2 > 0$ such that:
$$ \left\lVert x_1 \right\rVert \leq M_1 \textrm{ and } \left\lVert x_2 \right\rVert \leq M_2 , \forall x_1 \in S_1 , \forall x_2 \in S_2 . $$
Therefore for any $x \in S$, that is there exists $x_1 \in S_1$ and $x_2 \in S_2$, we have
$$ \left\lVert x \right\rVert \leq \left\lVert x_1 \right\rVert + \left\lVert x_2 \right\rVert \leq M_1 + M_2 $$
and thus S is bounded.
Similarly for the closedness. For any sequence $\left\lbrace x_{n} \right\rbrace _{n \geq 0} \in S$, there exists two sequences $\left\lbrace x_{1,n} \right\rbrace _{n \geq 0}$ and $\left\lbrace x_{2,n} \right\rbrace _{n \geq 0}$, which converge to some $x_{1,*} \in S_{1}$ and $x_{2,*} \in S_{2}$, respectively since they belong to some compact sets, such that $x_{n} =  x_{1,n} +  x_{2,n}$, and hence 
$$ x_{n} \to x_{1,*} + x_{2,*} $$
which is belongs to $S$. 
A: $f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, f(x,y) = x+y$ is continuous and 
$S_1 + S_2 = f[S_1 \times S_2]$. 
$S_1 \times S_2$ is compact by Tychonoff, e.g.
A: Yet another way to prove this is to use sequential compactness: suppose $y_n = x_n + x_n'$ is a sequence in the sum. There is then a subsequence of $(x_n)$ that converges in $S_1$, say $(x_{n_j})$, and then there is a subsequence of $(x_{n_j}')$ that converges in $S_2$, say $(x_{n_{j_l}}')$. Then certainly $y_{n_{j_l}}$ is a subsequence of $(y_n)$ that converges in $S_1+S_2$. 
