# 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.

• Use this definition of the compactness - another sequence contains an convergent subsequence Sep 17, 2017 at 10:14
• This Q is definitely a duplicate. Sep 17, 2017 at 13:03
• There is this (more general) question: Sum of closed and compact set in a TVS. I assume that something like that was asked here before - but it is probably not that easy to find. Sep 17, 2017 at 17:00
• An extention to this question: If $S_1$ and $S_1+S_2$ are compact, can we say that $S_2$ is also compact?
– amj
May 6, 2020 at 23:45
• @amj, we cannot. Take $S_1 = [0, 1]$ and $S_2 = [0, 1]\backslash\{1/2\}$ on the real line. Both $S_1$ and $S_1 + S_2 = [0, 2]$ are compact, yet $S_2$ is not compact. Sep 15, 2020 at 15:33

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 $S_1+S_2$.

$f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, f(x,y) = x+y$ is continuous and

$S_1 + S_2 = f[S_1 \times S_2]$.

$S_1 \times S_2$ is compact by Tychonoff, e.g.

• In case the OP is not familiar with Tychonoff, it's easy to see that $S_1\times S_2$ is closed and bounded.
– bof
Sep 17, 2017 at 10:26
• Tychonoff: overkill surely? Sep 17, 2017 at 10:38
• @LordSharktheUnknown Heine-Borel otherwise. Sep 17, 2017 at 10:53
• @Henno Brandsma Slick! Jan 7, 2018 at 15:49

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 $S$.