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.

  • $\begingroup$ Use this definition of the compactness - another sequence contains an convergent subsequence $\endgroup$
    – kotomord
    Sep 17, 2017 at 10:14
  • $\begingroup$ This Q is definitely a duplicate. $\endgroup$ Sep 17, 2017 at 13:03
  • $\begingroup$ 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. $\endgroup$ Sep 17, 2017 at 17:00
  • $\begingroup$ An extention to this question: If $S_1$ and $S_1+S_2$ are compact, can we say that $S_2$ is also compact? $\endgroup$
    – amj
    May 6, 2020 at 23:45
  • $\begingroup$ @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. $\endgroup$ Sep 15, 2020 at 15:33

3 Answers 3


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.

  • $\begingroup$ In case the OP is not familiar with Tychonoff, it's easy to see that $S_1\times S_2$ is closed and bounded. $\endgroup$
    – bof
    Sep 17, 2017 at 10:26
  • 2
    $\begingroup$ Tychonoff: overkill surely? $\endgroup$ Sep 17, 2017 at 10:38
  • $\begingroup$ @LordSharktheUnknown Heine-Borel otherwise. $\endgroup$ Sep 17, 2017 at 10:53
  • $\begingroup$ @Henno Brandsma Slick! $\endgroup$
    – WhySee
    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 $S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.