Prove that the sum of two compact sets in $\mathbb R^n$ is compact.

Question

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.

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 — kotomord, Sep 17 '17 at 10:14
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. — Martin Sleziak, Sep 17 '17 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 06 '20 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. — Nicolas Boumal, Sep 15 '20 at 15:33

score 17 · Answer 1 · answered Sep 17 '17 at 10:16

17

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

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

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

answered Sep 17 '17 at 10:16

Henno Brandsma

242,131

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 '17 at 10:26
5

Tychonoff: overkill surely? – Angina Seng Sep 17 '17 at 10:38
@LordSharktheUnknown Heine-Borel otherwise. – Henno Brandsma Sep 17 '17 at 10:53
@Henno Brandsma Slick! – WhySee Jan 07 '18 at 15:49

score 14 · Answer 2 · answered Sep 17 '17 at 11:32

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 $S_1+S_2$.

JKay · Answer 3 · 2017-09-17T11:16:11.243

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 $S$.

This is a wrong proof. It's needed to prove that $x_n = x_{1,n} + x_{2,n}$, for all $n$, in order for this to be true. This is not true. To correct this proof, write $x_n=x_{1,n}+x_{2,n}$. Then, extract a convergent subsequence of ${x_{1,n}}$. Let we say that ${x_{1,n_k}}k$ is such a convergent subsequence. After, extract a convergent subsequence of ${x{2,n_k}}k$, let us say ${x{2,n_{\ell_k}}}k$. Then, by definition, $x{n_{\ell_k}} = x_{1,n_{\ell_k}} + x_{2,n_{\ell_k}} $, which converges as a sum of convergent sequences. This proves that sum of sequentially compact is also SQ. — R. W. Prado, May 25 '23 at 10:33

Prove that the sum of two compact sets in $\mathbb R^n$ is compact.

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