# Sum of compact sets

Let $A,B$ two non-empty compact subsets of a normed space X. How can we prove that the set $S=A+B=\{a+b : a \in A, b \in B\}$ is compact?

Here's my reasoning:
Let $\Omega = \{\Omega_1, \Omega_2,…\}$ be an open cover of $S$. $\Omega$ induces two open cover $X,Y$ respectively of $A,B$, where

$X_i = \{a \in A : a+b \in \Omega_i~~for~some~b\}$
$Y_i = \{b \in B : a+b \in \Omega_i~~for~some~a\}$

in practice $\Omega_i = X_i + Y_i$.

Now my idea is build a finite subcover this way: consider a finite subcover $X_F = \{X_j : j \in J\}$ where $J$ is a finite set of indices. If $Y_F = \{Y_j : j \in J\}$ is a finite subcover of $B$, then we are done as $\{\Omega_j : j \in J\}$ is a subcover of $\Omega$. Otherwise we can keep adding indexes to set $J$ until $Y_F$ becomes a finite subcover of $B$.

This is not too formal but I don't want to make a simple question unreadable, I think the idea should be clear.

Does this works? Is there a better/easier way to get the same result?

Remember $+$ is a function defined as

$$+:\begin{cases} A\times B\to A+B \\ (a,b)\mapsto a+b\end{cases}$$

Since addition is continuous on normed spaces and $A+B$ is the image of the compact set $A\times B$ under this map, the image is compact.

Adam's proof is the cleanest. Here is an alternative proof:

Suppose $x_n \in A+B$ is Cauchy. We have $x_n = a_n+b_n$, with $a_n \in A$ and $b_n \in B$. Since $A,B$ are compact, we have $a_{n_k} \to a \in A$ and $b_{n_k} \to b \in B$ for some subsequence $n_k$. Hence $x_{n_k}=a_{n_k}+b_{n_k} \to a+b \in A+B$, and since $x_n$ is Cauchy, we have $x_n \to a+b$. Hence $A+B$ is complete.

Let $\epsilon>0$ and choose finite ${ 1 \over 2}\epsilon$-nets for $A,B$. That is, some finite collection $a_k,b_k$ such that for any $a \in A,b\in B$ there is some $k,k'$ such that $\|a-a_k\| < { 1 \over 2}\epsilon$, $\|b-b_{k'}\| < { 1 \over 2}\epsilon$. Let $x \in A+B$, then $x=a+b$ for some $a \in A,b\in B$. As above, there is some $k,k'$ such that $\|x-(a_k+b_{k'})\| = \|a+b-(a_k+b_{k'})\| \le \|a-a_k\|+ \|b-b_{k'}\| < \epsilon$. Hence the collection $a_k+b_{k'}$ form a finite $\epsilon$-net for $A+B$. Hence $A+B$ is totally bounded.

• how do you know the $n_k$ are the same for both $A$ and $B$? – bringingdownthegauss Aug 5 '15 at 21:54
• @bringingdownthegauss: First choose a convergent subsequence from $a_n$, then select a further subsequence of this subsequence from $b_n$. – copper.hat Aug 6 '15 at 5:37
• @copper.hat Why not start with an arbitrary sequence $(x_n)$ in $A+B$ and, arguing as in the first paragraph, conclude it has a convergent subsequence? Since $X$ is a normed (thus metric) space, sequential compactness and compactness are equivalent. Am I missing something? – Markus Apr 8 '16 at 3:14
• @Markus: That will work too. – copper.hat Apr 8 '16 at 3:33

$$(x_n)$$ be any sequence in $$A+B$$. Then we can write $$x_n=a_n+b_n$$ where $$a_n\in A$$ and $$b_n\in B$$. $$(a_n)$$ is a sequence in $$A$$, which is compact. Hence it has a subsequence $$(a_{n_k})$$ converging to some $$a\in A$$. $$(b_{n_k})$$ is a sequence in $$B$$ which is a compact space, so it also has subsequence $$(b_{n_{k_l}})$$ converging to $$b\in B$$. $$(a_{n_{k_l}})$$ also converges to $$a$$. Then $$(x_{n_{k_l}})$$ converges to $$a+b$$. So every sequence in $$A+B$$ has a convergent subsequence. Hence, it is compact.