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.

  • $\begingroup$ how do you know the $n_k$ are the same for both $A$ and $B$? $\endgroup$ – bringingdownthegauss Aug 5 '15 at 21:54
  • 1
    $\begingroup$ @bringingdownthegauss: First choose a convergent subsequence from $a_n$, then select a further subsequence of this subsequence from $b_n$. $\endgroup$ – copper.hat Aug 6 '15 at 5:37
  • 1
    $\begingroup$ @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? $\endgroup$ – Markus Apr 8 '16 at 3:14
  • $\begingroup$ @Markus: That will work too. $\endgroup$ – 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.


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.