# If $X$ is a Banach space, $B\subseteq X$ closed and bounded, $C\subseteq X$ compact, then $B+C$ is compact iff $B$ is compact.

If $X$ is a Banach space, $B\subseteq X$ closed and bounded, $C\subseteq X$ compact, then $$A\equiv B+C=\{x+y\mid x\in B, y\in C\}$$ is compact iff $B$ is compact The $(\impliedby)$ direction is easy, since we can define a sequence $a_n=b_n+c_n$ for $b_n\in B$ and $(c_n)\in C$, where each summand has a convergent subsequence.

However, the $(\implies)$ is not so clear.

Is the idea below correct?

For any sequence $(b_n)\in B$, if $(b_n)\to b$, then $b\in B$ since $B$ is closed. Since $A$ is compact, $(a_{n})\in A$ has a convergent subsequence, $(a_{n_k})\to a$. Define $a_n=b_n+c_n$. Since $C$ is compact, we know that $c_n$ has a convergent subsequence, too. So if $b_{n_k}=a_{n_k}-c_{n_k}$, then $(b_{n_k})$ is a convergent subsequence, since $(a_{n_k})$ and $(c_{n_k})$ converge. Hence, $B$ is compact.

• You defined $a_n$ as $b_n+c_n$. What is $c_n$? – José Carlos Santos Dec 7 '17 at 16:53
• You need the condition $C \neq \varnothing$. $B + \varnothing = \varnothing$ is compact whatever $B$ is. – Daniel Fischer Dec 7 '17 at 17:14

## 2 Answers

Another approach would be as follows - Assuming C is non-empty, since $C \subset X$ is compact, so is $-C$. Consider the function $f : X$ x $X \rightarrow X$ given by $f(a,b) = a + b$ . Then, since $f$ is continuous, (addition is a continuous function on a Banach space), and it is given that A is compact, hence A - C is also compact, thus proving that B is compact. ( since image of compact sets under a continuous map is compact)

With regards to your approach, let $(b_n)_{n \in \mathbb N}$ be any sequence in B. Define $(c_n)_{n \in \mathbb N}$ to be the constant sequence c for some $c \in C$. Then $a_n = b_n + c$ $\forall n \in \mathbb N$ and since A is compact, $(b_n) + c$ has a convergent subsequence $(b_{n_k})$ which converges to some $b + c \in B + C$ since B is closed & C is compact and thus closed. Now since we started with $c \in C$ we have that $b \in B$ which implies B is compact.

Suppose that $B$ is not compact. Then there is a sequence $(b_n)_{n\in\mathbb N}$ of elements of $B$ such that none of its subsequences converges to an element of $B$. Since $B$ is closed, that means that no subsequence converges at all. Fix some $c\in C$ (here, I am assuming that $C\neq\emptyset$). Then $(b_n+c)_{n\in\mathbb N}$ is a sequence of elements of $B+C$ and therefore some subsequence $(b_{n_k}+c)_{k\in\mathbb N}$ converges to some element $b'+c'$, with $b'\in B$ and $c'\in C$. But then $(b_{n_k})_{k\in\mathbb{N}}$ converges to $b'+c'-c$. Therefore, a contradiction was reached.

• @Shashi You are right, of course. I hope that all is well now. Please tell me what you think. – José Carlos Santos Dec 7 '17 at 17:30
• Now, it is fine, sir! – Shashi Dec 7 '17 at 17:54