If $A$ and $B$ are compact subset of $\mathbb R$ , then so is $A+B$.

Question

Prove the following:

If $A$ and $B$ are compact subset on $\mathbb R$ , then so is $A+B:= \{a+b\mid a\in A ,b\in B\}$.

I was actually thinking about first proving that if $A\subseteq \mathbb R$ is compact, then every sequence in $A$ has a convergent subsequence whose limit belong to $A$. After that, by using the addition of 2 convergent subsequence of $A$ and $B$, I want to prove that every sequence in $A+B$ also has convergent subsequence thus proving that $A+B$ is compact. Is this alright or does anyone have a better solution? Thx

Answer 1

Continuity preserves compactness. That is if $f \colon X \to Y$ is a continuous function betweeen topological spaces and $K \subset X$ is compact then the image $f(K)$ is compact.

Addition $(-+ -) \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous function and if $A$ and $B$ are compact then $A \times B$ is compact in $\mathbb{R} \times \mathbb{R}$. Hence the image $A + B$ is compact in $\mathbb{R}$.

Answer 2

I assume you are using the fact that $X$ is compact if every sequence in $X$ has a convergent subsequence with a limit in $X$.

If that is the case, then what you wrote is not completely correct. You cannot just take two sequences in $A$ and $B$ and show that their sum has a convergent subsequence. What you need to do is to take a sequence in $A+B$, then show that it has a convergent subsequence.

For that, take any sequence in $A+B$ and see if you can decompose it into elements from $A$ and $B$, then find convergent subsequences there. Try to play around a bit.

Answer 3

That does not quite work. Given a sequence $\{a_i+b_i\}$ in $A+B$, if $\{n_i\}$ and $\{m_i\}$ are two sub-sequences of indices so that $a_{n_i}$ converges and $b_{m_i}$ converges, we can't just add these sequences because we want a subsequence of $\{a_i+b_i\}$. $\{a_{n_i}+b_{m_i}\}$ is not in general a subsequence of $\{a_i+b_i\}$. We need $m_i=n_i$.

The trick is to take a subsequence of $a_i$ that converges, and then a sub-sequence of the corresponding subsequence of the $b_i$. That is, pick $n_i$ first, then $m_i$ as a subsequence of $n_i$.

Answer 4

First, a closed set plus a compact set is a closed set. Here $A, B$ are compact, so at least $A+B$ is closed. Second, bounded set plus bounded set is still a bounded set, so $A+B$ is bounded as well, $A+b$ is bounded and closed in $\mathbb{R}$ means $A+B$ is compact.

Answer 5

Of course Chris Cave's answer provides is a very elegant way to show the result. But if you want to work with convergent subsequences this is also possible.

Let $a_k\in A$, $b_k\in B$ be sequences. We have to show that $c_k:=a_k+b_k$ has a convergent subsequence in $A+B$. Since $A$ is compact we find a convergent subsequence $a_{k_i}$ of $a_k$ with limit $a\in A$. Since $B$ is compact we find a convergent subsequence $b_{k_{i_j}}$ of $b_{k_i}$ with limit $b\in B$.

Then we have $$\lim_{j\to\infty} c_{k_{i_j}} = \lim_{j\to\infty} a_{k_{i_j}} + \lim_{j\to\infty}b_{k_{i_j}} = a+b\in A+B.$$

I hope you proved already $\lim (a_k +b_k)=\lim a_k + \lim b_k$ for convergent sequences $a_k$ and $b_k$ (that's actually nothing less than using the continuity of the addition).