Topology in $\Bbb {R}^n$ Proof Exercise

Question

Given set A$\ \subset \Bbb {R}^n$ and set B$\ \subset \Bbb {R}^n$, we define A+B={$\text{ a+b such that a $\in$ A and b $\in$ B}\}$.Prove: $$\text{a) If A closed set, B compact set then A+B is a closed set}$$ $$\text{b) If A,B are compact sets then A+B is a compact set}$$

Answer 1

Given that $\mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let $\{y_n\}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_n\in A$, $b_n\in B$. First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n \rightarrow y$. Now because $B$ is compact there exists a $b\in B$ such that $b_n\rightarrow b$. Now $a_n \rightarrow y-b$ and for closedness of $A$ we deduce that $y-b\in A$. Call $a=y-b$, then $y=a+b$ where $a\in A$ and $b\in B$. So $y\in A+B$ implies that $A+B$ is closed. Second we want to show that if $\{y_n= a_n+b_n\}$ is a sequence of $A+B$, then it admits a converging subsequence. From $\{a_n\}\in A$ we can take a converging subsequence $\{a_{n_j}\}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From $\{b_{n_j}\}\in B$ we can take a converging subsequence $\{b_{n_{j_k}}\}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}\rightarrow y$, so $A+B$ is compact.

Answer 2

Hint for a)

Take any convergent sequence $\{a_k+b_k\}_{k=1}^{\infty}$ where $a_k\in A$ and $b_k\in B$.

We can pick out a convergent subsequence of $\{b_k\}_{k=1}^{\infty}$ denoted $\{b_{k_j}\}_{j=1}^{\infty}$ (why?). Say $b_{k_j}\rightarrow b$. What can you now say about $\{a_{k_j}\}_{j=1}^{\infty}$?