Difference of two compact sets is compact

Question

Consider this question:

Let $K_{1}$ and $K_{2}$ be compact sets. Define $A=\{||x_{1}-x_{2}||:x_{1} \in K_{1}, x_{2} \in K_{2}\}$. Show that $A$ is compact.

My atempt consist in prove that $A$ is compact using the notion of sequentially compact.

$\textbf{My answer}$: Let $K_{1}$ and $K_{2}$ be compact sets. Therefore, every sequence has convergent subsequence. Consider the sequences $(x_{n}^{1})_{n \in \mathbb{N}} \in K_{1}, \forall n \in \mathbb{N}$, and $(x_{n}^{2})_{n \in \mathbb{N}} \in K_{2}, \forall n \in \mathbb{N}$.

Hence, $\exists (x_{n_{j}}^{1})_{j \in \mathbb{N}} \in K_{1}:x_{n_{j}}^{1} \to x^{1}$, and $\exists (x_{n_{j}}^{2})_{j \in \mathbb{N}} \in K_{2}:x_{n_{j}}^{2} \to x^{2}$. Since both $K_{1}$ and $K_{2}$ are compact sets, $x^{1} \in K_{1}$ and $x^{2} \in K_{2}$.

Therefore, the sequence defined as $y_{n}=x_{n_{j}}^{1}-x_{n_{j}}^{2} \to (x^{1}-x^{2}) \in A$, hence, $A$ is compact.

Is this a valid proof?

Answer 1

It is alright, but you should define $y_{n}=x_{n}^{1}-x_{n}^{2}$, which has a subsequence of the form $y_{n_{j}}=x_{n_{j}}^{1}-x_{n_{j}}^{2}$, which, in turn, converges to $x_{1}-x_{2}$. Since a set is compact iff every sequence contained in the set has a convergent subsequence, $A$ is compact.

Answer 2

Hint : let $S_1,S_2$ are compact sets .$g,f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, f(x,y) = x+y$,$g(x,y)=xy$ are continuous and so

$S_1 + S_2 = f[S_1 \times S_2]$.

$S_1 \times S_2$ is compact by Tychonoff , and if $S$ is compact so $-S$ is compact.