Question about compact set in $\mathbb{R}^n$

Problem : Let $$K_1,K_2$$ compacts set in $$\mathbb{R}^n$$. Define : $$K_1*K_2= \bigcup_{x\in K_1,y \in K_2}[x,y]$$ where $$[x,y]=\{ \lambda x + (1-\lambda)y \vert \lambda \in [0,1] \}$$

Prove that $$K_1*K_2$$ is compact.

My proof :

Let $$\{a_k\} \in K_1*K_2$$, so , for each $$k \in \mathbb{N}$$ exists $$x_k \in K_1$$ and $$y_k \in K_2$$ such that : $$a_k \in [x_k,y_k]$$. By the definition of $$[x_k,y_k]$$ exists $$\lambda_k \in [0,1]$$ such that $$a_k = (\lambda_k) x_k + (1-\lambda_k)y_k$$. Because $$[0,1]$$ is compact exists a subsequence $$\{\lambda_k\}_{k\in \mathbb{N_1}}$$ such that :

$$\lim_{k\in \mathbb{N}_1} \lambda_k = \lambda \in [0,1]$$

Indeed, because $$K_1$$ is compact exists a subsequence $$\{x_k\}_{k\in\mathbb{N}_2\subseteq \mathbb{N}_1}$$ of $$\{x_k\}_{k \in \mathbb{N}_1}$$ such that :

$$\lim_{k\in \mathbb{N}_2} x_k = b \in K_1$$

and because $$K_2$$ is compact exists a subsequence $$\{y_k\}_{k\in\mathbb{N}_3 \subseteq \mathbb{N}_2}$$ of $$\{y_k\}_{k \in \mathbb{N}_2}$$ such that :

$$\lim_{k\in \mathbb{N}_3} y_k = c \in K_2$$

Finally :

$$\lim_{k \in \mathbb{N}_3}a_k = \lim_{k \in \mathbb{N}_3}(\lambda_k) x_k + \lim_{k \in \mathbb{N}_3}(1-\lambda_k)y_k = \lambda b + (1-\lambda)c=a \in [b,c]$$

So, for each sequence $$\{a_k\} \in K_1*K_2$$ exists a subsequence convergent to a point of $$K_1*K_2$$, then $$K_1*K_2$$ is compact.

I know that my proof is something technical, in several books they resume the part of the use of subsequence saying that :

For the compactness of $$K_1$$ for example, we can suppose, if it were necessary considering subsequence, that $$x_k$$ converges to $$b$$

Everything could be stated in terms of sequence : $$\lambda_k, x_k$$ and $$y_k$$ but I wanted to do something more "formal". My proof is correct?

• Don’t use \textbf in math mode to create boldface. If you surround text with a single asterisk, *text* you get italics. Two asterisks, you get boldface, **this** will produce this. And *you can do it to multiple words, with a single set*: you can do it to multiple words with a single set. – Arturo Magidin May 10 '19 at 3:42
• Thanks for the comment! – Juan Daniel Valdivia Fuentes May 10 '19 at 3:43
• Your proof looks fine. – copper.hat May 10 '19 at 3:57

Define $$f:[0,1]\times K_1 \times K_2$$ by $$f(t,x,y)=tx+(1-t)y$$. Then $$K_1*K_2$$ is simply the range of this function. Since $$f$$ is continuous and its domain is compact, so is the range.