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?

  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 10 '19 at 3:42
  • $\begingroup$ Thanks for the comment! $\endgroup$ – Juan Daniel Valdivia Fuentes May 10 '19 at 3:43
  • $\begingroup$ Your proof looks fine. $\endgroup$ – 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.


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