1. ## Compact sets

Originally Posted by Plato
I assume that we are in a metric space.
Here are traditional proofs for both properties.
Suppose that $\displaystyle x$ is a limit point of $\displaystyle K$ but $\displaystyle x\notin K$.
$\displaystyle \left( {\forall y \in K} \right)$ this is true $\displaystyle r_y = \frac{{d(x,y)}}{4} > 0$.
The collection $\displaystyle \left\{ {B(y;r_y )} \right\}_{y \in K}$ covers $\displaystyle K$.
So finite subcollection $\displaystyle K \subset \bigcup\limits_{j = 1}^n {B(y_j ;r_{y_j } )}$ also covers $\displaystyle K$.

But note that $\displaystyle x \in \bigcap\limits_{j = 1}^n {B(x ;r_{y_j } )}$. That is a open set that contains $\displaystyle x$ and no other point of $\displaystyle K$. Contradiction.

For bounded, there is finite collection $\displaystyle \bigcup\limits_{j = 1}^n {B(y_j ;1)}$ covering $\displaystyle K$.
Let $\displaystyle M = \max \left\{ {d(y_k ,y_j )} \right\} + 2$. It is easy to show that $\displaystyle M$ is a bound for$\displaystyle K$.
I don't understand the bolded part

2. ## Re: Compact sets

Hey Akatsuki.

Open sets don't contain the limit points of the set. Does that help?

3. ## Re: Compact sets

Originally Posted by Akatsuki
I don't understand the bolded part
This is the standard proof for compactness.
Cover each point with a ball and cover $x$ with a corresponding ball so the two ball do not intersect.
Now compactness gives a finite collection of balls.
Union of open sets is open and finite intersection of balls is open.
Thus we produce and open set that contains $x$ and on other point of $K$.