Does anyone have an elegant/easy to remember proof of this? I can prove bounded easily, and I have notes on why its closed, but they're kind of messy and confusing!
Suppose (for a contradiction) that for all $\displaystyle y\in K$ there exists $\displaystyle \epsilon _y$ such that $\displaystyle B_{\epsilon _y} (y)$ contains only a finite number of terms of a given sequence $\displaystyle (x_n)\subset K$ then since $\displaystyle K$ is compact, the cover $\displaystyle \{ B_{\epsilon _y} (y) : y\in K \}$ has a finite subcover ie. there exists $\displaystyle y_0,...,y_m$ such that $\displaystyle K \subset \cup_{i=1}^{m} B_{\epsilon _{y_i}} (y_i)$ but this is clearly a contradiction since each of these balls contain only finitely many terms. So we conclude that for any given sequence $\displaystyle (x_n)\subset K$ there exists a $\displaystyle y\in K$ such that for all $\displaystyle \epsilon >0$ $\displaystyle B_{\epsilon } (y)$ contains infinitely many terms of said sequence ie. there exists a subsequence $\displaystyle (x_{n_k})$ converging to $\displaystyle y$. This in turn proves that $\displaystyle K$ is closed trivially.
By how we defined them, each $\displaystyle B_{\epsilon _y} (y)$ contains only a finite number of terms of a sequence (a countable infinte set) so when we get our finite subcover we can get at most a finite number of terms of the sequence in the union of a finite number of balls, but these union contains $\displaystyle K$ which contains the sequence.
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$\displaystyle K$.
Here's what I think is the simplest proof that any compact set is bounded.
(Assuming that A is in a metric space, of course.)
Let p be any point in the compact set, A. Let $\displaystyle B_p(n)$ be the open ball centered at p with radius n. Certainly every point in A has some distance from p and there exist an integer larger than that distance. That is, the set of all such open balls is an open cover of A. Since A is compact, there is a finite subcover of A, so there is a largest "N". Show that, if x and y are any points in A, d(x,y)< 2N.
To show that any compact set, A, is closed, show that its complement is open. (Again, in a metric space.)
Let p be a point in the complement of A. For any q in A, let B(q) be the open ball, of radius 1/2 the distance from p to q, centered on q. Let C(q) be the open ball, of radius 1/2 the distance from p to q, centered on p (note that C(q), though centered on p, is still "indexed" by q). The set of all open balls, B(q), for all q in A, is an open cover for A. Since A is compact, there exist a finite subcover, $\displaystyle \{B(q_1), B(q_2), \cdot\cdot\cdot, B(q_n)\}$. Look at the corresponding collection of open sets $\displaystyle \{C(q_1), C(q_2), \cdot\cdot\cdot, C(q_n)\}$. Since p is in each of them it is in there intersection. Further, since this is a finite collection, its intersection is an open set. Finally, since every member of A is in one of the sets $\displaystyle B(q_i)$, it is not in the corresponding $\displaystyle C(q_i)$ and so not in there intersection.
That is, the intersection of the all the $\displaystyle C(q_i)$ is an open set, containing p, which contains no member of A. That means that p is an interior point of the complement of A and, since p could be any member of the complement of A, the complement of A is open and A itself is closed.