## Proposition 5.2.3: Compact means Closed and Bounded

### Proof:

First, suppose**S**is closed and bounded. Take a sequence in

**S**. Because

**S**is bounded, the sequence is bounded also, and by the Bolzano-Weierstrass theorem we can extract a convergent subsequence from . Using the theorem about closed sets, accumulation points and sequences, we know that the limit of the subsequence is again in

**S**. Hence,

**S**is compact.

Now assume that **S** is compact. We have to show that **S**
is bounded and closed.

Suppose **S** was not bounded. Then for every *n* there
exists a number
*a _{n} S*
with

*| a*. But then no subsequence of the sequence will converge to a finite number, hence

_{n}| > n**S**can not be compact. Thus,

**S**must be bounded.

Suppose **S** was not closed. Then there exists an accumulation
point *c* for the set **S** that is not contained in **S**.
Since *c* is an accumulation point, there exists a sequence
of elements
in **S** that converges to *c*. But then every subsequence
of that sequence will also converge to *c*, and since *c*
is not in **S**, that contradicts compactness. Hence, **S**
must be closed.