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For metric spaces there is another, perhaps more natural way of thinking about compactness. It is based on the following classical result.

**The Bolzano-Weierstrass theorem**

*Every bounded sequence in ***R*** has a convegent subsequence.*

This is attributed to the Czech mathematician Bernhard Bolzano (1781 to 1848) and the German mathematician Karl Weierstrass (1815 to 1897).

From this we are led to the generalisation:

**Definition**

A metric space is sequentially compact if every bounded infinite set has a limit point.

The main result is:

**Theorem**

A compact metric space is sequentially compact.

**Proof**

Let *A* be an infinite set in a compact metric space *X*. To prove that *A* has a limit point we must find a point *p* for whicch every open neighbourhood of *p* contains infinitely many points of *A*.

Suppose that no such point existed. Then every point of *X* has an open neighbourhood containing only finitely many points of *A*. These sets form an open cover of *X* and extracting a finite open cover gives a covering of *X* meeting *A* in only finitely many points. This is impossible since *A* *X* and *A* is infinite.

**Corollary**

In a compact metric space every bounded sequence has a convergent sub-sequence.

**Proof**

Given the above limit point *p*, take *x*_{i1}to be in a 1-neighbourhood of *p*, *x*_{i2}to be in a ^{1}/_{2}-neighbourhood of *p*, ... and we get a sub-sequence converging to *p*.

Together with the Heine-Borel theorem this implies the Bolzano-Weierstrass theorem.

**Remark**

In fact, a metric space is compact if and only if it is sequentially compact. The proof that sequentially compact comact is harder.

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