# Bolzano-Weierstrass Theorem

*(See MATH 409 Lecture 7#Bolzano-Weierstrass Theorem→)*

**Theorem.** Every bounded sequence of real numbers has a convergent subsequence.

*Proof.* Let be a bounded sequence of real numbers. We are going to build a nested sequence of intervals for , such that each contains infinitely many elements of and for all . The sequence is built inductively.

*Basis.* First we set to be any closed bounded interval that contains all elements of (such an interval exists because the sequence is bounded).

*Induction.* Now assume that for some the interval is already chosen and it contains infinitely many elements of . Then at least one of the subintervals and also contains infinitely many elements of . We set to be such an interval. By construction, and .

Since for all , it follows by induction that for all . As a consequence, as . By the Nested Intervals Property, the intersection of the intervals consists of a single number .

Next we are going to build a strictly increasing sequence of natural numbers such that for all . The sequence is built inductively:

*Basis.* First let .

*Induction.* Now assume that for some the number is already chosen. Since the interval contains infinitely many elements of the sequence , there exists such that . We set .

Now we claim that the subsequence of the sequence converges to . Indeed, for any , the points and both belong to the interval . Hence . Since as , it follows that as .