Theorem. All convergent sequences are bounded.
Proof. Suppose we have a sequence x_n L . Since it is given that the sequence is convergent (it has some limit L), it is true by hypothesis that for any arbitrary > 0 we can find an N such that | x_n - L | < any n > N. Let's allow our arbitrary to be equal to 1. Since the sequence is bounded after N by |x_n| < 1 + |L|, and there is a finite number of terms before N, we can allow the bound to be:

M = max{ x₁, x₂, x₃, ..., x_n, 1 + |L| }

Therefore, all convergent sequence are bounded.//

The converse of the theorem does not apply. A counterexample can be located easily, any kind of sequence that oscillates (such as something similar to sine or cosine).

Balzano-Weierstrass Theorem. Any sequence that is bounded has a convergent subsequence.
Proof. Let a_n be bounded such that |a_n| < M for all n . All terms of a_n belong within the interval I₀ = [-M, M]. Since the sequence is infinitely long, an infinite number of terms in the sequence exist within [-M, 0] or [0, M] (note: possibly both). Select one of the two intervals and denote it I₁. Likewise, we can split that interval into two halves, one or both of which must contain an infinite number of terms from the sequence. Since I₀ I₁ I₂ ... , by the Nested Interval Property we know that I_n .

Likewise, we can pick a_n terms from within each interval such that n₁ < n₂ < n₃ < n₄ < ... such that (a_nk)_k I_k. Since the length of I_k is M/2^k, and M/2^k = 0, I_n = {a}.

(a_nk)_k is a subsequence of (a_n)_n and (a_nk)_k a.//