Proposition. Any in a metric space is bounded.

Proof

If xnax_n\to a, then d(xn,a)<1d(x_n,a)<1 for all sufficiently large nn. The finitely many remaining values d(xn,a)d(x_n,a) are also bounded, so there is R<R<\infty such that d(xn,a)Rd(x_n,a)\le R for every nn. Thus the sequence is .