# Bounded set in a finite dimensional space is relatively compact

Currently I'm self studying functional analysis, namely finite rank operators. In the text, the author gives the following important remark:

Remark A: Every bounded operator of finite rank is a compact operator because it maps the unit ball to a bounded set in a finite dimensional subspace which is always relatively compact$$^{[1]}$$.

I know that an operator is compact if and only if it maps the unit ball to a relatively compact set. Thus, my questions really becomes fairly straightforward: why is a bounded set in a finite dimensional subspace always relatively compact?

$$^{[1]}$$ Note here that a set $$K$$ is said to be relatively compact if and only if every sequence $$x_n\in K$$ has a Cauchy subsequence $$x_{n_k}\in K$$.

• Jun 25, 2021 at 17:02

Because of Heine-Borel's theorem in $$\mathbb{R}^n$$: every closed and bounded set is compact.