1
$\begingroup$

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 $x_{n_k}\in K$.

$\endgroup$
1

1 Answer 1

Reset to default
2
$\begingroup$

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

$\endgroup$
2
  • 3
    $\begingroup$ Together with the fact that any two norms on a finite-dimensional real vector space are equivalent. $\endgroup$
    – José Carlos Santos
    Jun 25, 2021 at 17:08
  • $\begingroup$ The original question is norm-free; it is valid for general topological vector spaces. $\endgroup$
    – uniquesolution
    Jun 25, 2021 at 19:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .