# Mathematics Prelims

## July 18, 2008

### Compact Closed Unit Ball Implies Finite Dimension

Filed under: Analysis,Functional Analysis — cjohnson @ 2:32 pm

If $X$ is a normed space and the closed unit ball centered at zero is compact, then $X$ is finite dimensional.

Proof: Suppose $X$ is an infinite dimensional normed space and let $x_1$ be any point in $X$ with $\|x_1\| = 1$ and let $Y_1$ be the one-dimensional subspace of $X$ generated by $x_1$.  Recall that a finite dimensional subspace is always closed, and since $X$ is infinite dimensional, $Y_1$ is a proper subspace of $X$.  By Riesz’s Lemma, there exists an $x_2 \in X \setminus Y_1$ with $\|x_2\| = 1$ and $\|x_2 - y\| \geq \frac{1}{2}$ for all $y \in Y_1$.  Let $Y_2$ be the two-dimensional subspace generated by $x_1, x_2$.  There exists a $x_3 \in X \setminus Y_2$ such that $\|x_3\| = 1$ and $\|x_3 - y\| \geq \frac{1}{2}$ for all $y \in Y_2$.  Note that since $X$ is infinite dimensional, we can keep applying this procedure generating a sequence $(x_n)$ such that $\|x_n\| = 1$ but $\|x_m - x_n\| \geq \frac{1}{2}$ for all $m \neq n$This means that our sequence can not be Cauchy, and so it can not be convergent, and so the closed unit ball of radius one can not be compact. This means that, since all points in the sequence are at least distance $1/2$ from one another, no subsequence can be Cauchy, so no subsequence can be convergent.  Hence, if the closed unit ball of radius one is compact, then the space is finite dimensional.

## 1 Comment »

1. Note, also the converse implication should be true.

Comment by francescotudisco — September 19, 2011 @ 8:09 am

Blog at WordPress.com.