I am studying functional analysis, and in the setting of normed spaces I have seen the theorem that states that the unit ball is compact iff the space is finite dimensional.

I also saw an exercise:

Let $X$ be finite dimensional, prove that for any non-empty and closed $C$, $x\in C$ there exist $c\in C$ s.t $||x-c||=d(x,C)$.

The proof started with:

take $c\in C$ and consider $$ D=\overline{B(x,||x-c||)}\cap C $$

$D$ is closed and bounded hence, since $X$ is finite dimensional, is compact.

I would be glad if someone could help me with the following two questions:

1) It seems that the theorem I read about the compactness also extends to $X$ is finite dimensional iff $\{x,||x-x_{0}||\leq d\}$ is compact for any $x_{0}\in X,d\in\mathbb{R}^{+}$. is this correct ?

2) $C$ is closed, $\overline{B(x,||x-c||)}$ is also closed, hence so is the intersection. since $\overline{B(x,||x-c||)}$ is bounded then so is the intersection.

And so I get that $D$ is closed and bounded. Why can't I conclude that $D$ is compact without the assumption that $X$ is finite dimensional ?

  • 2
    $\begingroup$ Closed and bounded is not the same thing as compact in...well...infinite dimensions! The Heine-Borel theorem is truly a theorem about $\mathbb{R}^n$. $\endgroup$ – Alex Youcis Sep 17 '13 at 11:16

There is a lemma by Riesz, which states the following:

Let $X$ be a normed vector space, $U\subsetneq X$ a proper subspace, $\epsilon> 0$. Then there is $x \in X$ with $\|x\| = 1$ and ${\rm dist}(x, U) \ge 1 -\epsilon$.

If $X$ is infinite dimensional, you can use this Lemma with $\epsilon = \frac 12$ inductively to construct a sequence $(x_n)$ in $X$ such that $\|x_n\| = 1$ and $\|x_n - x_m\| \ge \frac 12$ for all $n \ne m$. Obviously $(x_n)$ cannot have any Cauchy subsequence. So the closed ball $\{x\in X \mid \|x\| \le 1\}$ is not compact.

The equivalence "compact $\iff$ closed and bounded" holds exactly iff $X$ is finite dimensional.

  • 1
    $\begingroup$ Could be worth noting that "compact $\implies$ closed and bounded" remains true even in infinite dimension. $\endgroup$ – Federico Sep 18 '13 at 18:22

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