1
$\begingroup$

Let $C[0, 1]$ be the set of continuous functions on $[0, 1]$ with the supremum metric $d(f, g) = sup_{x \in [0, 1]} |f(x) - g(x)|$.

The exercise asks to prove that the closed ball $\overline B(0, 1)$ defined as { $f | sup_{x \in [0, 1]} |f(x)| \leq 1$ } is not compact.

I have a few questions about this:

1) The definition of a compact set $S$ says: "if a collection of open sets $V_a$ satisfies $\cup{V_{a}} = S$, then we can take a finite sub-collection..."

So now how can we find a collection {$V_{a}$} satisfying $\cup{V_{a}} = \overline B(0, 1)$? Some open set $V_{a}$ must cover a constant function $1$, but then the set $V_{a}$ will include many functions f satisfying {f | $sup_{x \in [0, 1]}{f(x)} \gt 1$}, and so $\cup{V_{a}}$ will be larger than $\overline B(0, 1)$.

2) Thus I am going to assume that $\overline B(0, 1)$ is the whole space we have. Then consider a union $\cup_{f \in \overline B(0, 1)}{B(f, \frac{1}{2})}.$ It equals $\overline B(0, 1)$ but no finite sub-union covers the whole $\overline B(0, 1)$. Is this a reasonable proof?

$\endgroup$
  • 1
    $\begingroup$ If you are talking about a subspace, it is fine for a cover consisting of open sets in the ambient space to be larger than the subspace. That is pretty much inevitable in fact. When you restrict to the subspace, this goes away while the resulting intersections become open in the subspace topology. $\endgroup$ – Ian Nov 27 '17 at 10:27
  • $\begingroup$ I would not say your proof is complete because it is not clear to me, based on it, what the difference is between this case and the case of $[-1,1]$. In particular, why does no finite sub-union cover the whole ball? $\endgroup$ – Ian Nov 27 '17 at 10:28
  • $\begingroup$ @Ian, suppose you take an infinite number of open intervals to cover [-1, 1], most of them will be contained in one or a union of a few of those intervals. But that is not the case for the "intervals" $V_a$ that I defined. $\endgroup$ – mercury0114 Nov 27 '17 at 11:19
  • $\begingroup$ Why? What makes this ball so different from $[-1,1]$ and even $[-1,1]^d$? $\endgroup$ – Ian Nov 27 '17 at 14:30
0
$\begingroup$

It's a question that why the union you give has no finite sub-union covers $\bar B(0,1)$.

And there is a more general result:

A normed space $E$ is finite dimensional if and only if the closed unit ball is compact.

So it is sufficient and necessary to prove that $C[0,1]$ is infinite dimensional.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.