2
$\begingroup$

It is true that the closed unit ball centered at 0 in $l^{\infty}$ (set of bounded real sequences) is not compact which can be shown in multiple ways by exploiting results such as "compactness implies sequential compactness".

I would like to demonstrate the above by explicitly constructing an open cover of $\bar{B}(0,1)$ with no finite subcover, showing directly from the definition of compactness the result stated. However, I am having difficulty constructing such an open cover. Can you please provide a hint rather than a full solution for how to proceed?

Thank you.

1 Answers 1

1

The idea is to find a contructive proof of Heine-Borel theorem implies Bolzano-Weierstrass Theorem and transpose it.

Let $\langle f_j \rangle_{j=0}^\infty \subseteq B^-$ be a sequence without a convergent subsequence. Then for every $x \in B^-$, there exists an open ball $B_x$ centred at $x$ which contains only finitely many $f_j$s. The open cover $(B_x)_{x \in B^-}$ has no finite subcover. (Otherwise $\langle f_j \rangle$ has a convergent subsequence)

  • 0
    Using the second paragraph of your answer: I figured that if we pick $(B(x,1/4))_{x \in \bar{B}}$ to be the open cover and if $(f_j)$ is the sequence with 1 in the $j$-th position and 0 everywhere else, then each open ball in the cover contains at most one $f_j$ which means there doesn't exist a finite subcover. However, I did not understand how you were motivated to choose this approach. Can you explain a bit more why a constructive proof of "Heine-Borel theorem implies Bolzano-Weierstrass theorem" helps here? Thank you very much! – 2017-01-13
  • 0
    @user334263 I was motivated to use this approach since this proof is adapted from a proof of Bolzano-Weierstrass Theorem (compactness implies sequential compactness in metric spaces). – 2017-01-14