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Consider $\ell^\infty $ the vector space of real bounded sequences endowed with the sup norm, that is $||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$.

Prove that $B'(0,1) = \{x \in l^\infty : ||x|| \le 1\} $ is not compact.

Now, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof.

However, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence?

Does the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence?

I think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks.

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2 Answers 2

You have to find a sequence in $B'(0,1)$ which does not contain a convergent subsequence. Because the existenc of such a sequence implies (right from the definition) that $B'(0,1)$ isn't sequentially compact.

Hint Consider the sequence $(y^n)_n$ defined by $$y^n_k := \begin{cases} 1 & n=k \\0 & n \not= k \end{cases} \qquad (k \in \mathbb{N})$$ Then $y^n \in B'(0,1)$ for all $n \in \mathbb{N}$ and $\|y^n-y^m\|_{\infty}=1$ for $m \not= n$. This means that $(y^n)_n$ doesn't contain a Cauchy-subsequence, hence in particular no convergent subsequence.

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Let $E = \{ e_n \}_n$ where $e_n$ is the vector with $1$ in the $n$th position and zero everywhere else. Clearly $E \subset \overline{B}(0,1)$, hence it is bounded. Since $\|e_n - e_m\| = 1$ whenever $n \neq m$, it is clear that each point in $E$ is isolated. Hence $E$ is closed.

Now let $U_n = B(e_n, \frac{1}{2}$). Then $\{ U_n\}_n$ is an open cover of $E$ that had no finite subcover. Hence $E$ is not compact.

If you would rather use a sequential argument, choose the sequence $x_n =e_n$. As above, we have $\|x_n -x_m\| = 1$ whenever $n \neq m$, hence no subsequence can be Cauchy. Hence no subsequence can converge, hence $E$ is not sequentially compact.

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