Let $\ell^\infty$ denote the space of all bounded real sequences with the usual norm and let $A=\{0,1\}^\mathbb{N}$ denote the set of sequences taking values in $\{0,1\}$. It's easy to see that $A$ is closed and discrete in the norm topology of $\ell^\infty$. Does it remain closed or discrete in the weak topology of $\ell^\infty$ ?

If this example does not work, can we construct another uncountable subset of $\ell^\infty$ which is discrete and closed in the weak topology?

I have also a related question. In the space $L^\infty[0,1] $ of measurable essentially bounded functions, consider the subset $B$ of characteristic functions of intervals $[0,t]$ for $t\in[0,1]$. Here also, we can show that $B$ is closed and discrete in the norm topology. Is it still discrete in the weak topology?

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    $\begingroup$ Certainly it's weakly closed. If $f_n$ is the functional that returns the $n$th entry, then the set in question is $$\bigcap_{n \in \mathbb{N}} f_n^{-1}\lbrace 0, 1 \rbrace.$$ I'm not good enough with $(\ell^\infty)^*$ to comment if it's discrete or not, but I suspect it isn't. $\endgroup$ – Theo Bendit Jun 21 '18 at 4:09
  • $\begingroup$ Well thanks a lot. If this set is not discrete, can you exhibit an uncountable closed and discrete subset of $\ell^\infty$ (in the weak topology)? I know it must exist $\endgroup$ – Hicham Gebran Jun 21 '18 at 5:51

I don't know the answer to your related question and I haven't figured out whether or not $A$ is discrete in the weak topology (I also suspect it isn't), but there are fairly nice, explicitly describable subsets of $A$ that are definitely closed and discrete in the weak topology.

Fix a bijection $\gamma : \mathbb{N} \rightarrow 2^{<\omega}$, where $2^{<\omega}$ is the set of finite binary strings. Let $B=\{x\subset \mathbb{N}:\gamma (x)\text{ is the set of finite initial segments of some }\alpha \in 2^\omega\}$. To see that $B$ is closed, note that $A$ under the weak* topology is homeomorphic to Cantor space, i.e. $2^\mathbb{N}$ with the standard product topology (this is implied by Theo's argument that it is closed, since the $f_n$'s are all elements of $\ell^1$). $B$ is closed as a subset of $2^\mathbb{N}$ because it's equal to $\bigcap_{n=0}^\infty B_n$ where $$B_n = \{x \subseteq \mathbb{N} : \gamma(x)\cap 2^{<n}\text{ contains a string of each length }< n\text{ and is linearly ordered by extension}\},$$ which is a collection of clopen sets since each $B_n$ only depends on finitely many 'coordinates' ($2^{<n}$ is the set of finite binary strings of length $<n$). Since the weak* topology is coarser than the weak topology, $B$ is closed in the weak topology as well. Also clearly $B$ has the same cardinality as $2^\omega$ and in particular is uncountable.

To see that $B$ is discrete, note that as a family of subsets of $\mathbb{N}$, $B$ has the property that it is 'almost disjoint,' meaning that for any $x,y\in B$ with $x\neq y$, $x\cap y$ is finite: Suppose that $x$ corresponds to the sequence $\alpha \in 2^\omega$ and $y$ corresponds to the sequence $\beta \in 2^\omega$, then $\alpha \neq \beta$, so they can only have finitely many initial segments in common.

For each $x\in B$, we'll exhibit an element $f_x$ of the dual space $(\ell^\infty )^\ast$ such that $f_x(x)=1$ and $f_x(y)=0$ for all $y\in B$ with $x \neq y$. Let $\mathcal{F_x}$ be the Fréchet filter on $x$, i.e. $\mathcal{F_x}=\{z\in A:x\backslash z\text{ is finite}\}$. Extend $\mathcal{F}_x$ to an ultrafilter $\mathcal{U}_x$. Note that by construction $x\in \mathcal{U}_x$, but $y\notin\mathcal{U}_x$ for any $y\in B$ with $x\neq y$.

By the answer to this question, the finitely additive measure on $\mathbb{N}$ given by $\mathcal{U}_x$ (i.e. $\mu(Q)=1$ if $Q\in \mathcal{U}_x$ and $0$ otherwise) gives an element of $(\ell^\infty)^\ast$, $f_x$, with the property that for $y\in A$, $f_x(y)=1$ if $y\in\mathcal{U}_x$ and $f_x(y)=0$ if $y\notin\mathcal{U}_x$, so we have that $f_x(x) =1$ and $f_y(y)=0$ for all $y\in B$ with $x\neq y$ as required.

So now the set $B$ is discrete because each $x\in B$ has a weak neighborhood $V_x = \{z\in\ell^\infty:f_x(z)>\frac{1}{2}\}$ such that $V_x \cap B = \{x\}$.


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