6
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
1
+200
$\begingroup$

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 $V_x \cap B = \{x\}$.

$\endgroup$

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.