1

Let $S$ be a uncountable set. Consider function $f: S\to R$. Define the norm $$ \|f\|=\sum_{x\in S}|f(x)|:=\sup_{\mbox {all finite subset A$\subset$ S }} \sum_{x\in A}|f(x)| $$

Now consider the normed space $X=\{f: \|f\|<\infty\}$. I try to prove that this normed space is not separable.

I try to find a uncountable subset $L\subset S$ so that for any $x, y \in L$, $\|x-y\|\ge 1$.

I consider the similar example as in Why is $l^\infty$ not separable?.

For any subset $I\subset \mathbb{N}$, define $f^I(k)=a_k^I=1$ if $k\in I$ otherwise $f^I(k)=0$ for $k\notin I$.

Then for any $I\neq J$, $f^I\neq f^J$, and $$ \|f^I-f^J\|=\sup_{A} \sum_{k\in A}|a^I_k-a^J_k| $$

However, I am stuck here on how to say this sum is greater than 1?

Jose Avilez
  • 13,432
Hermi
  • 1,488
  • Let me switch to a more well-known, convenient notation called indicator function notation: $$\mathbf{1}_{I} (k) = \begin{cases} 1, & k \in I \ 0, & k \in I^c \end{cases} $$ Now, show that $I \neq J$ implies that there exists $k$ for which $|\mathbf{1}_I(k)- \mathbf{1}_J(k)| = 1$ holds. (In light of the definition of indicator function, this amounts to proving that $I\neq J$ implies the existence of $k$ for which $k \in I \setminus J$ or $k \in J \setminus I$, which should be almost trivial.) Using this, you can easily show that $$| \mathbf{1}_I - \mathbf{1}_J| \geq 1.$$ – Sangchul Lee Sep 27 '23 at 03:49
  • 1
    Note that for any $f:S \to \mathbb{R}$ and for any fixed $s\in S$, we have $$|f| \geq |f(s)|$$ because $|f(s)| = \sum_{k\in{s}} |f(k)|$ belongs to the set $$ \left{\sum_{k\in A} |f(k)| : A \text{ is a finite subset of } S\right} $$ of all possible finite sums on which the supremum is taken. This is the fact I am using here. – Sangchul Lee Sep 27 '23 at 05:05

1 Answers1

1

For $\alpha \in S$, let $e_\alpha$ be the function such that $e_\alpha(\alpha)=1$ and $e_\alpha (\gamma)=0$ for any $\gamma \neq \alpha$. The family of functions $\mathcal{F} = \{e_\alpha \, : \, \alpha \in S\}$ is an uncountable family such that for any two distinct $e_\alpha, e_\beta \in \mathcal{F}$ we have $$\|e_\alpha - e_\beta\| \geq |e_\alpha (\alpha) - e_\beta (\alpha)| = 1$$

Thus, your space is not separable.

Jose Avilez
  • 13,432
  • @Hermi You can show that $|e_\alpha | = 1 < \infty$ for each $e_\alpha \in \mathcal{F}$. – Jose Avilez Sep 27 '23 at 04:21
  • @Hermi yes, that is correct. You have uncountable many indicator functions supported at a single point, whose distance amongst them is $\geq 1$ – Jose Avilez Sep 27 '23 at 04:25
  • @Hermi careful, $I \setminus J$ might be empty. You probably want $k \in I \Delta J = (I \setminus J) \cup (J \setminus I)$. Otherwise, it's fine. – Jose Avilez Sep 27 '23 at 04:31
  • @Hermi, You can pick any $I \subseteq S$ so long as $I$ is finite so that $\mathbf{1}_I : S \to \mathbb{R}$ is a member of $X$. Our consideration reveals that any uncountable family of finite subsets of $S$ will suffice, including (but not limited to) Jose's choice ${ {\alpha} : \alpha \in S}$. – Sangchul Lee Sep 27 '23 at 05:01