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Let $(X,d)$ a metric space such that $\exists A\subseteq X$ uncountable and $\exists \epsilon\gt0$ such that $\forall x,y \in A,x\neq y\Rightarrow d(x,y)\gt\epsilon$. Prove that $X$ is not a separable space and and use this result to show that $(\ell^\infty,d_\infty)$ is not separable.

There is a result that says that in a separable metric space, every uncountable subset contains a point of accumulation. There is a result that says that in a separable metric space, every uncountable subset contains a point of accumulation. I started considering the result, which did not lead to anything. Then, by contradiction, I wanted to prove that $X$ is separable. In order to use that, I could see if $X$ has a countable base or a cover of open sets of $X$ has a countable subcover.

Prism
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Ariel Marcelo Pardo
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3 Answers3

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Let me show you another way of proving that $(\ell^{\infty}, d_{\infty})$ is not separable. The proof is specific to $\ell^{\infty}$ but it is a cute trick, nonetheless.

Assume, to the contrary, that there is a countable dense set $S$ in $\ell^{\infty}$. Enumerate $S$ into a list, i.e. $S=\{x_1, x_2, x_3, …\}$. Since $x_i\in\ell^{\infty}$, we can write $x_{i}=(x_{i1}, x_{i2}, x_{i3}, …)$ for each $i\in\mathbb{N}$. Define a sequence $y=(y_n)$ given by: $$ y_{n}=\begin{cases} x_{nn}+1 & \ \ \text{if } |x_{nn}|\leq 1 \\ 0 & \ \ \text{if } |x_{nn}|>1 \end{cases} $$ Since $|y_{n}|\leq 2$ for each $n\in\mathbb{N}$, we see that $y\in\ell^{\infty}$. By construction, $|y_{n}-x_{nn}|\geq 1$, and so $\lVert y-x_n \rVert_{\infty}\geq 1$ for each $x_n\in S$. So $S$ cannot be dense, because, for example $S$ cannot intersect the open ball of radius $1/2$ around the point $y$. Contradiction.

Prism
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  • What is $|x_{nn}|$? – Ariel Marcelo Pardo Jul 14 '15 at 14:15
  • @Ariel By definition, $x_{nn}$ is the $n$-th entry of the element $x_n\in\ell^{\infty}$. So the trick used in this answer is referred to as diagonalization. From $x_{1}$, you pick its first entry $x_{11}$, from $x_{2}$, you pick its second entry $x_{22}$, and so on. – Prism Jul 14 '15 at 15:17
  • $x_{nn}$ is a real number, right? – Ariel Marcelo Pardo Jul 14 '15 at 15:37
  • why $|y_{n}|\leq 2$? i mean, $y_n\in\ell^{\infty}$ – Ariel Marcelo Pardo Jul 14 '15 at 15:39
  • $x_{nn}$ is real number. And $y_n$ is also real number. It is not true that $y_{n}\in \ell^{\infty}$. It is $y\in \ell^{\infty}$. Yeah I agree, the notation is a bit weird, because $x_n\in\ell^{\infty}$ but $y_n$ is just a real number. – Prism Jul 14 '15 at 15:43
  • it could be improved? – Ariel Marcelo Pardo Jul 14 '15 at 15:46
  • @Ariel: Maybe by writing $x_{i}=(x_{i}^{1}, x_{i}^{2}, x_{i}^{3}, …)$ as opposed to $x_{i}=(x_{i1}, x_{i2}, x_{i3}, …)$. In any case, the main idea is diagonalization, which I hope is clear :) I construct a single sequence $y$ whose entries are $y_n$, and I force $y_n$ to be different from $n$-th entry of $x_n\in\ell^{\infty}$ (which is $x_{nn}$) – Prism Jul 14 '15 at 15:48
  • @Prism What is the idea to define $y_n$? Is it started from $|y_n−x^n_n|≥1$ for every $n$? So if we define $y_n=0$, then we have $|0−x^n_n|=|x^n_n|≥1$. And then for $|x^n_n|<1$, we can define $y_n=x^n_n+1$ to make sure that $|y_n|=|x^n_n+1|≤|x^n_n|+1<1+1=2$ which implies $\sup_{n \in \mathbb{N}} |y_n|<\infty$. – user136524 Jan 09 '22 at 06:05
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    @MathLearner Yes, that is correct: it is exactly how that sequence was reverse-engineered! – Prism Jan 09 '22 at 20:18
  • @Prism oh, I see. Thank you so much. – user136524 Jan 10 '22 at 04:34
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If $X$ is separable, then all uncountable subsets of $X$ contain a point of accumulation.

As a result, $A$ contains a point of accumulation, say $x_0$. Consider $B_d[x_0, \epsilon]$. Then, $\left( B_d[x_0, \epsilon] \setminus \{x_0\}\right) \cap A \neq \varnothing$ (since $x_0$ is a point of accumulation). Hence, there's $y_0 \in A$, $y_0 \neq x_0$, such that $d(x_0, y_0) \le \epsilon$. But that's impossible.

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Hint: if $D$ is dense in $X$, given $x \in A$, we must have $D \cap B(x, \epsilon/2) \neq \varnothing$. Take a point there and call it $f(x)$. We have a map $f: A \to D$, then. Prove that $f$ is injective, and conclude that $D$ is not countable. Since $D$ was an arbitrary dense set, $X$ is not separable.

Ivo Terek
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