$\mathbf l^\infty$ is not separable

hbghlyj

S.6.pdf
Definition S.6.6 A metric space is said to be separable if it contains a countable dense set.
...
Proposition S.6.10 $\mathbf l^\infty$ is not separable
Proof Consider the set $S \subseteq \mathbf l^{\infty}$ of all sequences whose entries are all zeros and ones. This is an uncountable set, for if we supposed it listed in a countable way, say $\boldsymbol{x}_1, \boldsymbol{x}_2, \boldsymbol{x}_3, \ldots$ then we could construct the 'awkward' element of $S$ whose $n$th entry is 0 if the $n$th entry in $\boldsymbol{x}_n$ is 1 and vice-versa - this element is not on our list, so $S$ can't be countable. Also, if $\boldsymbol{x}, \boldsymbol{y}$ are distinct elements of $S$ then $\|\boldsymbol{x}-\boldsymbol{y}\|_{\infty}=1$ (since $\boldsymbol{x}$ and $\boldsymbol{y}$ differ by 1 in at least one entry, but do not differ by more than 1 in any entry). We now prove that there cannot be a countable dense set $C \subseteq \mathbf l^{\infty}$. For suppose that $C$ is such a countable set. Since $C$ is dense, for each point $\boldsymbol{x} \in S$ there must exist a point $c(\boldsymbol{x}) \in C$ within distance $1 / 2$ of $\boldsymbol{x}$. But for distinct $\boldsymbol{x}, \boldsymbol{y} \in \mathbf{l}^{\infty}$, we must have $c(\boldsymbol{x}) \neq c(\boldsymbol{y})$, since if these were equal then
\[
\|\boldsymbol{x}-\boldsymbol{y}\|_{\infty} \leqslant\|\boldsymbol{x}-c(\boldsymbol{x})\|_{\infty}+\|c(\boldsymbol{x})-\boldsymbol{y}\|_{\infty}<1 .
\]
This would define an injection $c$ from the uncountable set $S$ into the countable set $C$, which is impossible, since if we followed $c$ by an injection of $C$ into $\mathbb{N}$ we would get an injection of $S$ into $\mathbb{N}$ and $S$ would be countable.

hbghlyj

$$C\subset\mathbf l^1\subset\mathbf l^2\subset\cdots\subset\mathbf l^\infty$$
Proposition S.6.11 Both $\mathbf{l}^1$ and $\mathbf{l}^2$ are separable.
Proof Let $C \subset \mathbf l^1$ be the set of sequences which are eventually zero (hence are absolutely convergent) and all of whose entries are in $\mathbb{Q}$. This is a countable set: to see this we note that if $C_m \subseteq C$ is the set of sequences $\left(x_n\right)$ such that $x_n=0$ for $n>m$ and $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$, then $C_m$ is in one-one correspondence with $\mathbb{Q}^m$ and hence is countable. Now $C$ is the union over all $m \in \mathbb{N}$ of the $C_m$, a countable union of countable sets, hence countable. (For facts about countability we refer to S.2.)

We shall prove that $C$ is dense in $\mathbf{l}^1$. For let $\boldsymbol{x}=\left(x_n\right) \in \mathbf{l}^1$ and let $\varepsilon>0$. We need to show that there is a point $\boldsymbol{c} \in C$ such that $\|\boldsymbol{x}-\boldsymbol{c}\|_1<\varepsilon$. Since $\sum\left|x_n\right|$ is convergent, there exists $N \in \mathbb{N}$ such that $\sum_{n=N+1}^{\infty}\left|x_n\right|<\varepsilon / 2$. For each $n=1,2, \ldots, N$, by Corollary 4.7 there exists a rational number $r_n$ such that $\left|x_n-r_n\right|<\varepsilon /(2 N)$. Let $\boldsymbol{c}=\left(r_1, r_2, r_3, \ldots, r_N, 0,0, \ldots\right)$. Then $c \in C$ and
\[
\|\boldsymbol{c}-\boldsymbol{x}\|_1=\sum_{n=1}^N\left|r_n-x_n\right|+\sum_{n=N+1}^{\infty}\left|x_n\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon
\]
The proof for $\mathbf{l}^2$ is very similar, using the same countable set $C$, which is also contained in $\mathbf{l}^2$.
Let $\boldsymbol{x}=\left(x_n\right) \in \mathbf{l}^2$ and let $\varepsilon>0$. By convergence of $\sum x_n^2$, there exists an $N \in \mathbf{N}$ such that $\sum_{n=N+1}^{\infty} x_n^2<\varepsilon^2 / 2$. For each $n=1,2,3, \ldots, N$ by Corollary 4.7 there exists a rational number $r_n$ with $\left|r_n-x_n\right|<\varepsilon / \sqrt{2 N}$. Put $\boldsymbol{c}=\left(r_1, r_2, r_3, \ldots, r_N, 0,0, \ldots\right)$. Then
\[
\|\boldsymbol{c}-\boldsymbol{x}\|_2=\left(\sum_{n=1}^N\left(r_n-x_n\right)^2+\sum_{n=N+1}^{\infty} x_n^2\right)^{1 / 2}<\left(\varepsilon^2 / 2+\varepsilon^2 / 2\right)^{1 / 2}=\varepsilon
\]

Czhang271828

证明过繁了, 我们只要在 $\ell^\infty$ 中构造不可数个两两不交且半径为 $1$ 的的开球即可, 例如取
$$
\{(x_1,x_2,\ldots, x_n,\ldots)\mid x_i\in\{\pm 2\}\}
$$
为这些开球的顶点集. 开球数量不可数, 因为这些顶点可以和 $[0,1]$ 间实数建立双射(考虑二进制表示).