Let $\ell_\infty=\{\bar{x}=(x_1,x_2,x_3,\dots) \,:\,x_n\in \mathbb{R}, \sup_{n\in\mathbb{N}} x_n<\infty, n\in \mathbb{N} \}$ and $$d_\infty(\bar{x},\bar{y})=\sup_{n\in\mathbb{N}} x_ny_n,$$ for all $\bar{x},\bar{y} \in \ell_\infty$.
Then, $(\ell_\infty, d_\infty)$ is not separable.
I want to prove $(\ell_\infty, d_\infty)$ is not separable by showing that there exists a subset of $\ell_\infty$ which is countable but not dense in $\ell_\infty$.
In the class, my lecture proved it in a similar way as in the question of this post: Show that $l_\infty$ is not separable.
I want to try with another subset of $\ell_\infty$.
This is my attempt.
Given a set of sequences $Y=\{\bar{y}=(y_1,y_2,\dots,y_n,0,0,0,\dots)\, :\, y_k\in \mathbb{Q}, 1\leq k\leq n, n\in\mathbb{N}\}$.
Hence, $Y$ is a countable set.
Then, take any $\bar{y}\in Y$, so $\bar{y}=(y_1,y_2,\dots,y_n,0,0,0,\dots)$ where $y_1,y_2,\dots,y_n\in \mathbb{Q}$. Since the sequence $\bar{y}$ is finite, then $$\sup_{n\in \mathbb{N}} y_n<\infty.$$
So, $\bar{y}\in \ell_\infty$.
Thus, $Y$ is a countable subset of $\ell_\infty$.
Next, I will show that $Y$ is not dense in $\ell_\infty$.
Choose $\varepsilon_0=\frac{1}{4}$ and $\bar{x}=(1,1,1,1,\dots)\in \ell_\infty$. Take any $\bar{y}\in Y$, then
\begin{align*}
d_\infty(\bar{x},\bar{y})&=\sup_{n\in \mathbb{N}} x_ny_n\\
&=\sup \{x_1y_1,x_2y_2,\dots,x_ny_n,x_{n+1}y_{n+1},x_{n+2}y_{n+2},\dots \}\\
&=\sup \{1y_1,1y_2,\dots,1y_n,10,10,\dots \}\\
&=\sup\{1y_1,1y_2,\dots,1y_n,1,1,\dots \}\\
&\geq \sup\{1y_1, 1y_2,\dots,1y_n,1,1,\dots \}\\
&=1>\frac{1}{4}=\varepsilon_0.
\end{align*}
Because $d_\infty(\bar{x},\bar{y})\geq\varepsilon_0$, so $Y$ is not dense in $\ell_\infty$.
Hence, $(\ell_\infty, d_\infty)$ is not separable.
Is there any mistakes of my proof?
Thanks for any help.
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3The negation of "there is a countable dense set" is "all countable sets are not dense" not "there is a countable set that is not dense" – Quimey Dec 10 '21 at 14:01

To expand on @Quimey: $\mathbb N$ is a countable subset of $\mathbb R$ which is not dense. But $\mathbb R$ (with the canonical topology) is separable. – Maximilian Janisch Dec 10 '21 at 14:02

https://math.stackexchange.com/questions/660418/whyislinftynotseparable – d.k.o. Dec 10 '21 at 14:04

@Quimey Ah, I often careless when negated a definition :( thanks for remind me. So, I must use arbitrary countable subsets of $\ell_\infty$ which are not dense, right? – user136524 Dec 11 '21 at 03:15