Let $Y$ be the subspace of $B(\Bbb N, \Bbb F)$ consisting of those sequences tending to zero. Show that $Y$ is separable

Question

Let $Y$ be the subspace of $B(\Bbb N,\Bbb F)$ consisting of those sequences tending to zero. Show that $Y$ is separable. Here $\Bbb F= \Bbb C$ or $\Bbb R$ (field of complex numbers or real numbers), $\Bbb N$ is the field of natural numbers, and $B(\Bbb N,\Bbb F):= \{f : \Bbb N → \Bbb F : f \text{ is bounded}\}$.

Answer 1

I assume $B(N,F)$ is a metric space with the metric $$d_\infty(x_n,y_n)=\sup_n \vert x_n-y_n \vert$$The set $$S:=\{y_n \in \Bbb Q: \text{finitely many $y_n$ are non zero }\}$$ is a countable dense subset of $Y$

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Countable part is easy(?). For the other part, take $x_n=(x_1,x_2,...)\in Y$. Then $\lim x_n=0$. That is, $$(\forall \varepsilon>0:\exists N \in \Bbb N): \vert x_n \vert< \varepsilon$$ whenever $n >N$. Choose rational $y_1,y_2,...,y_N$ so that $\vert x_i-y_i \vert < \varepsilon$ for $1 \leq i \leq N$. Define $y_n=(y_1,y_2,..,y_N,0,0,\cdots,0)$, we have $y \in S$ and $$d_\infty(x_n,y_n) < \varepsilon$$

Answer 2

Let $D$ be the set of sequences with rational values that have only at most finitely many non-zero values, This is essentially $\bigcup_n \mathbb{Q}^n$ so a countable set.

Then for any sequence $(x_n)$ tending to $0$ and any $\varepsilon>0$ find $N$ such that all $|x_n| < \frac{\varepsilon}{2}$ for $n \ge N$. Then note that $(x_1,\ldots,x_N)$ can be approximated in the sup-metric (on $N$ coordinates) by a rational vector $(q_1,\ldots,q_N)$ such that $\sup_{i=1,\ldots N} |x_i - q_i| < \varepsilon$.

Then $(q_1,\ldots, q_N,0,0,0,\ldots)$ is in $D$ and in $B((x_n),\varepsilon)$.

This shows density of $D$ in the set of sequences under consideration.