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I've been going over practice problems, and I ran into this one. I was wondering if anyone could help me out with the following problem.

Let $X$ be a metric space of all bounded sequences $(a_n) \subset \mathbb{R}$ with the metric defined by $$d( (a_n), (b_n)) = \sup\{ |a_n - b_n| : n = 1, 2, \ldots \}.$$ Let $Y \subset X$ be the subspace of all sequences converging to zero. Determine whether or not $X$ and $Y$ are separable.

Thanks in advance!

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up vote 8 down vote accepted

Suppose that $A=\{\alpha_n:n\in\Bbb N\}$ is a countable subset of $X$, where $\alpha_n$ is the sequence $\langle a_{n,k}:k\in\Bbb N\rangle$. Note that for any $x\in\Bbb R$ there is always a $y\in[-1,1]$ such that $|x-y|\ge 1$. Thus, we can construct a sequence $\beta=\langle b_k:k\in\Bbb N\rangle$ such that $b_k\in[-1,1]$ and $|b_k-a_{k,k}|\ge 1$ for each $k\in\Bbb N$. Clearly $\beta\in X$, but for each $n\in\Bbb N$ we have $\sup_{k\in\Bbb N}|b_k-a_{n,k}|\ge|b_n-a_{n,n}|\ge 1$, so the distance between $\beta$ and $\alpha_n$ is at least $1$. Thus, $A$ is not dense in $X$, and $X$ is not separable.

$Y$, on the other hand, is separable. Let $D$ be the set of sequences of rational numbers that are $0$ from some point on, i.e., that have only finitely many non-zero terms. Show that $D$ is both countable and dense in $Y$. For the latter, start with any $\langle a_k:k\in\Bbb N\rangle\in Y$ and any $\epsilon>0$ and construct an element of $D$ that is less than $\epsilon$ away from $\langle a_k:k\in\Bbb N\rangle\in Y$.

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To show that $X$ is not separable, take an arbitrary sequence of elements of $X$ (a sequence of sequences) and construct an element that differs at least $1$ under $d$ from every element in your sequence.

To show that $Y$ is separable, look at eventually zero sequences with rational values.

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