Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 6 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 $\langle a_k:k\in\Bbb N\rangle\in Y$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.