Vector space with countable dense subset

Question

An exercise gives the following,

let $\ell^{1}= \{ (x_n): \sum_{n=1}^{\infty}\vert x_n\vert < \infty\}$

be a vector space with norm $\Vert x \Vert_1=\sum_{n=1}^{\infty}\vert x_n\vert$

$D=\{ (q_n): q_n\in \Bbb Q \text{ for all n and } \exists N \in \Bbb N \text{ s.t } q_n=0 \text{ for all } n > N\}$
(Set of rational valued sequences that are eventually zero).

I need to show that

a) $D$ is a countable subset of $\ell^{1}$

b) $D$ is dense in $\ell^{1}$

These 2 will imply that $\ell^1$ is a separable space.

I can use the fact that $\ell^{1}$ is a normed space.

I can not get my head around showing that something is a countable subset. I understand that countable means we can number all the elements with the natural numbers but applying it to sets of sequences is a bit of a stretch for me. A dense set $A$ is one where any $x\in X\setminus A$ is a limit point of $A$. I'm not even sure where to start with this one. Any examples/hints would help.

Answer 1

You can write $$ D=\bigcup_m\{(q_n):\ q_j=0\ \text{ if } j>m\}. $$ As a countable union of countable is countable, now you only need to show that each set is countable. And the set $\{(q_n):\ a_j=0\ \text{ if } j>m\}$ is basically a finite cartesian product of countable, so countable.

To show that $D$ is dense, first show that given $\varepsilon>0$ and $x\in\ell^1$, you can find a sequence $y$, with finitely many terms nonzero, with $\|x-y\|<\varepsilon$ (for this, use that the tail of $x$ is arbitrarily small, since $x\in\ell^1$.

Finally, show that you can approximate the $y$ above with a $y'$ that only has rational entries.

Answer 2

Define $$D_n=\{ (q_1,\ldots,q_n,0,0,\ldots): q_i\in \Bbb Q \text{ for } 1\leq i\leq n\}$$ Now $D_n$ has the same cardinality as $\mathbb{Q}^n$ and the latter one is countable. Also, $D$ is a countable union of $D_n$, so $D$ is countable too.

Now use some $D_n$ to approximate an element of $\ell^1$.

Answer 3

Note that $D = \cup_n Q_n$, where $Q_n$ is the set of rational sequences $q_k$ whose value is zero for $k >n$. The set $Q_n$ is countable.

If the sets $A_1, A_2,...$ are countable then so is $\cup_n A_n$. It is painful to write down an explicit bijection or surjection, however it is easy to create a surjection informally. Start with the first element of the first set. For the next round, take the second element of the first set and the first element of the second set. For the next round, take the third element of the first set, the second element of the second and the first element of the third. Continuing this way we will visit every element of the union.

Hence $D$ is countable.

Regarding denseness, we can proceed in two steps:

First note that if we let $R^* = \{ x \in l_1 | x_k \text{ is non zero for at most a finite number of indices} \}$, then $R^*$ is dense in $l_1$. To see this, pick $x \in l_1$ and note that $\sum_k |x_k| < \infty$. Choose $\epsilon>0$ and find $N$ such that $\sum_{K > N} |x_k| < \epsilon$ and let $x^* = (x_1,...,x_N,0,0,...)$. Then $\|x-x^*\| <\epsilon$.

Second, if we let $R^N$ be the space of sequences whose value is zero for $k > N$, we see that $Q_N$ (defined above) is dense in $R^N$.

So, given $\epsilon>0$, pick $x^* \in R^*$ such that $\|x-x^*\| < {1 \over 2} \epsilon$, and pick $q \in Q_N$ (where $N$ comes from the sequence $x^*$) such that $\|q-x^*\| < { 1\over 2} \epsilon$. Then $\|x-q\| < \epsilon$.