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I was wondering that since $l^\infty$ is not separable, which means that there is not a countable dense set in it. However the set $\mathbb{Q}^\mathbb{N}$ is countable (am I right in this?). So what do we get if we take the closure of $\mathbb{Q}^\mathbb{N}$ in $l^\infty$? I guess we don't get $l^\infty$ because this would mean that $\mathbb{Q}^\mathbb{N}$ is dense.

Also I was wondering the same about $L^\infty$. The set of all characteristic functions of intervals with rational endpoints is countable, so what do we get if we take its closure in $L^\infty$?

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The space $\mathbb{Q}^{\mathbb{N}}$ is not a subspace of $\ell^{\infty}$. – Tomás Oct 24 '12 at 21:18
$|\Bbb Q^{\Bbb N}|\ge |2^{\Bbb N}|=|\Bbb R|$, not countable. – Berci Oct 24 '12 at 21:19
Perhaps you meant $\mathbb Q^{<\omega}$? This is countable and a subspace of $\ell^\infty$ – Thomas Andrews Oct 24 '12 at 21:21
$\def\Q{\mathbb Q}\def\N{{\mathbb N}}$$\Q^\N \cap \ell^\infty$ is dense in $\ell^\infty$: Given $x \in \ell^\infty$ and $\epsilon> 0$, for each $n \in \N$ choose $q_n\in \Q$ with $|x_n - q_n| < \epsilon$. Let $q := (q_n) \in \Q^\N \cap \ell^\infty$, then $\|q - x\|_\infty < \epsilon$. – martini Oct 24 '12 at 21:21
I would rather add a comment to your answer Thomas but for some reason I cannot. Anyway.Well I meant the space of all bounded sequences that take values in $\mathbb{Q}$. I was always confused by ordinals. Is this an uncountable set? Thank you for the reply. – tst Oct 24 '12 at 23:07
up vote 7 down vote accepted

As noted in two comments above, $\mathbb Q^{\mathbb N}$ is not countable and it is not a subset of $\ell^\infty$ since it contains unbounded items.

The set $T=\mathbb Q^{<\omega}$ - the set of sequences of rationals which are zero for all but finitely many values - is both countable and a subset of $\ell^\infty$. $\operatorname{cl}(T)$ is the set of elements of $\ell^\infty$ which converge to $0$ - that is, given $a=\{a_i\}_{i\in\mathbb N}\in\ell^\infty$, $a\in \operatorname{cl}(T)$ if and only if $\lim_{i\to\infty} a_i = 0$

Proof. "If." Assume $\lim_{i\to\infty} a_i =0$ and let $\epsilon>0$. Then choose $N$ such that $|a_i|<\epsilon$ for every $i>N$. Define $b_i=0$ for $i>N$, and, for $i\leq N$, find rational $b_i$ such that $|b_i-a_i|<\epsilon$. Then $b=\{b_i\}\in T$, and $|b_i-a_i|<\epsilon$ for all $i$, so $|b-a|<\epsilon$. Therefore, $a=\{a_i\}$ is in the closure of $T$.

"Only if." On the other hand, assume $a=\{a_i\}\in \operatorname{cl}(T)$. Let $\epsilon>0$. Then, by the definition of closure, there exists $b=\{b_i\}\in T$ such that $|b-a|<\epsilon$. Let $N$ be a number for which $b_i=0$ for all $i>N$ (which exists since $b\in T$.) Then $|a_i-b_i|=|a_i|<\epsilon$ for all $i>N$. But we've just show that for all $\epsilon>0$ there is an $N$ such that for all $i>N$ $|a_i|<\epsilon$. So $\lim_{i\to\infty}a_i = 0$.

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I'd very much like to see a proof of this closure. – Henrique Tyrrell Oct 24 '12 at 22:29
@HenriqueTyrrell Proof added. Fairly elementary. – Thomas Andrews Oct 25 '12 at 0:20

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