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I am trying to prove that $l^\infty$ is not separable. However, I have proved it is separable. Can you help find the flaw in my logic?

(In case my teacher uses different definitions from everyone else, $l^\infty$ is defined here to be the set of all bounded sequences, and the metric used is $d((x_n), (y_n)) = sup\{|x_n - y_n|\}$.)

My logic is consider $A$, the set of all bounded sequences with all terms in $\mathbb{Q}$. We show that all bounded sequences $(x_n)$ are in the closure of this set. To show this, we must show that for all $\epsilon$, an $\epsilon$-ball around $(x_n)$ intersects $A$. But since you can find a rational number arbitrarily close to any number in the reals, you can just find a rational sequence $(a_n)$ such that $|a_n - x_n| < \epsilon$ for all $n$. Thus $d((x_n), (a_n)) < \epsilon$, and $B_\epsilon((x_n)) \cap A \neq \emptyset$ for all $\epsilon$. Therefore $l^\infty$ is the closure of $A$, and since $A$ is countable, $l^\infty$ is separable.

Can you help find the flaw in my logic?

Thank you!

1 Answers 1

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$A$ is not countable. It contains all sequences of $0$s and $1$s, and there are already $2^{\aleph_0}>\aleph_0$ of those.

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    And quotients of "the set of all convergent sequences with all terms in $\mathbb Q$" form one of the two standard constructions of the reals, so it had better be uncountable for that reason.2011-10-24