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In this question thread, user940 answered and said that, "Assume that $A⊂l^∞$ is countable..."

I want to understand:

  1. why can we assume something like this? can assuming something like this make our result less general?

  2. can we do a similar proof without assuming that $A$ is countable, and instead by considering any sequence $a_k \in A$ such that $||a_k-b||>$ some $\epsilon?$

ali
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  • Why wouldn't you be able to assume that? If the assumption leads to something useful or not is a different matter. – Alborz Jan 04 '23 at 14:29
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    A space is separable if it has a countable dense subset. That user is proving that for each countable subset $A$ of $\ell^\infty$, $A$ is not dense in $\ell^\infty$. That is, they are proving exactly the negation of the statement "$\ell^\infty$ is separable". – Rhys Steele Jan 04 '23 at 14:55
  • @RhysSteele thank you! – ali Jan 04 '23 at 16:48

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