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## l-infinity is not separable

Denote by $\ell^\infty$ the space of all bounded complex(real) sequences. Prove that this space is not separable.

Proof: Consider the set of all sequences made of elements $0,1$. These sequences are all in $\ell^\infty$. If two such sequences are distinct, the distance between them is $1$. Moreover, the cardinal of this set of sequences is equal to the cardinal of $(0,1)$, which means the set is uncountable. The balls of radius $1/2$ centered in these points of $\ell^\infty$ do not intersect. If there would be a countable dense set, this set would need to have an element in every open ball, which is a contradiction, because we would have a injection from an uncountable set to a countable set.

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1. April 24, 2013 at 6:45 am

why the cardinal of this set of sequences is equal to the cardinal of (0,1)?

• April 24, 2013 at 11:49 am

The cardinal of the set $\{f: \Bbb{N} \to \{0,1\}\}$ is $2^{\aleph_0}$, i.e. the cardinal of the power set $\mathcal{P}(N)$ (the set of parts of $\Bbb{N}$). In fact you can construct a bijection between the two: to each function $f: \Bbb{N} \to \{0,1\}$ you can associate the set $X_f=\{ n \in \Bbb{N} : f(n)=1\}$, and this is a bijection.

To see that the power set of $\Bbb{N}$ has, in fact, the same cardinal number as $\Bbb{R}$ you can take a look at the corresponding Wikipedia page.

If you still can’t find a proof you can understand, let me know and I’ll post one.