Mathematics Prelims

July 16, 2008

l^\infty is not separable

Filed under: Analysis,Functional Analysis — cjohnson @ 6:21 pm

Recall that a a subset $M$ of a metric space $X$ is called dense if the closure of $M$ is the entire space; $\overline{M} = X$.  If $M$ is a dense subset of $X$, then every point of $X$ is a limit point of $M$, and so for every $x \in X$ and $r > 0$, the open ball $B_r(x)$ must contain a point of $M$ that is distinct from $x$.  We say that a space is separable if it has a countable dense subset.

The space $\ell^\infty$ is the set of all bounded sequences of real (or complex) numbers where the metric is given by $d(x, y) = \sup_{n \in \mathbb{N}} |x_n - y_n|$ where $x = (x_n)_{n \in \mathbb{N}}$ and similarly for $y$.

Consider the set of all sequences that whose entries are made up of zeroes and ones.  Obviously this is a subset of $\ell^\infty$.  Furthermore, each of these sequences corresponds to the binary representation of a number in $(0, 1)$, and every number in $(0, 1)$ has a binary representation, so a bijective mapping between $(0, 1)$ and our set exists.  This means that our set is uncountable.  Note that because of the metric on $\ell^\infty$, any two (distinct) elements in the set are distance one apart.  If we place a ball of radius $r < \frac{1}{2}$ around each point, then none of these balls will intersect.  This tells us that, since any dense subset of $\ell^\infty$ must have an element in each ball, any dense subset of $\ell^\infty$ must be uncountable, so $\ell^\infty$ is not separable.

1. I’m not sure how similar your prelim is to mine, but this is a nice exercise: Can you use what you just proved to show that $(\ell^\infty)^*\neq \ell^1$?

Comment by hilbertthm90 — July 16, 2008 @ 10:26 pm

2. This is not quite a bijection: the sequence (0,0,1,1,1,1,…) corresponds to the same point in (0,1) as the sequence (0,1,0,0,0,0,…). However, if you remove sequences with an infinite succession of ones, then what is left does biject with (0,1), and so that zero/one set of sequences must be uncountable because a subset of it bijects with (0,1).

Comment by Don — February 6, 2009 @ 12:22 pm

3. Oh, with regard to the previous comment, I think that’s quite simple. Separability is preserved by taking duals, and l^1 is separable, so it can’t be the dual of l^inf.

Comment by Don — February 6, 2009 @ 12:24 pm

4. I understand the proof above that says l^inf is not seperable. But I don’t understand how this is not a contradiction of the theorem: An infinite dimensional normed space is seperable iff it has a countable orthonormal basis. (proof is trivial and found almost anywhere on web).
What I don’t get is that surely l^inf is seperable because it has a countable orthonormal basis which is the set e_i = (0,0,0…1,…0,0,0) where the 1 is in the i-th place. The set e_1, e_2, e_3, …… etc is countable and forms a basis of l^inf so surely l^inf is seperable. Right? (well clearly I am wrong, but I am not sure how?)

Comment by Questionmark — June 6, 2011 @ 8:00 am

• The quick answer is that the set {e_1, e_2, e_3, …} is not a basis. It’s a subtle point, but if a set is a basis then everything in the space can be written as a _finite_ linear combination of things in the basis. The set {e_1, e_2, …} isn’t a basis because all the finite linear combinations of elements of that set will have infinitely many zeros.

Comment by cjohnson — June 6, 2011 @ 10:53 am

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