Kreiszig 1.3-9: l-infinity is not seperable

[Rasalhague]

Kreiszig: Introductory Functional Analysis with Applications, example 1.3-9, states that the metric space l^{\infty} is not seperable. He invites the us to consider the bijection between the interval [0,1] and the set Y of sequences, y = (\eta_1, \eta_2, ...), whose terms consist of no number other than a 1 or a 0:

\hat{y}\in \left [ 0,1 \right ] = \sum_{k=1}^{\infty} \frac{\eta_k}{2^k}.

The existence of this bijection, and the fact that [0,1] is uncountable, shows that Y is uncountable.

Consider the uncountably many balls of radius 1/3, each centered on one y in Y. (Whether open or closed is unstated, so I guess that's not relevant.) These balls don't intersect.

If M is any dense set in in l^{\infty}, each of these nonintersecting balls must contain an element of M.

Therefore M is uncountable. M was chosen arbitrarily. Therefore, he concludes, l^{\infty} has no dense subset which is countable. In other words, l^{\infty} is not seperable.

My question. Am I right in thinking this assumes that no ball can contain a limit point of M which is not in M? If so, why is that?

 

[micromass]

My question. Am I right in thinking this assumes that no ball can contain a limit point of M which is not in M? If so, why is that?

I'm a bit confused on why you would think of limit points here. You suppose M is dense, right? Well, in that case, every point of of \ell^\infty (which is not in M) is a limit point of M! So that's not what's going on here.

Can you tell us where you're stuck in the proof??

 

[micromass]

Also note that the function you mention (i.e. the one that sends \hat{y} to \sum{\eta_k/2^k} ) is NOT a bijection. It is merely a surjection! It is indeed true that a point in [0,1] can have two binary representations:

\frac{1}{2}=\frac{1}{2}+0+0+0+0+\hdots=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...

 

[Rasalhague]

Re. #3. Ah, yes, thanks for pointing that out. I thought, for some reason, he was saying that Y is uncountable because it has the same cardinality as [0,1], but I see now he actually only needs to show that it has at least as great a cardinality as [0,1]. (Incidentally, do you know if there exists a bijection between them?)

Re. #2. Got it now! (I hope.)

There are uncountably many balls, and each contains an element of l-infinity, namely each y in Y. In each ball, either its y is in M, or its y is a limit point of M. So if its y is not in M, then this y is a limit point of M, so every open ball around an element of Y, including the ball indicated, must contain an element of M. Therefore every ball of radius, say, 1/3 around some element of Y must contain an element of M. Therefore M too has at least as great a cardinality as [0,1], and so is uncountable.

I think what was confusing me was only that I'm new to some of the concepts and they hadn't bedded down yet in my brain. I just didn't follow through the reasoning far enough.

 

[micromass]

Yes, that is both correct.

And it is indeed true that there is a bijection between [0,1] and the subset of \ell^\infty that you mention. However, finding that explicit bijection is not easy...

 

[Landau]

Personally, I despise decimal expansions, and I try to avoid them whenever I can.

In this case, just take Y=\{0,1\}^\mathbb{N}, the set of sequences consisting of 0's and 1's. This is an uncountable subset of \ell\infty, with any pair distinct elements a distance of 1 from each other. The uncountability is even easier now: there is an obvious bijection between Y and the power set of \mathbb{N}.

Now suppose Z is a dense subset. By definition this means that any ball around any element meets Z. In particular, for every y in Y the ball B(y) at y of radius 1/2 contains an element z(y) of Z. But if y' in Y is another element, then B(y') and B(y) are disjoint (as ||y-y'||=1), so z(y)/=z(y'). Hence the map Y->Z, y\mapsto z(y) is an injection . As Y is uncountable, Z must be too.

(I find this argument also a bit clearer than talking about 'uncountably many balls')

 

[Rasalhague]

Thanks for the confirmation micromass, and thanks Landau for the extra insight. That's very helpful.