Talk:Stone–Čech compactification

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[edit] Awesome

I used this compactification just the other day and was able to reduce the volume of all things I won by 1.5 to 3 times. Thanks Wikipedia! —Preceding unsigned comment added by 70.162.83.30 (talk) 02:18, 17 September 2007 (UTC)

[edit] Overkill

The application to functional analysis strikes me as the most amazing piece of overkill and actually achieves very little. You can characterise the dual space of L^{\infty} in a trivial manner - it's just the space of finitely additive finite measures on the underlying measure algebra. This is about a five line proof. Bringing the Stone Cech compactification and the Riesz representation theorem into it just complicates the issue for no apparent gain. David MacIver 18:00 21st April

> What if we change it for "the computation of the dual space of C_b(X)"; C_b(X) being the space of continous and bounded scalar-valued functions over a completely regular topological space? Do you think that would be more interesting or is it also "trivial"?

It is true that you can caracterize the dual space of L^{\infty} in more elemental terms (though I wouldn't say "trivially") but I find it amazing that you can get a countably additive measure instead of a finitely additive one by enlarging its support. Anyway, I always find myself amazed by things other (smarter) mathematicians consider trivial.

Could you indulge me and write down that five line proof for me? J L 23:14, 22 April 2006 (UTC).

Sure. (Sorry for the delay. I don't have net access at the moment).

Define T : l^inf* -> { finitely additive measures on N } by Tf (A) = f(I_A) (where I_A is the indicator function of A). It's obvious that this is linear and gives a finitely additive measure. We just need to show that it's a bijection.

S, the set of simple functions (linear combinations of indicator functions) is dense in l^inf. If Tf = Tg then f and g agree on simple functions, and so by continuity they agree everywhere. Thus T is injective.

Now let m be a finitely additive measure. Without loss of generality we can assume it's positive. Then we can define f' on S by f'(I_A) = m(A) and extending linearly. Then (by positivity) we have ||f'|| <= m(N). So f' is continuous and can thus be extended to a linear functional on l^inf, say f. Obviously Tf = m.

QED

Ok, a bit more than five lines. Still short though.

David MacIver 10:15 9th May

I guess to capture l^inf* fully you also need to define some norm on the space of finitely additive measures on N and show that T and T−1 are continuous. AxelBoldt 22:23, 28 May 2006 (UTC)
I think ||mu|| = sup { sum_k |mu(A_k)| : A_k is a finite partition of N } works, by analogy with something similar in the countably additive case. I wouldn't swear to it though - I'll try and check the details later. I think in this case, T and T^{-1} should actually be isometries. David MacIver 14:25 1st June
I was reading Carothers's A short course on Banach Space Theory yesterday and happened to come across a section which has bearing on this. Firstly, my conjecture about the appropriate norm on the space of finitely additive measures was correct (I never got around to verifying the details myself). Secondly, this is of independent interest. You can essentially run this proof backwards - show that finitely additive measures on D correspond to countably additive measures on beta D, use this to deduce a (slightly weakened) form of the Riesz representation theorem for Stone Cech compactifications of discrete spaces, then use a pull back argument to get it for arbitrary compact hausdorff spaces. The only weakening this involves is that the measures are Baire measures rather than Borel (The Baire measurable sets are the smallest sigma algebra such that all continuous real valued functions are measurable). This isn't really much of a weakening, as in the cases where these two are different the Baire algebra is in some sense the more natural one to consider anyway. David MacIver 17:30 7th June

The page says "every noncompact metric space contains a copy of betaN", which confuses me: (0,1) is a noncompact metric space of cardinality 2^omega - how can it contain a "copy" of betaN, which is of cardinality 2^2^omega?

I'm removing the sentence about "copy of Bw in any noncompact metric space", because it's just plain wrong. (Another example: the discrete omega is a noncompact metric space, with the discrete metric. How could it contain a copy of betaomega?) Aug 25 Jan Stary

[edit] Question about Overkill

I have a question about the proof given at the "Overkill" section.

Concerning the statement:

"Now let m be a finitely additive measure. Without loss of generality we can assume it's positive."

Why is it that we can assume that the measure `m' is positive? Is there some sort of Hahn descomposition for finitely additive measures?

I will be grateful if someone could clarify on this point or correct the proof. —Preceding unsigned comment added by 24.232.44.135 (talk)

[edit] Is the title of the application appropriate?

A section is named: An application: the dual space of the space of bounded sequences of reals. But we are speaking about the dual space of l^\infty(\mathbb{N}), i.e., about integers, not reals, aren't we? --Kompik (talk) 15:21, 31 January 2008 (UTC)

Feeling ashamed of this silly mistake. Of course we are talking about reals and \mathbb{N} stands for sequences. --Kompik (talk) 14:13, 7 February 2008 (UTC)