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How can I solve this problem.

Let X be an uncountable set with the discrete topology. Show that the Baire $\sigma$-algebra of X differs from Borel $\sigma$-algebra of X.

Arturo Magidin
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Park
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  • $\LaTeX$ tip: If you put the - inside the $, it is interpreted like the operator $-$; you want it to be a simple hyphen, so it should be outside of the $\LaTeX$ – Arturo Magidin May 14 '12 at 00:24
  • In this case, both the Baire $\sigma$-algebra and the Borel $\sigma$-algebra have explicit descriptions. Can you think of some examples of Baire sets? Borel sets? – froggie May 14 '12 at 00:32
  • It is a problem that I can't find examples..... – Park May 14 '12 at 00:49
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    Start with: What are the open sets? What are the compact sets? – GEdgar May 14 '12 at 02:13
  • Since discrete topology, open sets are arbitrary – Park May 14 '12 at 02:28
  • Exactly. Which subsets of $X$ are compact, given the fact that every subset of $X$ (and so, in particular, every singleton ${x}$ with $x\in X$) is open? – Cameron Buie May 14 '12 at 02:59

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Since the topology is discrete, each subset of $X$ is open and the Borel $\sigma$-algebra is the collection of subsets of $X$.

We have to use Halmos' definition, with Dudley one the two $\sigma$-algebras coincide. The compact subsets of $X$ are finite (and $G_{\delta}$ since they are open), hence the smallest $\sigma$-algebra containing them contains all the countable subsets of $X$, and their complement. Since $$\{A\subset X, A\mbox{ or }X\setminus A\mbox{ is at most countable}\}$$ is a $\sigma$-algebra, it's actually the Baire $\sigma$-algebra of $X$.

$X$ contains a uncountable set of uncountable complement, which show that Borel and Baire $\sigma$-algebras are not the same.

We can also use @t.b. argument: to see that $|X\times X|=|X|$, apply Zorn's lemma to $$P:=\{(A,g), A\subset X, f\colon A\times A\to A\mbox{ is a bijection}\},$$ with partial order $(A_1,f_1)\leq (A_2,f_2)$ if and only if $A_1\subset A_2$ and $g_{\mid A_1\times A_1}=f$. It shows that $(X,f)$ is maximal for some $f$. Then take $x_0\in X$, $S:=\{x_0\}\times X$, which is uncountable, with uncountable complement. Then $f(\{x_0\}\times X)$ does the job.

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Davide Giraudo
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    I don't understand that last part. If $X$ is any uncountable set, the co-countable $\sigma$-algebra is not the power set of $X$. At least if we assume some choice. – Asaf Karagila Aug 04 '12 at 17:30
  • Can you give more details? (I assumed the large cardinality to avoid these problem, but it's probably not needed). – Davide Giraudo Aug 04 '12 at 17:31
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    Davide, if $X$ is uncountable it has a subset of size $\aleph_1$; we can partition this set into two uncountable sets (in fact to $\aleph_1$ uncountable sets). Add the rest of $X$ to one of these, then you have an uncountable set whose complement is also uncountable. – Asaf Karagila Aug 04 '12 at 17:33
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    You have $# X = #(X \times X)$ for every infinite set. If $X$ is uncountable then every slice ${x_0} \times X$ is uncountable and has uncountable complement. – t.b. Aug 04 '12 at 17:36
  • That's indeed what I have to assert (the fact that we can partition). Thanks for the link, it's exactly what I needed. – Davide Giraudo Aug 04 '12 at 17:36
  • I think that the link is not as relevant as you think. It gives a counterexample which may occur without the axiom of choice; while not giving an argument why we can partition if we have the axiom of choice. – Asaf Karagila Aug 04 '12 at 17:44
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    Here's a nice answer by Andres Caicedo giving a detailed explanation of Asaf's suggestion in his second comment. (It's always dangerous to answer these measure theoretic questions with all those set theorists around...) – t.b. Aug 04 '12 at 17:47