# Baire $\sigma$ -algebra

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.

-
$\LaTeX$ 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$– Arturo Magidin May 14 '12 at 0:24 In this case, both the Baire$\sigma$-algebra and the Borel$\sigma$-algebra have explicit descriptions. Can you think of some examples of Baire sets? Borel sets? – froggie May 14 '12 at 0:32 It is a problem that I can't find examples..... – Park May 14 '12 at 0:49 Start with: What are the open sets? What are the compact sets? – GEdgar May 14 '12 at 2:13 Since discrete topology, open sets are arbitrary – Park May 14 '12 at 2:28 ## 1 Answer Since the topology is discrete, each subset of$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. - 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 4 '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 4 '12 at 17:31 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 4 '12 at 17:33 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 4 '12 at 17:36
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 4 '12 at 17:47