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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.

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    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?2012-05-14

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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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    Here's a nice [answer by Andres Caicedo](http://math.stackexchange.com/q/97637/5363) 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...)2012-08-04