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.
 A: 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. 
