1
$\begingroup$

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.

  • 0
    $\LaTeX$ 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$2012-05-14
  • 0
    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?2012-05-14
  • 0
    It is a problem that I can't find examples.....2012-05-14
  • 1
    Start with: What are the open sets? What are the compact sets?2012-05-14
  • 0
    Since discrete topology, open sets are arbitrary2012-05-14
  • 0
    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

1 Answers 1

1

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.

  • 2
    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.](http://math.stackexchange.com/questions/17432/uncountable-subset-with-uncountable-complement-without-the-axiom-of-choice/146700#146700)2012-08-04
  • 0
    Can you give more details? (I assumed the large cardinality to avoid these problem, but it's probably not needed).2012-08-04
  • 1
    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.2012-08-04
  • 1
    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.2012-08-04
  • 0
    That's indeed what I have to assert (the fact that we can partition). Thanks for the link, it's exactly what I needed.2012-08-04
  • 0
    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.2012-08-04
  • 3
    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