1
$\begingroup$

Let $X$ be a metric space, and $\mathcal{B}_X$ denote the $\sigma$-algebra generated by the open sets. I have to prove that $\mathcal{B}_X$ is the smallest collection of subsets of $X$ that contain the open sets of $X$ and is stable under countable union and countable intersection. (note the complement is missing!)

I have proved before that every closed set is a $G_{\delta}$. So, I proved that the closed sets are in the collection. But I don't know how to proceed and prove that every complement is in the $\sigma$-algebra. Any help?

$\endgroup$
  • $\begingroup$ If some set $S$ is a $G_\delta$, it is in $\mathcal B$. Hence, $\mathcal B$ contains the closed sets, which are complements of... $\endgroup$ – Pedro Tamaroff Sep 5 '14 at 22:47
  • $\begingroup$ Yes, but I have to prove that complements of everything of the collection, not only open sets, are there. $\endgroup$ – Aloizio Macedo Sep 5 '14 at 22:49
  • $\begingroup$ Once you know the complements of open sets are there, you're done. The set is generated by open sets! Recall that complementation changes intersections and unions, so there's no fuzz there. $\endgroup$ – Pedro Tamaroff Sep 5 '14 at 22:53
  • $\begingroup$ But the gerenerating processes are different. On one, I can take complements. On the other, I can't. How can you guarantee that they generate the same things? Sorry, it may be evident, but I'm not getting it. $\endgroup$ – Aloizio Macedo Sep 5 '14 at 22:56
1
$\begingroup$

Define a sequence of families $S_\alpha$ of subsets of $X$ by transfinite induction, for $\alpha \leq \omega_1 = \aleph_1$, as follows.

$S_0 = \{\text{closed and open subsets of X}\}$

$S_{\alpha + 1} = \{\text{countable unions and intersections of elements of $S_\alpha$}\}$

$S_{\beta} = \cup_{\alpha < \beta} S_{\alpha}$, for $\beta$ a limit ordinal.

Then it can be proved that the collection mentioned in your question is $S_{\omega_1}$. The key point is that any countable sequence of ordinals less than $\omega_1$ is bounded above by an ordinal below $\omega_1$ (in addition to the fact you alluded to that closed sets are countable intersections of open ones).

Now you should be able to prove by induction on $\alpha$ that $S_\alpha$ is closed under complements.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What is perhaps easier is to only do what you are proposing up to $\omega$, i.e. Define $\mathcal{A} := \bigcup_n S_n$. Show that this is an algebra and invoke the monotone class theorem (en.wikipedia.org/wiki/Monotone_class_theorem). This removes the need for transfinite recursion/induction. $\endgroup$ – PhoemueX Sep 6 '14 at 4:19
  • $\begingroup$ Isn't $S_2$ already an algebra? $\endgroup$ – Dave Sep 6 '14 at 4:30
  • $\begingroup$ That could be, I just wanted to be sure :) Also, it would of course suffice to only take finite unions and intersections in the definition of the $S_n$ for the monotone class argument to work. $\endgroup$ – PhoemueX Sep 6 '14 at 4:34
  • $\begingroup$ Oh wait, I'm not so sure $S_2$ is an algebra now. It would be if we only took finite unions and intersections. In any case, your argument works. $\endgroup$ – Dave Sep 6 '14 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.