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" The intersection of every sequence of open subsets of R is a borel set.However set of all such intersections is not the borel set because it is not closed under countable unions"

I cannot see why the set is not closed under countable unions. Any countable union of such sets can be expressed as the countable intersection of opensets using the fact that countable union of countable sets is countable. More specifically if the first set is the intersection of all O_1n Second set O_2n and so on... Then union of all such sets can be written as intersection of the sets where each set is the union of a possible combination of O _1(n_1),O_2(n_2) and so on...certainly number of such possibilities is countable. I dont understand where i'm going wrong.

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  • $\begingroup$ Try to actually write out your idea in detail. Let's say $U_i^j$ is open for each $i,j\in\mathbb{N}$ and you want to rewrite $\bigcup_{j\in\mathbb{N}}(\bigcap_{i\in\mathbb{N}}U^j_i)$ as a countable intersection of open sets - what exactly would those open sets be? $\endgroup$ Commented Aug 10, 2020 at 21:44

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As it stands, your argument is all hand-waving; if, as was suggested in the comments, you try to make it precise, you’ll find yourself running into difficulties. In the meantime, here is an actual example of a countable union of $G_\delta$ sets that is not itself a $G_\delta$ set.

Let $P=\Bbb R^+\setminus\Bbb Q$, the set of positive irrationals, and let $Q=\Bbb Q\setminus\Bbb Q^+$, the set of non-positive rationals. Then $P$ is a $G_\delta$ in $\Bbb R$, and so is $\{q\}$ for each $q\in Q$, so $P\cup Q$ is a countable union of $G_\delta$ sets. It is not, however, a $G_\delta$ set itself: if it were one, $Q$ would also be one, and the Baire category theorem shows that $Q$ is not a $G_\delta$.

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Let $ K = \mathbb N \times \mathbb N \times \mathbb N \times \dots $. $$ \bigcup _ { j \in \mathbb N } \left( \bigcap _ { i \in \mathbb N } U ^ j _ i \right) = \bigcap _ { ( n _ 1 , n _ 2 , \dots ) \in K } \left( \bigcup _ { j \in \mathbb N } U ^ j _ { n _ j } \right) \text . $$ Whats wrong with the notation used here?

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  • $\begingroup$ How big is $\mathcal{K}$? $\endgroup$ Commented Aug 11, 2020 at 0:06
  • $\begingroup$ Isn'tt it countable? $\endgroup$
    – mathgirl
    Commented Aug 11, 2020 at 0:08
  • $\begingroup$ You should check that. (HINT: what do you know about the set of all infinite binary sequences? How does that relate to $\mathcal{K}$?) $\endgroup$ Commented Aug 11, 2020 at 0:08
  • $\begingroup$ Ah right! Thats silly. mistook cartesian product to be union.Thank you! $\endgroup$
    – mathgirl
    Commented Aug 11, 2020 at 0:10

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