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My book states a version of Baire's Theorem as

If $\{G_1,G_2,G_3,\ldots\}$ is a countable collection of dense, open sets, then the intersection $\bigcap_{n=1}^\infty G_n$ is not empty.

I am having trouble understanding why this is the case. If $G_1=\mathbb{Q}$, $G_2=\mathbb{I}$, and $G_n=\mathbb{R}$ when $n\geq 3$, then $\bigcap_{n=1}^\infty G_n=\emptyset$. This obviously contradicts the theorem.

Is my example invalid because $\mathbb{I}$ and $\mathbb{R}$ are uncountable? The book uses the phrase "countable collection of dense, open sets" rather than "collection of countable, dense, open sets", which leads me to believe that the sets $G_n$ do not need to be countable.

Additionally, to prove this theorem the book says to

construct a nested sequence of closed intervals $I_1\supseteq I_2\supseteq I_3\supseteq \ldots$ satisfying $I_n\subseteq G_n$.

Then it says to use the nested interval property to conclude that $\bigcap_{n=1}^\infty G_n\neq\emptyset$. But for $G_1=\mathbb{Q}$ there is no closed interval $I_1=[a,b]$ where $a\neq b$ such that $I_1\subseteq G_1$. This is because the interval $I_1$ will contain some irrational numbers, which are not in $\mathbb{Q}$. Thus $I_1\nsubseteq G_1$. Can someone explain and prove this theorem to me?

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  • $\begingroup$ You need open dense subsets. $\endgroup$ – Sangchul Lee Dec 31 '15 at 3:44
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    $\begingroup$ Precise statement in your direction is in complete metric space, countable intersection of open dense subsets is also open dense subset. $\endgroup$ – p Groups Dec 31 '15 at 4:06
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The theorem does not hold in all spaces; I’m guessing from your discussion that here it’s specifically for $\Bbb R$, which is a space in which it does hold. Your example is not a counterexample: $\Bbb Q$ and $\Bbb I$ are dense, but they are not open, so they are irrelevant to the theorem.

You are correct in thinking that countable refers to the size of family of dense open sets, not to the sizes of the sets themselves. Indeed, every non-empty open set in $\Bbb R$, dense or not, is uncountable: it has the same cardinality as $\Bbb R$ itself.

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  • $\begingroup$ So I can visualize $G_n$ as the entire real line with a given amount of points removed, or $G_n=\mathbb{R} \backslash E_n$ where $E_n$ is a nowhere dense set, correct? $\endgroup$ – cheesyfluff Dec 31 '15 at 3:58
  • $\begingroup$ @cheesyfluff: Yes, provided that $E_n$ is a closed nowhere dense set (since you want $G_n$ to be open). $\endgroup$ – Brian M. Scott Dec 31 '15 at 4:00
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Your example is invalid because neither $\mathbb{Q}$ nor $\mathbb{I}$ are open sets, which is an important assumption in the theorem. You are correct that the sets $G_n$ need not be countable; in fact, since they are non-empty and open, they are never countable.

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