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Every dense $G_\delta$ subset of $\Bbb R$ is uncountable. I know that I have to use Baire's Theorem but I don't know how.

Thanks!

Martin Sleziak
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user113576
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  • Hm... what's $G_\delta$ mean? And do you remember that a set $E \subset \mathbb{R}$ is dense iff $\overline{E} = \mathbb{R}$? Why don't you start using those two facts and tell us what you conclude? – Jesse Madnick Mar 11 '14 at 23:27
  • @JesseMadnick $G_\delta $ set is a set that can be written as the countable infinite intersection of open sets. – user113576 Mar 11 '14 at 23:29
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    If the set were countable, it would be of first category. You can also show, using denseness of the set, that its complement is of first category. – David Mitra Mar 11 '14 at 23:41
  • @DavidMitra,could you give me some help by starting the proof? – user113576 Mar 11 '14 at 23:47

2 Answers2

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Let $A$ be a dense $G_\delta$. Then $A =\bigcap\limits_{i=1}^\infty O_i$ where each $O_i$ is open. So, $A^C=\bigcup\limits_{i=1}^\infty O_i^C $.

Each $O_i^C$ is closed. Show, using the denseness of $A$, that each $O_i^C$ is nowhere dense.

Once you do this, it follows by definition that $A^C$ is of first category. It then follows that $A$ must be uncountable. Otherwise, $\Bbb R=A\cup A^C$ would be of first category (this is where Baire comes into play).

David Mitra
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Let $A$ be a dense $G_{\delta}$ set. Let $A=\cap_{n\in \Bbb N}U_n$ where each $U_n$ is open. Each $U_n$ is dense because $U_n\supset A$ and $A$ is dense.

Let $B=\{b_n:n\in \Bbb N\}$ be any non-empty countable subset of $\Bbb R.$ (It does not matter whether $b_m=b_n$ for some distinct $m,n.$) Let $V_n=U_n\setminus \{b_n\}$. Then $V_n$ is dense and open.

We have $$A \setminus B=\cap_{n\in \Bbb N}(\;U_n\setminus \{b_n\}\;)= \cap_{n\in\Bbb N}V_n.$$ By the Baire Category Theorem, $A\setminus B$ is dense. Therefore $A\not \subset B.$ In particular $A\ne B.$

DanielWainfleet
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