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I need help in understanding the proof of the theorem from the this question: Countable and uncountable number: Answer by Andrés E. Caicedo

Third, there are $G_δ$ sets $G$ such that $G$ and its complement have uncountable intersection with any open set. For instance, let ($q_n:n≥0$) enumerate the rationals. For $i,j∈N$ let $$I_{i,j}=\left(q_i-\frac1{2^{i+j}},q_i+\frac1{2^{i+j}}\right)$$ $$G_j=\bigcup_i I_{i,j}$$ $$G=\bigcap_j G_j$$ But it’s obvious that $$I_{1,1}\supset I_{2,1} \supset I_{3,1} \supset \ldots$$ $$I_{1,2}\supset I_{2,2} \supset I_{3,2} \supset \ldots$$ $$\ldots$$ Then $G_1 \supset G_2 \supset \ldots \supset G_n \supset \ldots $ . But $$G=\bigcap_j G_j=Q$$ And then all the other arguments of the author of the answer are meaningless. Where is the mistake?

Eric Wofsey
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CleverLever
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  • What do you mean by $\bigcap_jG_j=G_j$ ? –  Jan 29 '20 at 22:19
  • Or, to be more clear, what do you mean by $\bigcap_wG_w= G_j$ ? –  Jan 29 '20 at 22:21
  • It's not all that clear to me why $G$ should be a subset of $\Bbb Q$. –  Jan 29 '20 at 22:33
  • I mean, you can't possibly be using the non-existing identity $\bigcap_i \bigcup_j A_{i,j}=\bigcup_j\bigcap_i A_{i,j}$, can you? –  Jan 29 '20 at 22:39
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Jan 29 '20 at 23:25
  • (Thanks to this post I noticed the argument at the end that $G$ is comeager needed an extra line. Fixed now.) – Andrés E. Caicedo Jan 29 '20 at 23:28
  • @CleverLever You did not, but you did not prove that $\bigcap_j G_j\subseteq\Bbb Q$ either. That looked like something you might have done, though, seeing the consideration at the start and the one at the end. –  Jan 30 '20 at 12:05
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    Can you clarify what part is not clear? First, are you familiar with the technical terms of nowhere dense, meager and comeager? One shows easily that the complement of each $G_j$ is nowhere dense. This needs a little argument but it is truly easy, by design. Once you have this, that $G$ is comeager follows from the definitions. And once we have that, all that is needed is to apply the Baire category theorem for complete metric spaces. – Andrés E. Caicedo Jan 30 '20 at 14:00

1 Answers1

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Although the $G_j$ are decreasing, their intersection is far from just being the rationals. As indicated in the original post, in fact $G$ meets each open interval in uncountably many points.

A nowhere dense subset of $\mathbb R$ is a set $X\subseteq \mathbb R$ with the property that any nonempty open set $U\subseteq\mathbb R$ contains a nonempty open subset $V$ with $V\cap X=\emptyset$.

One can easily check (by design) that, for each $j$, the complement $G_j^c$ of $G_j$ is nowhere dense: given $U$, there is an open interval $I$ contained in $U$. Let $x$ be the center of $I$ and pick $n$ large enough that $(x-2/2^{n+j},x+2/2^{n+j})\subseteq I$. Pick $i>n$ such that $q_i\in (x-1/2^{n+j},x+1/2^{n+j})$ and check that $(q_i-1/2^{i+j},q_i+1/2^{i+j})\subseteq I$. The interval $(q_i-1/2^{i+j},q_i+1/2^{i+j})$ is contained in $G_j$ and therefore disjoint from $G_j^c$. It is contained in $U$ by construction, and it follows that $G_j^c$ is nowehere dense.

By definition, this means that $G^c=\bigcup_j G_j^c$ is meager and therefore $G$ is comeager. In fact, for any nontrivial closed interval $J\subseteq \mathbb R$, the same argument gives that $G\cap J$ comeager in $J$. It follows from standard formulations of the Baire category theorem that $G\cap J$ is actually uncountable. For instance, if it were countable, say equal to $\{s_n:n\in\mathbb N\}$, then $G\cap J\cap\bigcap_n(J\smallsetminus\{s_n\})$ would be an empty comeager subset of $J$, a contradiction.

Andrés E. Caicedo
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