# Help with proof of existence of $G_\delta$ sets $G$ such that both $G$ and its complement have uncountable intersection with any open interval

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?

• What do you mean by $\bigcap_jG_j=G_j$ ? – Gae. S. Jan 29 at 22:19
• Or, to be more clear, what do you mean by $\bigcap_wG_w= G_j$ ? – Gae. S. Jan 29 at 22:21
• It's not all that clear to me why $G$ should be a subset of $\Bbb Q$. – Gae. S. Jan 29 at 22:33
• 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 at 23:25
• 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 at 14:00

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.