# Uncountability of dense $G_\delta$ sets in $\Bbb R$

I am not able to understand this proof of the uncountability of dense $$G_\delta$$ sets in $$\Bbb R$$.

If $$(G_n)$$ is a sequence of open dense sets in $$\Bbb R$$ and if $$\bigcap_{n=1}^\infty G_n = \{x_1,x_2,...\}$$ then the sets $$\tilde G_n = G_n\setminus \{x_n\}$$ are still open and dense, but $$\bigcap_{n=1}^\infty \tilde G_n = \emptyset$$, contrary to Baire's theorem. Thus, $$\bigcap_{n=1}^\infty G_n$$ is uncountable.

1. What is a dense $$G_\delta$$ subset of $$\Bbb R$$? I think it should be $$G_\delta = \bigcap_{n=1}^\infty G_n, G_n \text{ open}$$ such that $$\overline{\bigcap_{n=1}^\infty G_n} = \Bbb R$$ Then why does the author insist that $$G_n$$ be dense too? Isn't it enough for $$G_n$$ to be open by definition of $$G_\delta$$?

2. Why $$\tilde G_n$$ are open and dense? I see that $$\tilde G_n$$ are open because deleting a point from an open set preserves its openness, but what about density?

3. Why is $$\bigcap_{n=1}^\infty \tilde G_n = \emptyset$$? If this is true, then I see the contradiction. Need help showing that $$\bigcap_{n=1}^\infty \tilde G_n = \emptyset$$ though.

Merci!

1. A dense $$G_\delta$$ set is a $$G_\delta$$ set that is dense. A $$G_\delta$$ set is the intersection of a sequence of open sets. If the intersection is dense, then each of the sets you are intersecting is also dense.

2. To see that $$\widetilde{G}_n$$ is dense in $$\mathbb R$$, consider any nonempty open set $$U$$ in $$\mathbb R$$. Since $$G_n$$ is dense in $$\mathbb R$$, $$G_n \cap U$$ is nonempty: say $$x \in G_n \cap U$$. Since $$G_n$$ and $$U$$ are open, $$G_n \cap U$$ contains an open interval $$(x-\epsilon, x+\epsilon)$$. Now $$\widetilde{G}_n \cap U$$ contains all but at most one point of this interval. Since the interval has infinitely many points, $$\widetilde{G}_n \cap U$$ is nonempty. By definition, that says $$\widetilde{G}_n$$ is dense in $$\mathbb R$$.

3. Your assumption was that $$\bigcap_n G_n = \{x_1, x_2, \ldots\}$$. You have removed each $$x_n$$ from $$\widetilde{G}_n$$, and therefore from the intersection. So there is nothing left.

• Thanks! Can you be more formal about (1)? Apr 6, 2021 at 17:06
• Also in (3) maybe you want to say: "you have removed each $x_n$ from $G_n$...." and not $\tilde G_n$. Although I understand the argument. Apr 6, 2021 at 17:21

The topological nature of the sets involved (open, $$G_\delta$$, etc.) is actually playing no role in your questions (1)-(3) at all; instead, the relevant points are general facts about density and intersections of sets.

For (1), the point is that a non-dense set cannot become dense by removing more elements - or, to put it another way, if the intersection of a bunch of sets is dense then each of the original sets had to be dense as well.

For (2), you need to argue that removing a single point from a dense set leaves a dense set. This isn't actually true in all topological spaces (think about a discrete space for instance) but it is true in $$\mathbb{R}$$; do you see why?

For (3), the point is that the $$x_n$$s are the only possible elements of the set $$\bigcap_{n\in\mathbb{N}}\tilde{G}_n$$, but by construction they aren't in that intersection - so that intersection must be empty. In more detail, we can't have $$x_1$$ in that intersection since it's not in $$\tilde{G}_1$$; we can't have $$x_2$$ in that intersection since it's not in $$\tilde{G}_2$$; and so on. But now think about what $$\tilde{G}_n\subseteq G_n$$ tells us about the relationship between the corresponding infinite intersections:

$$\bigcap_{n\in\mathbb{N}}\tilde{G}_n\subseteq\bigcap_{n\in\mathbb{N}} G_n=\{x_1,x_2,...\}$$.

So the only things which could possibly be elements of $$\bigcap_{n\in\mathbb{N}}\tilde{G}_n$$ are the $$x_n$$s, which we already know aren't in that intersection. So there's nothing in that intersection at all.

• For (2), how is a proof by contradiction? I use that a set is dense if and only if every open set intersects it. $G_n$ is dense, so $G_n \cap O \ne \emptyset$ for all open sets $O$ and all $n$. Suppose $G_n\setminus \{x_n\}$ is not dense. Then, there is some open set $O$ such that $G_n\setminus \{x\} = \emptyset$. In that case, $G_n \cap O = \{x_n\}$. Why is this a contradiction? Apr 6, 2021 at 17:12
• @analysis123 That's not a contradiction yet, you have to do more work. Think about finding a smaller (but still nonempty) open set than $O$ which has empty intersection with $G_n$. (HINT: what happens when you remove a single element from an open set?) Apr 6, 2021 at 17:25
• $X = O\setminus \{x\}$ is open and $X \cap G_n = \emptyset$. I think that completes it, right? Apr 6, 2021 at 17:32
• For the uncountability, we can say more. $|\cap G_n| \geq 2^{\aleph_0}$. Apr 6, 2021 at 18:26
• @SungjinKim And we can even upgrade that to $=2^{\aleph_0}$. Apr 6, 2021 at 19:13