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.



2 Answers 2

  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.

  • $\begingroup$ Thanks! Can you be more formal about (1)? $\endgroup$ Apr 6, 2021 at 17:06
  • $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$ Apr 6, 2021 at 17:12
  • $\begingroup$ @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?) $\endgroup$ Apr 6, 2021 at 17:25
  • $\begingroup$ $X = O\setminus \{x\}$ is open and $X \cap G_n = \emptyset$. I think that completes it, right? $\endgroup$ Apr 6, 2021 at 17:32
  • $\begingroup$ For the uncountability, we can say more. $|\cap G_n| \geq 2^{\aleph_0}$. $\endgroup$ Apr 6, 2021 at 18:26
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    $\begingroup$ @SungjinKim And we can even upgrade that to $=2^{\aleph_0}$. $\endgroup$ Apr 6, 2021 at 19:13

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