Problem about $G_\delta$ sets in $\mathbb R$.

Question

Which of the following cannot be written as the intersection of countably many open sets in $\mathbb R$?

$\mathbb Z$

$\mathbb Z'$

$\mathbb Q$

$\mathbb Q'$

$[0,1)$

My attempt: I know that the sets $\mathbb Z',\mathbb Q',[0,1)$ are $G_\delta$ and $\mathbb Q$ is not $G_\delta$ and because $\mathbb Z$ is closed in $\mathbb R$ so it is $F_\sigma$, not $G_\delta$.

Am I right?

Q' means complement of Q in R, similarly Z' is the complement of Z in R — Nannes, Jun 11 '14 at 10:55
Where you wrote $F_\delta$ I guess you meant $F_\sigma$, because $F_\delta=F$. Anyway, $\mathbb Z$ is $G_\delta$, because closed sets in $\mathbb R$ are $G_\delta$. You're right that $\mathbb Q$ is not $G_\delta$ (it is $F_\sigma$), and $[0,1)$ is both $G_\delta$ and $F_\sigma$. I don't know what you mean by $\mathbb Z'$ and $\mathbb Q'$. — bof, Jun 11 '14 at 10:55
OK, then you're right about $\mathbb Z'$ and $\mathbb Q'$ being $G_\delta$. — bof, Jun 11 '14 at 10:58
But you said that Z is G delta because it is closed, is this a theorem? Can you tell me some book from which i find this result? — Nannes, Jun 11 '14 at 11:00
$U_n=\bigcup_{z\in\mathbb Z}(z-\frac1n,z+\frac1n)$ is open, so $\mathbb Z=\bigcap_{n=1}^\infty U_n$ is $G_\delta$. The more general result, that every closed set is $G_\delta$, is proved in a similar way. It's not true in all topological spaces, but it is true in all metric spaces. I'll try and find an online reference for you. — bof, Jun 11 '14 at 11:08

score 1 · Accepted Answer · answered Jun 11 '14 at 12:45

$\mathbb{Z}$ is closed in a metric space so a $G_\delta$. To see this define for a set $A \subset X, r>0$: $B(A,r) = \{y \in X: d(y, A) < r \}$, show this is open, and show that $\overline{A} = \cap_{n \in \mathbb{N}} B(A, \frac{1}{n})$, so when $A$ is closed, we have written it as an intersection of countably many open sets.

Of course, this is overkill in general (but nice to know), and we can just define $U_n = \cup_{a \in \mathbb{Z}} (a - \frac{1}{n}, a+\frac{1}{n})$, which is open as a union of open sets, and show that $\cap U_n = \mathbb{Z}$ (same idea of course).

$\mathbb{Z}'$ is open, so certainly a $G_\delta$.

$\mathbb{Q}$ is not a $G_\delta$ by the Baire category theorem.

$\mathbb{Q}' = \cap_{q \in \mathbb{Q}} \{q\}'$, which is a $G_\delta$ as the rationals are countable, and complements of singletons are open.

$[0,1) = \cap_n (-\frac{1}{n}, 1)$, so a $G_\delta$ as well. It's also an $F_\sigma$, as $[0,1) = \cup_n [0,1-\frac{1}{n}]$, but neither open or closed.

score 0 · Answer 2 · answered Jun 11 '14 at 13:35

Hint: The $G_\delta$ sets in $\mathbb R$ are precisely the sets of continuity points of a function. So, for each of the sets you're given, try to find a function that is continuous exactly in the set. This may be easier than using the definition of $G_\delta$ set.

Problem about $G_\delta$ sets in $\mathbb R$.

2 Answers2