Real Analysis Topology Problem Here is a problem I have been working on for the past 2 weeks, but to no avail:
Suppose that $\mathcal{F}$ is a family of open sets in $\mathbb{R}$, with the following property: for every open set $A\subset \mathbb{R}$ such that $\mathbb{Z}\subset A$, there exists $F\in\mathcal{F}$ such that $\mathbb{Z}\subset F\subset A$. Show that $\mathcal{F}$ must be an uncountable family.
Here the open sets are just the open sets in the metric space $(\mathbb{R},d$), where $d(x,y)=|x-y|$ is the Euclidean metric.
 A: This kinda reminds of the proof that $\mathbb{R}$ is uncountable. Suppose $\mathcal{F}$ is countable. Then the set of sets from $\mathcal{F}$ which contain $\mathbb{Z}$ as a subset (let's call it $\mathcal{G}$) is also countable. Let's write $\mathcal{G}=\{F_n\}_{n=1}^\infty$. Now we are going to define a new sequence of sets. 
First of all we define $A_0=(-\infty, \frac{1}{2})$. 
We know that $1\in F_1$. Since $F_1$ is open there is $0<\epsilon<\frac{1}{2}$ such that $(1-\epsilon,1+\epsilon)\subseteq F_1$. So now define $A_1=(1-\frac{\epsilon}{2},1+\frac{\epsilon}{2})$. 
Next, as we know $2\in F_2$. Since it is an open set there is $0<\epsilon<\frac{1}{2}$ such that $(2-\epsilon,2+\epsilon)\subseteq F_2$. So now define $A_2=(2-\frac{\epsilon}{2},2+\frac{\epsilon}{2})$. 
Continue this way by induction and finally define $A=\cup_{n=0}^\infty A_n$. Then $A$ is an open set as a union of open sets, and it contains $\mathbb{Z}$. Hence it must contain a set from $\mathcal{G}$. However, does it contain $F_1$? The answer is no, because $1+\frac{2\epsilon}{3}$ is an element in $F_1$, but not in $A$. And in a similar way you can show that $A$ doesn't contain any of the sets $F_n$. This is a contradiction. 
A: This is not really much different than the answer already posted by @Mark. 
(1). The result is essentially the same as saying that the quotient space, when we identify all points of $\Bbb Z$ into one point,say $p$, then the quotient space does not have a countable local base at the point $p$. 
(2). As a technicality, I would choose a slightly different presentation. 
Without loss of generality we may assume that $\Bbb Z\subseteq F$ for every $F\in\mathcal F$ and, towards a contradiction, that 
$\mathcal F=\{F_n:n=1,2,3,...\}$ is countable. 
For every $n$ pick $x_n\in F_n\cap(n-\frac12,n+\frac12)$ (there is no induction involved), and let $C=\{x_n:n=1,2,3,...\}$, let $A=\Bbb R\setminus C$. Then $C$ is closed, $A$ is open, and by construction there is no $n$ with $F_n\subseteq A$ since $x_n\in F_n\setminus A$ for each $n$. This contradiction shows that $\mathcal F$ cannot be countable. 
One may try to prove directly that $C$ as constructed above is closed (it won't be difficult). But this also follows from a general result the the union of any locally-finite family of closed sets is closed. The family $\{\{x_n\}:n=1,2,3,...\}$ is locally-finite and each member $\{x_n\}$ is a singleton, and hence closed. Thus 
$C=\cup\{\{x_n\}:n=1,2,3,...\}$ is a closed set. 
This also shows that we could have picked $x_n$ in some other (similar) ways, without changing the proof much. For example, we could have picked 
$x_n\in(n-1,n+1)\cap F_n$, or $x_n\in(n,n+1)\cap F_n$. 
Note that (unlike the families $\{(n-\frac12,n+\frac12):n=1,2,3,...\}$ 
and $\{(n,n+1):n=1,2,3,...\}$) the family 
$\{(n-1,n+1):n=1,2,3,...\}$ is not a disjoint family. Yet, it is locally-finite, so the resulting set $C=\{x_n:n=1,2,3,...\}$ would be closed even if we pick $x_n\in(n-1,n+1)\cap F_n$. 
A: Let $F=\{F_n: n\in \Bbb N\}$ be a countable open family in $\Bbb R$ such that $\Bbb Z\subset F_n$ for each $n\in\Bbb N.$ Then there exists an open $A\subset \Bbb R$ with $\Bbb Z \subset A, $  such that $\neg (F_n\subset A)$ for every $n\in \Bbb N.$
Proof. For each $n\in \Bbb N$ let $\frac {1}{2}>b_n>0$ such that $(n-b_n,n+b_n)\subset F_n.$ Let $A=(-\infty, \frac {1}{2})\cup (\;\cup_{n\in \Bbb N}(n-\frac {1}{2}b_n, n+\frac {1}{2}b_n)\,).$
For each $n\in \Bbb N$ we have  $A\cap (n-\frac {1}{2},n+\frac {1}{2})= (n-\frac {1}{2}b_n,n+\frac {1}{2}b_n)$ but $F_n \cap (n-\frac {1}{2},n+\frac {1}{2})\supset (n-b_n,n+b_n),$ so $\neg (F_n\subset A).$
