Closed sets intersection of countable open sets

[Dustinsfl]

Prove that every closed set in $\mathbb{R}$ is the intersection of a countable collection of open sets.

Let $G_n$ be a countable collection of open sets.
Then we would have 2 cases either $x\in\bigcap G_n$ which is a point which is closed.
Or we could $(a,b)$ in all $G_n$ but how to show that would be closed?

 

[tarantella]

I don't understand your reasoning: you start by "let $\{G_n\}$ a countable collection of open sets: that's not what is asked. We want to show that each closed subset $F$ of the real line can be written as a countable intersection of open sets. To see that, define $G_n:=\bigcup_{x\in F}B(x,n^{-1})$.

Note that this property is true for all metric spaces.

 

[Dustinsfl]

I don't understand your reasoning: you start by "let $\{G_n\}$ a countable collection of open sets: that's not what is asked. We want to show that each closed subset $F$ of the real line can be written as a countable intersection of open sets. To see that, define $G_n:=\bigcup_{x\in F}B(x,n^{-1})$.

Note that this property is true for all metric spaces.

I don't see how that union works.

 

[tarantella]

An open ball is open, and an arbitrary union of open sets is open hence so is $G_n$. Each $G_n$ contains $F$, and so does their intersection. If $x\in G_n$ for each $n$, then $d(x,x_n)\leq n^{-1}$ for some $x_n\in F$. We deduce that $\{x_n\}$ converges to $x$ and since this sequences lies in a closed set the limit still is in this closed set.

 

[Dustinsfl]

An open ball is open, and an arbitrary union of open sets is open hence so is $G_n$. Each $G_n$ contains $F$, and so does their intersection. If $x\in G_n$ for each $n$, then $d(x,x_n)\leq n^{-1}$ for some $x_n\in F$. We deduce that $\{x_n\}$ converges to $x$ and since this sequences lies in a closed set the limit still is in this closed set.

Why does each $G_n$ contain $F$? How is that guaranteed?

 

[tarantella]

Why does each $G_n$ contain $F$? How is that guaranteed?

Because $x\in B(x,n^{-1})$.

 

[Dustinsfl]

Because $x\in B(x,n^{-1})$.

Why is $F$ in there? Is $F=\{x\}$?

 

[tarantella]

No, just an element of $F$. We have $\{x\}\subset B(x,n^{-1})$, then take the union over $x\in F$.

 

[Dustinsfl]

No, just an element of $F$. We have $\{x\}\subset B(x,n^{-1})$, then take the union over $x\in F$.

I still don't understand how we can ensure $F$ is a closed set just by taking unions of open sets.

 

[Dustinsfl]

Is it true that all closed sets have a subcover? Is that the jist of the question?

 

[tarantella]

Actually, it seems that you believe that you have to show that a countable intersection of open sets is closed: it's not what you have to show (fortunately, as it's not true, taking $G_n=O$ where $O$ is open and not closed).

What you have to show is that each closed subset of $\Bbb R$ can be written as a countable intersection of open sets.