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closed sets intersection of countable open sets

dwsmith

Well-known member
Feb 1, 2012
1,673
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?
 

girdav

Member
Feb 1, 2012
96
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

girdav

Member
Feb 1, 2012
96
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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?
 

girdav

Member
Feb 1, 2012
96

dwsmith

Well-known member
Feb 1, 2012
1,673

girdav

Member
Feb 1, 2012
96
No, just an element of $F$. We have $\{x\}\subset B(x,n^{-1})$, then take the union over $x\in F$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Is it true that all closed sets have a subcover? Is that the jist of the question?
 

girdav

Member
Feb 1, 2012
96
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.