Closed sets intersection of countable open sets

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Discussion Overview

The discussion revolves around the question of whether every closed set in $\mathbb{R}$ can be expressed as the intersection of a countable collection of open sets. Participants explore various approaches and reasoning related to this concept, touching on definitions, properties of open and closed sets, and the implications in metric spaces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests starting with a countable collection of open sets but is challenged on this approach, as it does not directly address the requirement to express closed sets as intersections of open sets.
  • Another participant proposes defining $G_n$ as the union of open balls centered at points in a closed set $F$, questioning how this union guarantees that $F$ is included in the resulting intersection.
  • There is a discussion about the properties of open balls and unions of open sets, with some participants asserting that each $G_n$ contains $F$ and others seeking clarification on this point.
  • Participants express confusion about how to ensure that a closed set can be represented as a countable intersection of open sets, with some questioning the necessity of showing that a countable intersection of open sets is closed.
  • One participant emphasizes that the goal is to show that closed subsets of $\mathbb{R}$ can be represented as countable intersections of open sets, rather than proving that intersections of open sets yield closed sets.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the methods to demonstrate the original claim. There are competing views on how to approach the problem, and several participants express uncertainty about the implications of their reasoning.

Contextual Notes

Some participants highlight the need for clarity regarding the definitions and properties of open and closed sets, as well as the implications of using unions and intersections in this context. There are unresolved questions about the guarantees provided by the proposed definitions and constructions.

Dustinsfl
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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?
 
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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.
 
girdav said:
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.
 
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.
 
girdav said:
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?
 
dwsmith said:
Why does each $G_n$ contain $F$? How is that guaranteed?

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

Why is $F$ in there? Is $F=\{x\}$?
 
No, just an element of $F$. We have $\{x\}\subset B(x,n^{-1})$, then take the union over $x\in F$.
 
girdav said:
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.
 
  • #10
Is it true that all closed sets have a subcover? Is that the jist of the question?
 
  • #11
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.
 

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