Is this Set Open? - Analysis of X and A in the Real Number System

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Homework Help Overview

The discussion revolves around the set A defined as the union of intervals in the real number system, specifically examining whether this set is open or closed under a given metric. Participants are analyzing the implications of infinite intersections and unions of open sets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of the set A, questioning whether it is open or closed based on the behavior of the intervals as n approaches infinity. There is confusion regarding the distinction between infinite intersections and unions.

Discussion Status

Some participants have provided hints and clarifications regarding the definitions of open and closed sets, particularly in relation to finite versus infinite intersections. There is an ongoing exploration of the implications of these definitions on the set in question.

Contextual Notes

There is a noted misunderstanding about the nature of the set A, with some participants initially interpreting it as an intersection rather than a union. The implications of this misunderstanding are being discussed.

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Homework Statement



Take X = [tex]\Re[/tex] with the metric d(x,y) = |x-y|.

Let A = [tex]\bigcup^{\infty}_{n=1} \left( \frac{-1}{n},\frac{n+1}{n} \right)[/tex]

Determine whether this set is open or closed.

Homework Equations



The intersection of a finite number of open subsets of X is open.

The Attempt at a Solution



Well, I am pretty sure that it's open, as it seems like the intersection goes to (0,1) as n goes to infinity. However, I'm unsure because it's an infinite intersection, and the rule only applies to finite intersections...
 
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0 isn't in the open interval (0,1). Is it in the intersection?
 
Last edited:
Well, as n goes to infinity, -1/n goes to 0, and it's approaching 0 from below. So I would say that yes, it is in the intersection...meaning that the intersection is going to be [0,1] rather than (0,1), meaning that it's closed. Is that correct?
 
Spriteling said:
Well, as n goes to infinity, -1/n goes to 0, and it's approaching 0 from below. So I would say that yes, it is in the intersection...meaning that the intersection is going to be [0,1] rather than (0,1), meaning that it's closed. Is that correct?

That would be correct. The intersection of an infinite number of open sets is not necessarily open.
 
Yeah, I knew that it was only open for sure if it was a finite intersection.

Thank you for the hint!
 
Spriteling said:

Homework Statement



Take X = [tex]\Re[/tex] with the metric d(x,y) = |x-y|.

Let A = [tex]\bigcup^{\infty}_{n=1} \left( \frac{-1}{n},\frac{n+1}{n} \right)[/tex]

Determine whether this set is open or closed.

Homework Equations



The intersection of a finite number of open subsets of X is open.

The Attempt at a Solution



Well, I am pretty sure that it's open, as it seems like the intersection goes to (0,1) as n goes to infinity. However, I'm unsure because it's an infinite intersection, and the rule only applies to finite intersections...

What you have posted is an infinite union, not an infinite intersection.

[tex]\bigcup^{\infty}_{n=1} \left( \frac{-1}{n},\frac{n+1}{n} \right)=(-1,\,2)\cup(-1/2,\,3/2)\cup(-1/3,\,4/3)\cup\dots=(-1,\,2)[/tex]

Did you mean to post the following instead?

[tex]\text{A}=\bigcap^{\infty}_{n=1} \left( \frac{-1}{n},\frac{n+1}{n} \right)[/tex]

This infinite intersection is not the same as (0, 1). ‒1/n < 0 < (n+1)/n for all positive integers, n. So, 0 is in the infinite intersection. Similarly, 1 is also in this intersection.
 

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