Infinite Intersections of Infinite Sets

  • Thread starter Thread starter MrBeezer
  • Start date Start date
  • Tags Tags
    Infinite Sets
Click For Summary

Homework Help Overview

The discussion revolves around the nature of infinite intersections of infinite sets, specifically questioning the validity of the statement that if a sequence of sets \( A_{1} \supseteq A_{2} \supseteq A_{3} \supseteq \ldots \) contains an infinite number of elements, then their infinite intersection \( \bigcap^{\infty}_{n=1} A_{n} \) is also infinite.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants explore proof by induction to show that finite intersections are infinite but question the leap to infinite intersections. Counter-examples are suggested to illustrate potential flaws in the original statement. Specific examples of nested sets are discussed to demonstrate both infinite and empty intersections.

Discussion Status

The discussion is active, with participants providing insights and counter-examples. Some have identified gaps in reasoning and are exploring different scenarios to understand the implications of the original statement. There is no explicit consensus, but productive lines of inquiry are being pursued.

Contextual Notes

Participants are considering the implications of using proper subsets and the nature of the sets involved. The discussion includes examples that lead to both infinite and empty intersections, highlighting the complexity of the topic.

MrBeezer
Messages
5
Reaction score
0

Homework Statement



Decide if the following represents a true statement about the nature of sets. If it does not, present a specific example that shows where the statement does not hold:

If A[itex]_{1}[/itex][itex]\supseteq[/itex]A[itex]_{2}[/itex][itex]\supseteq[/itex]A[itex]_{3}[/itex][itex]\supseteq[/itex]A[itex]_{4}[/itex][itex]\supseteq[/itex]...A[itex]_{n}[/itex] are all sets containing an infinite number of elements, then the intersection [itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite as well.



Homework Equations





The Attempt at a Solution



I decided to attempt this using proof by induction.. although I'm a little unsure of how to do this using sets, here it goes anyways:

Step 1:


Show that [itex]\bigcap[/itex][itex]^{k}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite when k=1

[itex]\bigcap[/itex][itex]^{1}_{n=1}[/itex]A[itex]_{n}[/itex]=A[itex]_{1}[/itex]

A[itex]_{1}[/itex] is infinite.

Therefore,

[itex]\bigcap[/itex][itex]^{k}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite when k=1

Next Step:

Let,
[itex]\bigcap[/itex][itex]^{k}_{n=1}[/itex]A[itex]_{n}[/itex] be infinite.

Inductive Step:

Show that [itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite

[itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex]A[itex]_{n}[/itex]=[itex]\bigcap[/itex][itex]^{k}_{n=1}[/itex]A[itex]_{n}[/itex][itex]\cap[/itex]A[itex]_{k+1}[/itex]

[itex]\bigcap[/itex][itex]^{k}_{n=1}[/itex]A[itex]_{n}[/itex][itex]\cap[/itex]A[itex]_{k+1}[/itex]=A[itex]_{k+1}[/itex]

A[itex]_{k+1}[/itex] is infinite.


Therefore,


[itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite.

[itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite.





Are there any glaring errors here?

Any input would be greatly appreciated.

Thanks!

-Mike
 
Physics news on Phys.org
MrBeezer said:
[itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite.

[itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]A[itex]_{n}[/itex] is infinite.

That's a huge step here. I agree that induction shows that [itex]\bigcap_{k=1}^n{A_k}[/itex] is infinite. But you cannot just conclude that the same holds for an infinite intersection!

If I were you, I'd start looking for a counter-example.
 
Ok, that's what I was afraid of. Thank you for pointing out the gap there. I think I know a counter example, but my flawed inductive proof influenced me not to try it.

Let A[itex]_{1}[/itex]= {1,2,3,4...}
A[itex]_{2}[/itex]={2,3,4,5...}
A[itex]_{3}[/itex]={3,4,5,6...}

Assume there is an element, k[itex]\in[/itex][itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]

k[itex]\notin[/itex][itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex] A[itex]_{n}[/itex]

therefore, [itex]\bigcap^{\infty}_{n=1}[/itex] A[itex]_{n}[/itex]= (empty set).

Ok, so this specific example shows an infinite intersection of infinite sets that equals the empty set( I think), thus disproving the statement above.

is there a way to show that this is true for all infinite intersections of infinite nested subsets? Or can you get different results depending on the type of infinite sets..
 
MrBeezer said:
Ok, that's what I was afraid of. Thank you for pointing out the gap there. I think I know a counter example, but my flawed inductive proof influenced me not to try it.

Let A[itex]_{1}[/itex]= {1,2,3,4...}
A[itex]_{2}[/itex]={2,3,4,5...}
A[itex]_{3}[/itex]={3,4,5,6...}

Assume there is an element, k[itex]\in[/itex][itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]

k[itex]\notin[/itex][itex]\bigcap[/itex][itex]^{k+1}_{n=1}[/itex] A[itex]_{n}[/itex]

therefore, [itex]\bigcap^{\infty}_{n=1}[/itex] A[itex]_{n}[/itex]= (empty set).

Ok, so this specific example shows an infinite intersection of infinite sets that equals the empty set( I think), thus disproving the statement above.

is there a way to show that this is true for all infinite intersections of infinite nested subsets? Or can you get different results depending on the type of infinite sets..

Good job. That's an infinite intersection of infinite sets that's empty. And sure, the result depends on the sets. Can you give an example where the infinite intersection of infinite sets is infinite?
 
Well, don't have to use proper subsets so technically we could have nested subsets where A[itex]_{1}[/itex]=A[itex]_{2}[/itex]=A[itex]_{3}[/itex]=A[itex]_{4}[/itex]... right? I know the infinite intersection of those sets would be infinite, but that's no fun.

What if we had a infinite intersection of sets where A[itex]_{n}[/itex]=N-[itex]\sum[/itex][itex]^{n}_{1}[/itex]2n

So,

A[itex]_{1}[/itex]={1,3,4,5,6,7,8...}
A[itex]_{2}[/itex]={1,3,5,6,7,8...}
A[itex]_{3}[/itex]={1,3,5,7,8...}

In this case,

[itex]\bigcap[/itex][itex]^{\infty}_{n=1}[/itex]={1,3,5,7,9,11...}

So the infinite intersection equals an infinite amount of odd natural numbers.
 
MrBeezer said:
Well, don't have to use proper subsets so technically we could have nested subsets where A[itex]_{1}[/itex]=A[itex]_{2}[/itex]=A[itex]_{3}[/itex]=A[itex]_{4}[/itex]... right? I know the infinite intersection of those sets would be infinite, but that's no fun.

What if we had a infinite intersection of sets where A[itex]_{n}[/itex]=N-[itex]\sum[/itex][itex]^{n}_{1}[/itex]2n

So,

A[itex]_{1}[/itex]={1,3,4,5,6,7,8...}
A[itex]_{2}[/itex]={1,3,5,6,7,8...}
A[itex]_{3}[/itex]={1,3,5,7,8...}

In this case,

[itex]\bigcap^{\infty}_{n=1}[/itex]An = {1,3,5,7,9,11...}

So the infinite intersection equals an infinite amount of odd natural numbers.
No doubt you mean: [itex]\displaystyle A_{n}=\mathbb{N}-\bigcup_{k=1}^n\{2k\}[/itex]
 
Yes I do, thank you!
 
One proof of a version of Cauchy's theorem actually uses the fact that this is not the case. You take the intersection of some specific nested triangles in the complex plane, which shrink to exactly one point. It seems like you had countably infinite sets in mind, but this proof is too pretty to not mention.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
1K