Infinite Intersections of Infinite Sets

In summary, the statement that the intersection \bigcap^{\infty}_{n=1}A_{n} is infinite if A_{1}\supseteqA_{2}\supseteqA_{3}\supseteqA_{4}\supseteq...\supseteq A_{n} are all sets containing an infinite number of elements is false, as shown by the counter example of nested sets A_{1}={1,2,3,4...}, A_{2}={2,3,4,5...}, A_{3}={3,4,5,6...}. However, this statement may hold for certain types of infinite sets, such as the infinite intersection of nested subsets where A
  • #1
MrBeezer
5
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
 
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  • #2
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.
 
  • #3
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..
 
  • #4
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?
 
  • #5
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.
 
  • #6
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]
 
  • #7
Yes I do, thank you!
 
  • #8
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.
 

1. What is an infinite intersection of infinite sets?

An infinite intersection of infinite sets is a mathematical concept where the intersection of an unlimited number of sets is taken. This means that the resulting set will contain all the elements that are common to all the infinite sets being intersected.

2. How is an infinite intersection of infinite sets different from a finite intersection?

An infinite intersection of infinite sets differs from a finite intersection in that it involves an unlimited number of sets, whereas a finite intersection only involves a specific number of sets. Additionally, an infinite intersection may result in an infinite set, while a finite intersection will always result in a finite set.

3. What is the significance of studying infinite intersections of infinite sets?

The study of infinite intersections of infinite sets is important in mathematics as it allows for the analysis of complex and infinite systems. It also has applications in fields such as topology, set theory, and calculus.

4. Can any infinite set be intersected with any other infinite set?

No, not all infinite sets can be intersected. For an infinite intersection to be possible, the sets being intersected must have at least one element in common. If there are no elements in common, the resulting intersection will be an empty set.

5. Are there any real-world applications of infinite intersections of infinite sets?

Yes, there are real-world applications of infinite intersections of infinite sets. For example, in probability theory, infinite intersections are used to calculate the likelihood of events occurring in systems with an infinite number of components. They are also used in computer science to analyze the runtime complexity of algorithms.

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