Basic Set Theory (Indexed Collection of Sets)

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SUMMARY

The discussion centers on constructing an indexed collection of sets {A_{\alpha} : \alpha\in\Delta} where each A_{\alpha} is defined as A_{\alpha}=(0, \frac{1}{\alpha}) for \alpha in the natural numbers. It is established that while A_{\alpha} and A_{\beta} intersect for any alpha and beta, the overall intersection of all A_{\alpha} results in an empty set, as the limit of 1/n approaches 0. This highlights the properties of strictly decreasing chains of open sets and clarifies misconceptions regarding infinite sets.

PREREQUISITES
  • Understanding of indexed collections of sets
  • Familiarity with the concept of intersections of sets
  • Knowledge of limits and their properties in calculus
  • Basic principles of set theory, particularly regarding open sets
NEXT STEPS
  • Study the properties of open sets in topology
  • Explore the concept of limits in calculus, focusing on sequences
  • Learn about the intersection of infinite sets and their implications
  • Investigate strictly decreasing sequences and their behavior in set theory
USEFUL FOR

Students of mathematics, particularly those studying set theory and calculus, as well as educators seeking to clarify concepts related to indexed collections and intersections of sets.

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Homework Statement


Give an example of an indexed collection of sets {A_{\alpha} : \alpha\in\Delta} such that each A_{\alpha}\subseteq(0,1) , and for all \alpha and \beta\in\Delta, A_{\alpha}\cap A_{\beta}\neq \emptyset but \bigcap_{\alpha\in\Delta}A_{\alpha} = \emptyset.

Homework Equations


None.

The Attempt at a Solution


I've found a solution that is:
Let A_{\alpha}=(0, \frac{1}{\alpha}), where \alpha\in\Delta=\mathbb{N}

and my main problem is that I don't understand how this is possible.

I understand that A_{\alpha}\cap A_{\beta}\neq \emptyset for any alpha/beta because the intersection will always be (0, 1/max(alpha,beta)).

But I don't understand how \bigcap_{\alpha\in\Delta}A_{\alpha} = \emptyset is true. Wouldn't every set in the family have the smallest real in it, because every set would be (0, a)? I feel like I'm probably not thinking about this the right way.

Thanks for any help!
 
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By definition, if x is in the intersection then x is in (0,1/n) for every n in N. Equivalently, 0<x<1/n. As n-> infty, 1/n -> 0. Using the N-epsilon definition of limits we know that if we let epsilon=x then there is a natural N such that 1/N<x. Therefore, x is not in A_N. Contradiction.

The intersection of any finite number of A_\alpha will be (0,min{1/n_1,1/n_2,...,1/n_k}) but if we want to to intersect all of the sets, {1,1/2,1/3,1/4,...} has no minimum.

Essentially we are probing at the properties of strictly decreasing chains of open sets.
 
Thanks for the help.

When I was working on this, I initially figured that the intersection over all of the sets is empty like you proved (because n-> infinity and 1/n -> 0).

But, if you were to choose that set "at infinity" (which makes \bigcap empty) as your A_\alpha, then wouldn't its intersection with any A_\beta be empty too?

Can you explain to me why this is wrong? I understand how this works for any finite amount of sets, but my intuitive understanding of infinite sets feels quite weak.
 
I don't understand your problem. Maybe you are thinking of this in terms of limits?

\mbox{lim}_{k\to\infty}\cap_{n=1}^k\left(0,\frac{1}{n}\right)=\mbox{lim}_{k\to\infty}\left(0,\frac{1}{k}\right)=(0,0)=\emptyset

Severe abuse of notation.
 
I think it's just my grasp of infinite sets that is weak.

I suppose I need to do more work with families of sets.

Thanks again.
 

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