How Do You Prove Subset Relationships Within Intersecting Indexed Sets?

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SUMMARY

The discussion centers on proving subset relationships within indexed sets, specifically that the intersection of sets Ai (for all i in I = {1, 2, 3, ... n}) is a subset of A1 whenever i <= j. The user successfully demonstrates that if x is an element of the intersection of Ai, then x must also be in A1, as A1 is contained in all Ai. Additionally, they establish that A1 is a subset of the intersection of Ai by showing that any element of A1 is also an element of each Aj for all j >= 1. The solution is confirmed to be correct, with a suggestion to utilize induction for further proof.

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  • Familiarity with indexed sets and their properties
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  • Basic proficiency in logical reasoning and proof construction
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  • Study the principles of set theory, focusing on intersections and subsets
  • Learn about indexed sets and their applications in mathematics
  • Explore mathematical induction and its use in proofs
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Homework Statement


Show that the intersection of Ai (for all i in I = {1, 2, 3, ... n } = A1. Ai is a subset of Aj whenever i <= j.

Homework Equations


The Attempt at a Solution


Show:
***I'm having trouble showing part 1***1. that the intersection of Ai is a subset of A1, and
2. A1 is a subset of the intersection of Ai.

This is my attempt: 1. Let x be an element of the intersection of Ai. Then x is in Ai for all i in I. Since A1 is contained in all Ai, then x is contained in A1.

2. Let x be an element of A1, then as A1 is a subset of Aj, for all j >= 1, x is an element of Aj. Thus, x is an element of the intersection of Ai.
 
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I think that's completely correct. Except maybe that you don't need A1 is a subset of Ai for the first part. If x is in the intersection of the Ai, it's certainly in A1.
 
"This is my attempt: 1. Let x be an element of the intersection of Ai. Then x is in Ai for all i in I. [STRIKE]Since A1 is contained in all Ai, then x is contained in A1.[/STRIKE]"

Since x ∈ Ai for all i ∈ I, then clearly, x ∈ A1, because 1 ∈ I .

(Not that what you had was incorrect, but I think this is more direct.)

You could do (2.) by induction.
 

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