Is the Intersection of a Family of Sets Equal to the First Set in the Family?

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SUMMARY

The discussion centers on proving that the intersection of a family of sets, denoted as \(\bigcap_{n=1}^{\infty} A_n\), is equal to the first set \(A_1\) under the condition that \(A_n \subseteq A_{n+1}\) for all \(n \geq 1\). The proof utilizes mathematical induction, starting with the base case \(A_1 \subseteq A_1\) and assuming \(A_1 \subseteq A_n\) holds for all \(n\). The conclusion is that since every \(A_n\) contains \(A_1\), any element in the intersection must also belong to \(A_1\), thus confirming \(\bigcap_{n=1}^{\infty} A_n = A_1\).

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Homework Statement


Let [tex]\left\{A_n | n \in N\right\}[/tex] be a family of sets satisfying [tex]A_n \subseteq A_{n+1}[/tex] for all n >= 1.

(a) Write a proof by mathematical induction that [tex]A_1\subseteq A_n[/tex] for all n.

(b) Use part a to prove that [tex]\bigcap[/tex] from n=1 to infinity of [tex]A_n = A_1[/tex]

The Attempt at a Solution


(i) [tex]A_1\subseteq A_1[/tex] by some theorem in my book. Any set is a subset of itself.
(ii) Assume [tex]A_1\subseteq A_n[/tex] for all n >= 1
Then we know that[tex]A_n\subseteq A_{n+1}[/tex] by the given description of the family of sets.
Then [tex]A_1\subseteq A_n[/tex] is true by inductive hypothesis, therefore [tex]A_1\subseteq A_{n+1}[/tex] for all n>= 1 by induction.

For part b:

I think it seems very obvious but I'm kind of burned out from working the first one. So I have so far just written down that since [tex]A_1\subseteq A_{n+1}[/tex], then the family of sets from n=1 to infinity include A_1, thus the intersection from said limits of A_n = A_1

But I'm sure there must be some formalism I'm not catching.
 
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If there's an element in the interesection it must also be in A1, cause the intersection contains all the points in this sequence of sets which are common to all the sets.

Now if there's an element in A1, because A1 is a subset of every An with n>1, this element must be in the intersection as well.

You obviously can formalise this, but that's the gist of this question.
 

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