Is this proof correct?

  • Thread starter Raziel2701
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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.
 

Answers and Replies

  • #2
MathematicalPhysicist
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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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