Can you use induction on n cases (as opposed to infinity)?

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Homework Help Overview

The original poster is exploring the use of mathematical induction in a proof involving the equality of two sets, each defined as a union of Borel sets from 1 to n. They express uncertainty about applying induction in this finite context, as they are accustomed to its use in infinite scenarios.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of using induction for finite cases, with some affirming that it is acceptable. The original poster attempts to establish a base case and an inductive step, while others suggest simplifying the presentation of the proof.

Discussion Status

The discussion is active, with participants providing feedback on the proof's clarity and confirming that using induction on a finite set is valid. There is no explicit consensus on the best approach, but several participants offer constructive comments and suggestions for improvement.

Contextual Notes

Some participants note the importance of clarity in mathematical presentation, particularly regarding the use of LaTeX for better readability. There is also mention of the complexity introduced by certain logical operations in the proof.

bennyska
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Homework Statement


this is probably a dumb question, but I'm doing this proof where i have to show two sets are equal, where each set is a union from 1 to n sets. this is pretty easy to show with induction, i think, but I'm used to using induction when i have an infinite amount of things, so I'm not sure I'm allowed to use induction. any thoughts?

specifically, it goes like this:
Suppose that A_1, ..., An are Borel sets, that is they belong to ß. Define
the following sets: B_1 = A_1, B_n = A_n ∩ (A_1∪ ... ∪ A_n-–1)^c (^c is complement), and let S equal the universal set. Show that

U_i=1 to n A_i = U_i=1 to n B_i.

Homework Equations





The Attempt at a Solution



U_1 to 1 A_i = A_1 = U_1 to 1 B_i = B_1. So we have a base case. So assume it's true for n=k. Then we have that U_i=1 to k A_i = U_i=1 to k B_i.
Then we have that U_i to k B_i U B_k+1 = U_i to k A_i U (A_k+1 ∩ (A_1∪ ... ∪ A_k)^c
=U_i to k A_i U (A_k+1 ∩ A_1^c ∩ A_2^c...∩A_k^c)...
Let A_1^c ∩ A_2^c...∩A_k^c = D, and let U_i to k A_i = E
Then we have U_i to k B_i U B_k+1 = E U (A_k+1 ∩ D)
= (E U D) ∩ (E U A_k+1) = S ∩ (U_i to k A_i U A_k+1) = U_i=1 to k+1 A_i.

god that looks hideous. hopefully it makes sense. any comments would be appreciated.
 
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That is hideous to read. The basic idea is that E U (A_k+1 n E^c)=E U A_k+1. Right? You can certainly use induction on a finite set of premises, no problem with that. It looks ok to me. Cleaning up the presentation certainly wouldn't hurt. Using TeX wouldn't hurt either. But I think you've got one way to do it.
 
The purpose of induction is to show that if a statement is true for some value k, it has to be true for k+1.

It's up to you how far you want to extend your conclusion, so it's perfectly fine to use it on a finite set.
 
alright, sorry i was a bit lazy on the latex, i didn't think it would be that bad originally, and i haven't used latex in a while.

i've attached a pdf. how does that look?
 

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bennyska said:
alright, sorry i was a bit lazy on the latex, i didn't think it would be that bad originally, and i haven't used latex in a while.

i've attached a pdf. how does that look?

Fine. The deMorgan stuff is a little unnecessarily complicated. Just use that K U (L n K^c)=K U L. That's true, right?
 
I'm not sure where the "finiteness" of your question is. Also, to answer your original question, it is perfectly fine to use induction where the variable you are inducting on has finite range. It is a common technique used in real analysis. In fact, the original definition of induction imposes no restriction that the variable has to have infinite (countably infinite) range.
 
awesome, thanks you guys.
 

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