ZFC .... Axioms of Foundation .... and Infinity ....

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SUMMARY

The discussion centers on Proposition 1.7.5 from D. J. H. Garling's "A Course in Mathematical Analysis: Volume 1: Foundations and Elementary Real Analysis," specifically regarding the implications of the foundation axiom in set theory. The foundation axiom asserts that for any set \( f(\mathbb{Z}^+) \), there exists an element \( b \) such that \( b \cap f(\mathbb{Z}^+) = \varnothing \), leading to the conclusion that there exists \( n \in \mathbb{Z}^+ \) such that \( f(n) = b \). This establishes that infinite regress is not possible, as demonstrated by the impossibility of an infinite sequence of sets where \( f(n^+) \notin f(n) \).

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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume 1: Foundations and Elementary Real Analysis" ... ...

I am at present focused on Part 1: Prologue: The Foundations of Analysis ... Chapter 1: The Axioms of Set Theory ...

I need help with an aspect of the proof of Proposition 1.7.5 ...

Proposition 1.7.5 reads as follows:https://www.physicsforums.com/attachments/7003In the above proof we read the following:

"By the foundation axiom, there exists $$n \in \mathbb{Z}^+$$ such that no member of $$f(n)$$ is in $$f( \mathbb{Z}^+ )$$. ... ... "

Can someone please explain how/why the foundation axiom implies that there exists $$n \in \mathbb{Z}^+$$ such that no member of $$f(n)$$ is in $$f( \mathbb{Z}^+ )$$. ... ... ?

PeterNOTE:

To enable readers to follow the above post I am providing Garling's text on the foundation axiom and the axiom of infinity ... ...View attachment 7004
View attachment 7005
 
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In the foundation axiom, set $f(Z^{+}) = A$. The foundation axiom asserts that there is some member $b\in f(Z^{+})$ such that $b\cap f(Z^{+}) = \varnothing$. But since $b\in f(Z^{+})$, it follows that there exists some $n\in Z^{+}$ such that $f(n)=b$. And there you have the assertion in Garling. Does that answer your question?
 
Ackbach said:
In the foundation axiom, set $f(Z^{+}) = A$. The foundation axiom asserts that there is some member $b\in f(Z^{+})$ such that $b\cap f(Z^{+}) = \varnothing$. But since $b\in f(Z^{+})$, it follows that there exists some $n\in Z^{+}$ such that $f(n)=b$. And there you have the assertion in Garling. Does that answer your question?
Hi Ackbach ... sorry to be slow in answering... had to travel interstate ...

Thanks so much for the help ... very clear answer ... thanks

Peter
 
Peter said:
Hi Ackbach ... sorry to be slow in answering... had to travel interstate ...

Thanks so much for the help ... very clear answer ... thanks

Peter
Hi again, Ackbach ...

Just a further question you may be able to help with ...

Just prior to Proposition 1.7.5 Garling writes:

"Let us use the foundation axiom to show that infinite regress is not allowed."

I do not fully understand how Proposition 1.7.5 demonstrates that infinite regress is not possible ... can you explain ...?

Peter
 
Peter said:
Hi again, Ackbach ...

Just a further question you may be able to help with ...

Just prior to Proposition 1.7.5 Garling writes:

"Let us use the foundation axiom to show that infinite regress is not allowed."

I do not fully understand how Proposition 1.7.5 demonstrates that infinite regress is not possible ... can you explain ...?

Peter

Hmm. Is there some text in Garling that's missing in your scan? If so, could you please supply that? I'm thinking in particular between the statement you quoted and Proposition 1.7.5.
 
Ackbach said:
Hmm. Is there some text in Garling that's missing in your scan? If so, could you please supply that? I'm thinking in particular between the statement you quoted and Proposition 1.7.5.
Hi Ackbach,

The original post showed the complete text from the start of the foundation axiom through (and including) Proposition 1.7.5.

There is no missing text ...

Peter
 
Peter said:
Hi Ackbach,

The original post showed the complete text from the start of the foundation axiom through (and including) Proposition 1.7.5.

There is no missing text ...

Peter

I see. Well, what about after? Maybe the author's not done showing that infinite regress isn't allowed. Is there an indicator sentence anywhere later? Maybe after the next theorem? Or the one after that? I mean by "indicator sentence" something along the lines of, "And this shows that infinite regress is not allowed." In other words, I'm wondering if Theorem 1.7.5 is a lemma, and not the "final theorem" demonstrating that infinite regress is not allowed. I would agree with you that it's not obvious to me that Theorem 1.7.5 does the trick.
 
Ackbach said:
I see. Well, what about after? Maybe the author's not done showing that infinite regress isn't allowed. Is there an indicator sentence anywhere later? Maybe after the next theorem? Or the one after that? I mean by "indicator sentence" something along the lines of, "And this shows that infinite regress is not allowed." In other words, I'm wondering if Theorem 1.7.5 is a lemma, and not the "final theorem" demonstrating that infinite regress is not allowed. I would agree with you that it's not obvious to me that Theorem 1.7.5 does the trick.
Thanks for that Ackbach ...

I have concluded that Garling (a bit like Rudin) can be very terse and cryptic...

But ... thanks again for your support and help ...

Peter
 
Infinite regress would mean the existence of an infinite sequence $a_1\ni a_2\ni a_2\ni\dots$ of sets. Since $Z^+$ can be considered as the set of natural numbers (which is explained after Proposition 1.7.5), such sequence can be viewed as a function from $Z^+$, namely, $a_n=f(n)$, where $n$ in the right-hand side is $\emptyset$ with $n$ pluses. The proposition says that such sequence is impossible: there exists an $n\in Z^+$ for which $f(n^+)\notin f(n)$.
 
  • #10
Evgeny.Makarov said:
Infinite regress would mean the existence of an infinite sequence $a_1\ni a_2\ni a_2\ni\dots$ of sets. Since $Z^+$ can be considered as the set of natural numbers (which is explained after Proposition 1.7.5), such sequence can be viewed as a function from $Z^+$, namely, $a_n=f(n)$, where $n$ in the right-hand side is $\emptyset$ with $n$ pluses. The proposition says that such sequence is impossible: there exists an $n\in Z^+$ for which $f(n^+)\notin f(n)$.
Thanks for the explanation/clarification Evgeny ...

I appreciate your help and support ...

Peter
 

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