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

In summary: However, the proposition does not say what happens if we try to take the limit as $n\rightarrow\infty$.
  • #1
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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 \(\displaystyle n \in \mathbb{Z}^+\) such that no member of \(\displaystyle f(n)\) is in \(\displaystyle f( \mathbb{Z}^+ )\). ... ... "

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

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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  • #2
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?
 
  • #3
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
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
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.
 
  • #8
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
 
  • #9
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
 

FAQ: ZFC .... Axioms of Foundation .... and Infinity ....

1. What is ZFC and why is it important in mathematics?

ZFC stands for Zermelo-Fraenkel set theory with the Axiom of Choice. It is an axiomatic system used as the foundation of modern mathematics. It provides a rigorous and consistent framework for defining and manipulating mathematical objects, such as sets, numbers, and functions.

2. What are the Axioms of Foundation in ZFC?

The Axioms of Foundation, also known as the Axiom of Regularity, state that every non-empty set contains an element that is disjoint from it. This ensures that sets do not contain any "loops" or self-referential elements, which could lead to logical contradictions.

3. How does ZFC handle the concept of infinity?

ZFC includes the Axiom of Infinity, which states that there exists an infinite set. This allows for the existence of an infinite number of mathematical objects, such as the set of natural numbers.

4. What are some criticisms of ZFC?

Some mathematicians argue that ZFC is too limited and does not capture all of mathematics. For example, ZFC cannot prove the existence of large cardinals, which are infinite sets with properties beyond those of standard infinite sets.

5. Are the Axioms of ZFC necessary for all of mathematics?

There is ongoing debate about whether the Axioms of ZFC are necessary for all of mathematics. While some mathematicians believe that ZFC provides a complete and consistent foundation, others argue that there may be alternative axiomatic systems that could serve as a foundation for different branches of mathematics.

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