Axiom of Infinity & Garling, Th. 1.7.4 & the successor set

In summary, Garling discusses the concept of a successor set, defined as a set that contains the empty set and its successors. He then introduces the notation ##a^+## to denote the successor of a set ##a##. The set ##Z^+## is defined as the intersection of all sets ##B_i## in the power set ##P(S)## that are successor sets. In an example, it is shown that ##Z^+## contains infinitely many elements. The reader asks for clarification on the construction of ##Z^+##, providing their own example. It is clarified that the example is incorrect and a new example is given. The reader also notes that Garling's "successor set" is often called
  • #1
Math Amateur
Gold Member
MHB
3,990
48
I am reading D. J. H. Garling: "A Course in Mathematical Analysis: Volume I Foundations and Elementary Real Analysis ... ...

I am currently focused on Garling's Section 1.7 The Foundation Axiom and the Axiom of Infinity ... ...

I need some help with Theorem 1.7.4 ... and in particular with the notion of a successor set ##Z^+## ... ...... ... the relevant text from Garling is as follows:
?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png

In the above text we read the following:

" ... ... Suppose that ##S## is a successor set. Let

##Z^+ = \cap \{ B \in P(S) : B \text{ is a successor set } \} ## ... "
Note also that Garling defines a successor set as follows:

" ... ... A set ##A## is called a successor set if ##\emptyset \in A## and if ##a^+ \in A## whenever ##a \in A## ... ... "and

Garling defines ##a^+## as follows:

" ... ... If ##a## is a set, we define ##a^+## to be the set ##a \cup \{ a \}## ... ... "
Now my problem is that I do not understand the definition of ##Z^+## ... ... in each example I construct I seem to get ##Z^+ = \emptyset## ... ... and this cannot be right ...For example ...

Suppose that a successor set S is such that:

##S = \{ \emptyset , a , a \cup \{ a \} , \{ a , b \}, \{ a , b \} \cup \{ \{ a , b \} \} \}##

... then we have ...

##Z^+ = \cap B_i ## where ##B_i \in P(S)## and each ##B_i## is a successor set ...

then we have ..

##B_1 = \{ \emptyset, a , a \cup \{ a \} \}##

##B_2 = \{ \emptyset, \{ a , b \} , \{ a , b \} \cup \{ \{ a , b \} \} \}##

and

##B_3 = S##Indeed,

##B_1, B_2, B_3## seem to me to be the only subsets of ##P(S)## that are successor setsand##\cup B_i = \emptyset##BUT ... surely this cannot be right ...Can someone clarify this issue and show me how ##Z^+## is meant to be constructed ...

Hope someone can help ...

Peter====================================================

In order to enable readers to get a better understanding of Garling's notation and approach I am providing the first two pages of Section 1.7 ... as follows:
?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png

?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png
 

Attachments

  • Garling -  Successor Sets.png
    Garling - Successor Sets.png
    91.8 KB · Views: 925
  • Garling - 1 - Section 1.7 - Foundation Axiom and Axiom of Infinity - Part 1 ... ... .png
    Garling - 1 - Section 1.7 - Foundation Axiom and Axiom of Infinity - Part 1 ... ... .png
    35.1 KB · Views: 905
  • Garling - 2 - Section 1.7 - Foundation Axiom and Axiom of Infinity - Part 2.png
    Garling - 2 - Section 1.7 - Foundation Axiom and Axiom of Infinity - Part 2.png
    39.6 KB · Views: 732
Physics news on Phys.org
  • #2
Math Amateur said:
##Z^+ = \cap B_i ## where ##B_i \in P(S)## and each ##B_i## is a successor set ...

then we have ..

##B_1 = \{ \emptyset, a , a \cup \{ a \} \}##

[itex]B_1[/itex] is not a successor set. Maybe you are confused by the use of the variable [itex]a[/itex] in the definition of successor set. The definition means "for any set [itex]a[/itex], if [itex]a[/itex] is in [itex]A[/itex], then [itex]a^+[/itex] is in [itex]A[/itex]".

So if [itex]A[/itex] is a successor set, then [itex]0[/itex] is in [itex]A[/itex], and so is [itex]0^+[/itex], and so is [itex]0^{++}[/itex], etc. So every successor set has an infinite number of elements.
 
  • Like
Likes Math Amateur
  • #3
Thanks Steven ... Based on what you have said, I'll now try to reformulate my example ...

I am assuming ##S## should be as follows ... ...

##S = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ , a, a^+, a^{++}, a^{+++}, \ ... \ ... \ ... \ , \{ a , b \}, \{ a , b \}^+, \{ a , b \}^{++}, \{ a , b \}^{+++}, \ ... \ ... \ ... \ \} ##Then Garling defines ##Z^+## as follows:##Z^+ = \cap B_i## where ##B_i \in P(S)## and each ##B_i## is a successor set ...... ... then we find ... ...##B_1 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ \}####B_2 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ a, a^+, a^{++}, a^{+++}, \ ... \ ... \ ... \ \}####B_3 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ , \{ a , b \}, \{ a , b \}^+, \{ a , b \}^{++}, \{ a , b \}^{+++}, \ ... \ ... \ ... \ \}##

and

##B_4 = S##... so ... ...##B_1, B_2, B_3, B_4 ## now seem to me to be the only subsets of ##P(S)## that are successor setsand we find that ...##\cap B_i = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ \}##Is that correct?

PeterNOTE - just by the way, I think that what Garling is calling a "successor set" is often called an "inductive set" ... can you confirm that this is the case?
 
Last edited:

1. What is the Axiom of Infinity?

The Axiom of Infinity is one of the axioms of Zermelo-Fraenkel set theory, a fundamental branch of mathematics. It states that there exists an infinite set, meaning a set with an infinite number of elements. This axiom is important in constructing mathematical theories and proofs.

2. What is the Garling Theorem 1.7.4?

Garling Theorem 1.7.4 is a theorem in mathematical logic that is closely related to the Axiom of Infinity. It states that if a set contains an infinite number of elements, then it also contains a successor set, which is a set that contains the next element after each element in the original set.

3. How does the Axiom of Infinity relate to the concept of infinity?

The Axiom of Infinity is one of the fundamental principles used to define the concept of infinity in mathematics. It provides a way to formally state the existence of infinite sets, which are crucial in many mathematical theories and proofs.

4. Why is the Axiom of Infinity important in mathematics?

The Axiom of Infinity is important because it allows for the construction of mathematical theories and proofs involving infinite sets. Without this axiom, many important mathematical concepts, such as the natural numbers and real numbers, would not be well-defined.

5. Can the Axiom of Infinity be proven?

No, the Axiom of Infinity, like other axioms in mathematics, cannot be proven. It is accepted as a basic assumption in Zermelo-Fraenkel set theory and is consistent with other axioms and theorems in mathematics. However, it is possible to prove statements using the Axiom of Infinity as a starting point.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
2
Replies
62
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
665
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Topology and Analysis
Replies
2
Views
1K
Back
Top