Axiom of Infinity and Garling, Theorem 1.7.4 - the successor set Z^+

[Math Amateur]

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:
View attachment 6153
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/construction of $$Z^+$$ ... ... in each example I construct I seem to get $$Z^+ = \emptyset$$ ... ... and this cannot be right ...
For example ...
Suppose that

$$S = \{ \emptyset , a , a \cup \{ a \} , \{ a , b \}, \{ a , b \} \cup \{ \{ a , b \} \} \}$$$$Z^+ = \cap B_i$$ where $$B_i \in P(S)$$ and each $$B_i$$ is a successor set ...... ... then we find ... ...$$B_1 = \{ \emptyset, a , a \cup \{ a \} \}$$

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

$$B_3 = S$$ ... so ... ...$$B_1, B_2, B_3$$ seem to me to be the only subsets of $$P(S)$$ that are successor sets and we find that ...$$\cup B_i = \emptyset $$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:View attachment 6154
View attachment 6155

 

[Evgeny.Makarov]

Suppose that

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

This $S$ is not a successor set because $(a \cup \{ a \})^+\notin S$ (at least, not for al $a$).

 

[Math Amateur]

This $S$ is not a successor set because $(a \cup \{ a \})^+\notin S$ (at least, not for al $a$).

Thanks Evgeny ... 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 sets and 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?

 

[Evgeny.Makarov]

$$B_1, B_2, B_3, B_4 $$ now seem to me to be the only subsets of $$P(S)$$ that are successor sets

$P(S)$ contains infinitely many successor sets. All of them include $B_1$, but elements of the form $a^{+\dots}$ may start with, say, $a^{+++}$ and not with $a$. The intersection of all these sets is indeed $\{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ...\}$.

I think that what Garling is calling a "successor set" is often called an "inductive set" ... can you confirm that this is the case?

Yes.

 

[Math Amateur]

$P(S)$ contains infinitely many successor sets. All of them include $B_1$, but elements of the form $a^{+\dots}$ may start with, say, $a^{+++}$ and not with $a$. The intersection of all these sets is indeed $\{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ...\}$.

Yes.

Thanks so much Evgeny... most helpful indeed!

Peter