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

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The discussion focuses on understanding Garling's Theorem 1.7.4 regarding the successor set Z^+. The initial confusion arises from the construction of Z^+ leading to the conclusion that it equals the empty set, which is deemed incorrect. A revised approach to defining the set S is proposed, incorporating elements that represent the successor operation more clearly. Participants confirm that Garling's "successor set" aligns with the concept of an "inductive set," and they clarify that the intersection of all identified successor sets results in a non-empty set containing elements like {∅, ∅^+, ∅^{++}, ...}. The conversation emphasizes the importance of correctly identifying successor sets to accurately construct Z^+.
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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
 
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Peter said:
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 $a$).
 
Evgeny.Makarov said:
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?
 
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Peter said:
$$B_1, B_2, B_3, B_4 $$ 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 $\{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ...\}$.

Peter said:
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.
 
Evgeny.Makarov said:
$P(S)$ 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
 
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