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 \(\displaystyle Z^+\) ... ... ... ... the relevant text from Garling is as follows:
View attachment 6153
In the above text we read the following:

" ... ... Suppose that \(\displaystyle S\) is a successor set. Let

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

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

Garling defines \(\displaystyle a^+\) as follows:

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

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

\(\displaystyle B_2 = \{ \emptyset, \{ a , b \} , \{ a , b \} \cup \{ \{ a , b \} \} \}\)

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

\(\displaystyle S = \{ \emptyset , a , a \cup \{ a \} , \{ a , b \}, \{ a , b \} \cup \{ \{ a , b \} \} \}\)

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 ... ...

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

and

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

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]

\(\displaystyle B_1, B_2, B_3, B_4 \) now seem to me to be the only subsets of \(\displaystyle P(S)\) 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