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

1. Nov 4, 2016

### 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:

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 sets

and

$\cup B_i = \emptyset$

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:

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2. Nov 4, 2016

### stevendaryl

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

So if $A$ is a successor set, then $0$ is in $A$, and so is $0^+$, and so is $0^{++}$, etc. So every successor set has an infinite number of elements.

3. Nov 5, 2016

### Math Amateur

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 sets

and we find that ...

$\cap B_i = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ \}$

Is that correct?

Peter

NOTE - 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: Nov 5, 2016
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