I am reading a course in mathematical analysis vol 1 by J.H. Garling.

He defines a successor set as one that (1) contains $\emptyset$, and (2) contains $a^+$ whenever it contains $a$ (where $a^+$ is defined as the set $a\cup \{a\}$. He then states as a Theorem that there exists a successor set $Z^+$ such that any successor set $T$ must contain $Z^+$. Unfortunately I cannot understand the proof which is as follows:

"Note that if $A$ is a set, all of whose elements are successor sets, then it follows immediately from the definitions that the intersection of all elements of $A$ is also a successor set. Suppose that $S$ is a successor set. Let $Z^+=\cap\{B\in P(S):B\text{ is a successor set}\}$. Then if $T$ is a successor set, $T\cap S$ is a successor set, so $Z^+\subseteq T\cap S\subseteq T$."

In particular, I do not see why it is obvious that the intersection of successor sets is always a successor set, I see why that is the case if the intersection is non-empty, but i can't see why it couldn't be the empty set itself.

  • 2
    $\begingroup$ There are some sources that seem to use the term "successor set" about what is usually called "inductive sets", namely a set that contains $0$ (or $1$, depending on the author) and contains the successor of each of its elements. But with that definition, $0$ is a member of every successor set, and therefore also of any intersection of those sets. $\endgroup$ – Henning Makholm Jun 25 '16 at 16:03
  • $\begingroup$ So, if the intersection is empty, how can it be itself an inductive set? $\endgroup$ – Michele Galli Jun 25 '16 at 16:32
  • $\begingroup$ If the intersection would be empty then it would not be an inductive set. But as @Henning points out: If $A\neq\varnothing$ and all elements of $A$ are inductive sets then the intersection $\cap\{B\mid B\in A\}$ is not empty since it will contain element $0$. $\endgroup$ – drhab Jun 25 '16 at 18:14
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    $\begingroup$ @MicheleGalli Are you happy with those edits? By the way, I have deleted most of my comments as no longer needed - too many comments tend to be offputting. A common complaint is that things are put in comments that should be in the question itself. $\endgroup$ – almagest Jun 26 '16 at 20:43
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    $\begingroup$ $\emptyset^+ = \emptyset \cup \{\emptyset\} = \{\emptyset\}$, and $\{\emptyset\} \neq \emptyset$, because $\emptyset \in \{\emptyset\}$. $\endgroup$ – Daniel Fischer Jun 27 '16 at 21:57

Especially the first part of this answers your question. I read in the comments that there were more difficulties so decided to give a more complete answer.

Let $S$ be a successor-set.

If $\mathcal A:=\{B\in\wp(S)\mid B\text{ is a successor set}\}$ then $S\in\mathcal A$.

This guarantees that $\cap\mathcal A$ is a well defined subset of $S$.

Secondly every element of $\mathcal A$ is a successor set so that $\varnothing\in B$ is true for every $B\in\mathcal A$.

Consequently $\varnothing\in\cap\mathcal A$ showing that the intersection cannot be empty.

Let $\omega:=\cap\mathcal A$. We will prove that $\omega$ is a successor set.

As stated above we have $\varnothing\in\omega$. If $a\in\omega$ then for every $B\in\mathcal A$ we have $a\in B$. Every $B\in\mathcal A$ is a successor set, so we conclude that $a^+\in B$ for every $B\in\mathcal A$. That means exactly that $a^+\in\omega$ and proved is now that $\omega$ is a successor set.

Here a proof that $\omega\subseteq T$ is true for every $T$ that is a successor set.

Let $T$ be a successor set. Then $T\cap S\in\wp(S)$ is a successor set so that $T\cap S\in\mathcal A$.

Then $\omega\subseteq T\cap S\subseteq T$.

Final remark:

Essential is here the existence of a successor set $S$. Without that the reasoning could not have been made. The statement that a successor set exists is the so-called axiom of infinity.

  • $\begingroup$ I am still struggling to find out how MSE is meant to work, so I was really pleased with this outcome. Many thanks (I have already upvoted once, so can't do any more there!). $\endgroup$ – almagest Jun 29 '16 at 12:15

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