In Halmos Naive set theory, there is the following passage (excuse my french) in his section introducing natural numbers :

In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $\omega$.

I have trouble seeing why one should feel the need to state the bolded part in order to derive the existence of $\omega$. Would he not have arrived at the same conclusion had he decided to consider directly the intersection of every inductive set included in A?

Would he not also have done so had he decided to state that the intersection of every inductive set was also an inductive set instead of invoking families of inductive sets?

  • $\begingroup$ The word should be "inductive" rather than "successor". $\endgroup$ – Andrés E. Caicedo Jun 27 '13 at 20:32
  • $\begingroup$ It's me that should have added the sentence preceding these : "We shall say, temporarily, that a set A is a successor set if [...]" $\endgroup$ – TheCoconutChef Jun 27 '13 at 20:35
  • $\begingroup$ The intersection of an inductive set is empty, not an inductive set. When you say "the intersection of every inductive set", you mean which of the following two options?: 1) Consider the class $\mathcal A$ of all inductive sets, and take the intersection of all sets in $\mathcal A$, or 2) Let $A$ be an inductive set, and consider $\bigcap A$, the intersection of $A$ (that is, the intersection of the members of $A$). $\endgroup$ – Andrés E. Caicedo Jun 27 '13 at 20:38
  • $\begingroup$ Anyway, the answer is no: Take any inductive set $A$ strictly larger than $\omega$. Note that $\{A\}$ is non-empty, and its intersection is $A$, which is different from $\omega$. This answers the question in your title, which is not the question in the body of the text. $\endgroup$ – Andrés E. Caicedo Jun 27 '13 at 20:42
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    $\begingroup$ More to the point, most likely Halmos's framework has only defined intersections of set-sized families. But the collection of all inductive sets is not a set, so he cannot directly form its intersection. This answers the last paragraph, if you mean option 1 of my alternatives listed above. If you mean option 2, then you do not even get an inductive set. $\endgroup$ – Andrés E. Caicedo Jun 27 '13 at 20:45

The intersection of all happy sets is always a good candidate for the smallest happy set. However, you do need to show that such an intersection is still happy (otherewise ist just a rather small set, but not the smallest happy set). Also, we run into some logical problems if we want to take the intersection of all happy sets as the happy sets may form a proper class! This can be circumvented by grabbing an arbitrary happy set, taking the intersection of all its happy subsets and showing that the same intersection is obtained if one starts with any other happy set. All this follows if one has that the intersection of any nonempty family of happy sets is happy - and that there exists at least one happy set.

With happy=inductive, you have the situation of your question: The existence of at least one inductive set is the Axiom of Infinity; the intersection property is the bold part questioned by you.

  • $\begingroup$ Thanks, I needed that. $\endgroup$ – user12802 Apr 21 '18 at 16:38

Note that the definition of inductive is this:

  1. $\varnothing\in A$;
  2. $\forall x(x\in A\rightarrow x\cup\{x\}\in A)$.

The first property is definitely preserved by intersections. To see that the second property holds for intersections, if $A_i$ is a set of inductive sets then $x\in A_i$ for all $i$ means that $x\cup\{x\}\in A_i$ for all $i$. Therefore both properties must hold when taking intersections.

Generally speaking one can replace the second condition by any $\forall x\exists y(\ldots)$ where $\ldots$ define $y$ to be some unique set (e.g. $y=x\cup\{x\}$).


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