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Consider sets $A$, $B$ and $C$. $A$ is being contained in $C$, that is $A \subseteq C$.

Consider a function $f(x)$, such that $f:C \rightarrow C$.

Set $B$ is inductive if:

  1. $A \subseteq B$
  2. $\forall x(x \in B \leftrightarrow f(x) \in B)$

Now, what is the intersection of all inductive sets? It seems to me that a question like that is meaningless, since there can only be one inductive set. In other words, there is only one set that contains $A$ and is closed under $f(x)$. Am I correct?

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2 Answers 2

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No, there can be many inductive sets. For instance, the ordinal $\omega+\omega$ contains the empty set and is closed under successor operation. However the smallest such set is $\omega$ (and this is the intersection of all inductive sets relative to the empty set and the successor operation).

Your intuition for why there’s s unique one is probably just the intuition for why there’s a minimal one.

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  • $\begingroup$ Your answer is too advanced for me. But no worries, I figured it out by myself in simpler way. I just assumed it's too hard and posted it here without properly trying to solve the problem. $\endgroup$
    – Hanlon
    May 14, 2018 at 18:43
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No, because $B$ needs to include $A$ but it can contain other elements too. Starting from $A$ and applying $f(x)$ to get new elements can lead to one inductive set, but if we start from $B$ that contains elements of $A$ but other elements too, by applying $f(x)$ we may get another inductive set.

Intersection of all these inductive sets should give us the smallest inductive set, the one that starts only from $A$.

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