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Definition. A set $I$ is called inductive if $$(\emptyset\in I)\wedge\forall x\in I(x\cup\{x\}\in I)$$ This concept appears in the Axiom of Infinity (in ZF), which claims such a set exists. According to this definition, I think if a set is inductive, then it must contain all the finite ordinals $$\emptyset,\quad\{\emptyset\},\quad\{\emptyset,\{\emptyset\}\},\quad\cdots$$ and it should can be proved by induction. In other words, $$\forall I((\emptyset\in I)\wedge\forall x\in I(x\cup\{x\}\in I))\quad\to\quad\omega\subset I$$ Is my thought correct or somewhere wrong? Or is this corollary meaningless?

Also a question on the definition: In an inductive set, except the infinite "chain" that starts with $\emptyset$, there may exist chains that start with other elements as well. For instance, let $I$ be an inductive set and $$\{\{\emptyset\}\}\ \in\ I$$ then a new chain with the "root" $\{\{\emptyset\}\}$ belongs to $I$, $$\{\{\emptyset\}\},\quad\{\{\emptyset\},\{\{\emptyset\}\}\},\quad\{\{\emptyset\},\{\{\emptyset\}\},\{\{\emptyset\},\{\{\emptyset\}\}\}\},\quad\cdots$$ which is disjoint with the chain with the root $\emptyset$. But the definition above specifies the root of one chain in an inductive set necessarily as $\emptyset$. So isn't it more natural to remove the condition $\emptyset\in I$? That is to define a set $I$ as inductive if $$\forall x\in I(x\cup\{x\}\in I)$$ (If we want to ensure there's at least one element in $I$ to be started with, then we can add $\emptyset\neq I$.) This definition avoids that an inductive set always contains all the finite ordinals, which can just be considered as a special case.

We can also define the "set of the roots" of an inductive set as $$R=\{r\in I:\nexists s\in I(s\cup\{s\}=r)\}$$ (The $\in$-minimal elements of the chains in $I$.)

Lutalli
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By definition, $\omega$ is the intersection of all inductive sets. You can show that this is a set in $\sf ZF$ by showing that $$ \omega=\{x\in I:\forall J [J \text{ inductive}\to x\in J]\} $$ where $I$ is any inductive set.

In particular, the implication "$I$ inductive$\implies\omega\subseteq I$" is trivially true.

You could start inductive sets at other "roots", but note that the implication $$ I \text{ inductive}\wedge a\in I \implies \{a\}\in I $$ fails in general. Indeed, all you know is that $a\cup \{a\}\in I$, which coincides with $\{a\}$ only when $a=\emptyset$. So, if you want your inductive sets to have elements of every finite cardinality, you need to either start with the empty set.

On another note, the precise definition of $\omega$ in set-theoretic terms is a foundational concern. As such, it should be as simple as possible, so starting off with the simplest thing (the empty set) seems like the right move.

Reveillark
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  • Oops, about $a\cup{a}\neq{a}$ when $a={{\emptyset}}$ - There I made a typing problem. I have edited the question. – Lutalli Aug 25 '21 at 17:41
  • In the book Set Theory from Thomas Jech $\omega$ is defined as the least nonzero limit ordinal (but indeed it can be proved that it equals to the intersection of all inductive sets). And I was concentrating just on the definition of inductive sets themselves - the statement of Axiom of Infinity can just be kept. It can say, "there exists an inductive set that contains $\emptyset$ as a root." The definition without a specification of the root makes the concept simpler as well... – Lutalli Aug 25 '21 at 17:48
  • Then the intersection of all inductive sets is empty, though – Lutalli Aug 25 '21 at 18:06