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 $I$.)
Best Answer
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.