MHB Why does it suffice to show that X is an inductive set?

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To demonstrate that the elements of natural numbers are indeed natural numbers, it suffices to show that the set X, defined as {n ∈ ω: ∀y ∈ n (y ∈ ω)}, is inductive. This is based on the principle that if a subset X of ω contains 0 and is closed under the successor operation, then X must equal ω. Since X is a subset of ω, proving its inductiveness confirms that all elements n in ω satisfy the condition ∀y ∈ n (y ∈ ω). Thus, the conclusion follows that all elements of natural numbers are natural numbers. The argument effectively utilizes the properties of inductive sets to establish this relationship.
evinda
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Hi! (Smile)

We want to show that the elements of the natural numbers are natural numbers, i.e. $(n \in \omega \wedge x \in n) \rightarrow x \in \omega$

Could you explain me why, in order to show this, it suffices to show that $X=\{ n \in \omega: (\forall y \in n)(y \in \omega)\}$ is an inductive set? (Worried)
 
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I am looking again at the proof..

It suffices to show that $X=\{ n \in \omega: (\forall y \in n) (y \in \omega)\}$ is inductive because we know that $X \subset \omega$ and because of the following sentence:

For each subset $X$ of $\omega$ it holds the following:

$$0 \in X \wedge (\forall n (n \in X \rightarrow n' \in X)) \rightarrow X=\omega$$
we conclude that $\{ n \in \omega: (\forall y \in n) (y \in \omega)\}=\omega$ and thus $(\forall y \in n) (y \in \omega)$ holds for all $n \in \omega$, i.e. the elements of natural numbers are natural numbers, right? (Smile)
 

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