MHB How have we concluded that X is an inductive set?

[evinda]

Hello! (Nerd)

I am looking at the proof of the following sentence:

For any natural numbers $m,n$ it holds that:

$$n \in m \rightarrow n \subset m$$

Proof:

We define the set $X=\{ n \in \omega: \forall m (m \in n \rightarrow m \subset n)$ and it suffices to show that $X$ is an inductive set.
Then, $\varnothing \in X$.
Let $n \in X$.
We want to show that $n'=n \cup \{ n \} \in X$.
We pick a $m \in n'=n \cup \{ n \}$.
Then $m \in n$ or $m \in \{ n \}$.

• If $m \in n$ and since $n \in X$, we have that $m \subset n$
• If $m \in \{ n \} \rightarrow m=n \rightarrow m \subset n$

So, $n' \in X$ and therefore $X$ is an inductive set.I haven't understood why, having shown that $m \subset n$, we have concluded that $n' \in X$. (Worried)
Could you explain it to me? (Thinking)

 

[evinda]

In order to show that $n' \in X$ don't we have to show that $m \in n' \rightarrow m \subset n'$ ? Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

In both cases in the proof, we have $m\subseteq n$. This trivially implies $m\subseteq n'$.

 

[evinda]

In both cases in the proof, we have $m\subseteq n$. This trivially implies $m\subseteq n'$.

I see (Nod) Thank you very much! (Happy)