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)

 

[blue_raver22]

Sure! Let's break down the proof and see how we can conclude that $X$ is an inductive set.

First, we define the set $X$ as the set of all natural numbers $n$ such that for any natural number $m$, if $m$ is an element of $n$, then $m$ is also a subset of $n$. This definition sets the foundation for our proof.

Next, we show that the empty set, denoted by $\varnothing$, is an element of $X$. This is because the empty set has no elements, so the statement "for any natural number $m$, if $m$ is an element of $\varnothing$, then $m$ is also a subset of $\varnothing$" is automatically true.

Then, we assume that $n$ is an element of $X$. This means that for any natural number $m$, if $m$ is an element of $n$, then $m$ is also a subset of $n$.

Now, we want to show that $n' = n \cup \{n\}$ is also an element of $X$. In order to do this, we need to show that for any natural number $m$, if $m$ is an element of $n'$, then $m$ is also a subset of $n'$.

We have two cases to consider:
1. If $m \in n$, then by our assumption that $n$ is an element of $X$, we know that $m \subset n$.
2. If $m \in \{n\}$, then that means $m$ is equal to $n$. And since we already established in the previous case that $m \subset n$, we can conclude that $m \subset n'$ as well.

Therefore, we have shown that for any natural number $m$, if $m$ is an element of $n'$, then $m$ is also a subset of $n'$. This means that $n'$ satisfies the definition of $X$, and so we can conclude that $n' \in X$.

By repeating this process, we can show that any natural number $n$ is an element of $X$. This is what it means for $X$ to be an inductive set - every natural number is an element of $X$, and for any natural number $n$, $n \in X