# $m,n\in\mathbb{N}$. $m<n\implies m\subset n$.

I am reading Logic in mathematics and set theory by Kazuyuki Tanaka and Toshio Suzuki.

Lemma 2.5:
For any $$n\in\mathbb{N}$$, each element of the set $$n$$ belongs to $$\mathbb{N}$$.

Definition 2.3:
$$m,n\in\mathbb{N}$$.
$$m if and only if $$m\in n$$.

Theorem 2.6:
$$l,m,n\in\mathbb{N}$$.
If $$l and $$m, then $$l.

Corollary 2.7:
$$m,n\in\mathbb{N}$$.
$$m.

Proof of Corollary 2.7 by the authors:
It is obvious by Lemma 2.5 and Theorem 2.6.

How to use Lemma 2.5 and Theorem 2.6 to prove Corollary 2.7?

The following is my proof:

Let $$m\in\mathbb{N}$$.
If $$n=0$$, then $$m does not hold.
So, $$m holds.

Let $$m,n\in\mathbb{N}$$.
Suppose $$m holds.

If $$m, then this means $$m\in n\cup\{n\}$$.
If $$m\in n$$, then this means $$m.
By inductive hypothesis, $$m\subset n$$.
And $$n\subset n\cup\{n\}$$.
So, $$m\subset n\cup\{n\}$$.
If $$m\in\{n\}$$, then $$m=n$$.
Then, $$m=n\subset n\cup\{n\}$$.
So, by induction, Corollary 2.7 holds.

• What in Definition 2.3 is getting defined? Commented Aug 1 at 13:34

If I'm not wrong, there is no need for induction and you can do it much simpler like this: let m<n, then: $$\ell\in m \iff \ell Then, if $$\ell\in m$$, we have $$\ell and $$m, and $$\ell. So we conclude that $$\ell\in m\implies \ell\in n$$, so $$m\subset n$$.
• I used it because I applied Theorem 2.6 to $\ell,m,n$, so it must be that since $\ell\in m$, by Lemma 2.5 $\ell\in\mathbb N$, if not I cannot apply Theorem 2.6. Commented Aug 1 at 13:36
• You have not shown that $m$ is a proper subset of $n$. This is where you need the induction. If we have the result $n \notin n$, then we can use this to show the proper subset. Otherwise to show $n \notin n$ we need induction. Commented Aug 1 at 20:50
• Yes, I think by the definition $m\notin m$ because it is not true that $m<m$; then we conclude that $m\in n$ which we have because $m<n$ and we are done. Commented Aug 2 at 21:03