I'm not sure if my below proof contains subtle errors that I'm unable to recognize. Please have a check on it!
Furthermore, I found that my proof is long. Please suggest me any simpler or shorter proof. I believe that looking at your proofs help me learn a lot of valuable things.
Thank you so much!
Let $\mathbb N$ is the set of natural numbers by Von Neumann construction (https://www.wikiwand.com/en/Natural_number#/Formal_definitions).
Theorem: $\forall k,n \in\mathbb N:k\in n\iff k\subsetneq n$.
Lemma 1: $\forall k,n \in\mathbb N:k\in n\implies k\subseteq n$.
Let $T=\{n\in\mathbb N\mid \forall k \in\mathbb N:k\in n\implies k\subseteq n\}$. It's clear that $0=\varnothing\in T$ by vacuity. Assume $n\in T$, then $\forall k \in\mathbb N:k\in n\implies k\subseteq n$. For $k\in n+1=n\cup \{n\}$. There are two cases in total.
If $k=n$, then $k\subseteq n\cup\{n\}=n+1$, hence $k\subseteq n+1$.
The only other possibility is that $k\in n$, then $k\subseteq n$ [Since $n\in T$], hence $k\subseteq n+1$.
Either way, $k\in n+1\implies k\subseteq n+1$. Thus $n+1\in T$. By principle of induction, $T=\mathbb N$, or equivalently Lemma 1 is proved.
Lemma 2: $\forall k,n \in\mathbb N:k\in n\implies n\not\subseteq k$.
Let $T=\{n\in\mathbb N\mid\forall k \in\mathbb N:k\in n\implies n\not\subseteq k\}$. It's clear that $0=\varnothing\in T$ by vacuity. Assume $n\in T$, then $\forall k \in\mathbb N:k\in n\implies n\not\subseteq k$. For $k\in n+1=n\cup \{n\}$. There are 2 cases in total.
If $k=n$, then $n+1=n\cup \{n\}\not\subseteq k$ [If $n\cup \{n\}\subseteq k=n$, then $n\in n$, hence $n\not\subseteq n$, which is a contradiction].
The only other possibility is that $k\in n$, then $n\not\subseteq k$ [Since $n\in T$], hence $n\cup \{n\}\not\subseteq k$, or $n+1\not\subseteq k$.
Either way, $k\in n+1\implies n+1\not\subseteq k$. Thus $n+1\in T$. By principle of induction, $T=\mathbb N$, or equivalently Lemma 2 is proved.
As a corollary of Lemma 2, $n\subseteq n\implies n\notin n$.
Lemma 3: $\forall k,n \in\mathbb N:k\in n\implies k\subsetneq n$.
$\forall k,n \in\mathbb N:k\in n\implies k\subseteq n$ [From Lemma 1] and $n\not\subseteq k$ [From Lemma 2], so $k\in n\implies k\subsetneq n$.
Lemma 4: $\forall k,n \in\mathbb N:k\subsetneq n\implies k+1\subseteq n$.
Let $T=\{n\in\mathbb N\mid \forall k \in\mathbb N:k\subsetneq n\implies k+1\subseteq n\}$. It's clear that $0=\varnothing\in T$ by vacuity. Assume $n\in T$, then $\forall k \in\mathbb N:k\subsetneq n\implies k+1\subseteq n$. For $k\subsetneq n+1=n\cup\{n\}$. There are two cases in total.
If $n\in k$, then $n\subsetneq k\subsetneq n\cup\{n\}\subseteq k$. Thus $k\subsetneq k$, which is impossible to happen. So this case is not possible.
If $n\notin k$, then $k\subseteq n$. If $k=n$, then $k+1=n+1$, hence $k+1\subseteq n+1$. If $k\subsetneq n$, then $k+1\subseteq n$ [Since $n\in T$], hence $k+1\subseteq n\cup\{n\}=n+1$.
Either way, $\forall k \in\mathbb N:k\subsetneq n+1\implies k+1\subseteq n+1$. Thus $n+1\in T$. By principle of induction, $T=\mathbb N$, or equivalently Lemma 4 is proved.
Now we use above Lemmas to prove the Theorem. The statement $(\forall k,n \in\mathbb N:k\in n\implies k\subsetneq n)$ is true by Lemma 3. Next, we prove the converse i.e. $\forall k,n \in\mathbb N:k\subsetneq n\implies k\in n$. We have that $k\subsetneq n\implies k+1=k\cup\{k\}\subseteq n$ [By Lemma 4] $\implies\{k\}\subseteq n\implies k\in n$. In short, $k\subsetneq n\implies k\in n$. Thus the converse statement is proved too. Finally, the Theorem is proved. $\Box$
Best Answer
$\Bbb N$(or $\omega$ or $\aleph_0$) is a set of ordinals(transitive set that is stricly well ordered under $\in$) by definition.
So: $k\in n\implies\forall a\in k (a\in n)\implies k\subseteq n$
Because $n$ is an ordinal so is $n+1$, hence, because $\in$ is strictly well ordered on $n+1$ and $n\in n+1$, $n\notin n$.
The other direction: if $a,b$ are ordinals such that $b\subsetneq a$ then take $A=a\setminus b$ and take the least element in that set and call it $c$, clearly $c\in a$, I'll now show that $c=b$:
Given $d\in c$ we have $d\in b$, otherwise $d$ is a smaller than $c$ and in $A$ and we get contradiction, so $c\subseteq b$
Given $d\in b$ then $d\in c$, otherwise we have(from the well ordering) $c\in d$ that implies $c\in b\implies c\notin A$. Hence $b\subseteq c$
And thus $b=c\in a$