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An ordinal number is a set $alpha$ with the following properties:

(a) If $x,y in alpha,$ then either $xin y$, $yin x$, or $x=y.$

(b)If $yin alpha$ and $xin y$, then $xin alpha$.

Let $alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $alpha$, define $leq$ on $alpha$ by $xleq y$ iff $x=y$ or $xin y$.

Theorem:-
Let $alpha$ is an ordinal number. Then $(alpha,leq)$ is a well ordered set.

Proof:-

I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $alpha$.

Doubt 1. anti-symmetric
$xleq y$ and $yleq x$. Our claim is $x=y.$

Case 1 $xleq y$ and $yleq x$ means $x=y$ and $yin x implies xin x$. Which is a paradox. How do I prove the anti-symmetry?

Case 2 $xleq y$ and $yleq x$ means $x=y$ and $y= x implies x= y$.

Case 3 $xleq y$ and $yleq x$ means $xin y$ and $y= x implies xin x$.

Case 4 $xleq y$ and $yleq x$ means $xin y$ and $yin x implies x= y$.

Also How do I prove there is a minimal elemnt for each subset of $alpha$?

Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.

As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.

Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.

I preassume that the axiom of regularity is accepted in this context.

Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.

Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.

Our conclusion is then that we are dealing with case 2 as was to be shown.

Proved is now that $leq$ is an anti-symmetric relation on $alpha$.

Let $A$ be a non-empty subset of $alpha$.

According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.

So if $xin A$ then we do not have: $xin a$.

But what we do have is: $xin avee x=avee ain x$.

So what remains is: $x=avee ain x$ or equivalently $aleq x$.

This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.