Let $X$ be a set, and let $\Omega$ be the space of all pairs $(Y,\leq)$,where $Y$ is a subset of $X$ and $\leq$ is a well ordering of $Y$.If $(Y,\leq)$,$(Y',\leq')$ are elements of of $\Omega$,we say that $(Y,\leq)$ is an initial segment of $(Y',\leq')$ if there exists an $ x\in Y'$ such that $ Y:=\{y\in Y':y<'x\}$(so in particular $ Y\subsetneq Y'$）,and for any $ y,y'\in Y$,$ y\leq y'$ if and only if $ y\leq'y'$.Define a relation $ \preceq$ on $ \Omega$ by defining $ (Y,\leq)\preceq (Y',\leq')$ if either $ (Y,\leq)=(Y',\leq')$ or $ (Y,\leq)$ is an initial segment of $ (Y',\leq')$.Show that $ \preceq$ is a partial ordering of $ \Omega$.
Proof:1.Reflexivity:$\forall (A,\leq)\in \Omega$,$(A,\leq)=(A,\leq)$,so $(A,\leq)\preceq(A,\leq)$.2.Anti-symmetry:If $(A,\leq)\preceq(B,\leq')$,and $(B,\leq')\preceq(A,\leq)$,then if $(B,\leq')=(A,\leq)$,done.Otherwise,$(A,\leq)$ is an initial segment of $(B,\leq')$,and $(B,\leq')$ is an initial segment of $(A,\leq)$,this contradicts So $(A,\leq)=(B,\leq')$.3.Transitivity:If $(A,\leq)\preceq(B,\leq')$,and $(B,\leq')\preceq (C,\leq'')$,then $(A,\leq)\preceq (C,\leq'')$.This is because when $(A,\leq)\preceq(B,\leq')$,if $(A,\leq)=(B,\leq')$,then of course $(A,\leq)\preceq (C,\leq'')$.If $(A,\leq)$ is an initial segment of $(B,\leq')$,then $A=\{y\in B:y<'x\}$,because $(B,\leq')\preceq (C,\leq'')$,if $(B,\leq')=(C,\leq'')$,then of course $(A,\leq)\preceq (C,\leq'')$.If $(B,\leq')$ is an initial segment of $(C,\leq'')$ which means that $B=\{y\in C:y<''c\}$.Then it is obvious that $(A,\leq)$ is an initial segment of $(C,\leq'')$.$\Box$
There is exactly one minimal element of $\Omega$,what is it?
Proof:
The minimal element of $\Omega$ is $(\emptyset,\leq)$.There is no smaller.Using Zorn's lemma,conclude the well ordering principle:Every set $X$ has at least one well-ordering.Proof:See
Use the well ordering principle to prove the axiom of choice.Proof:From the well ordering principle we know that $I$ can be well ordered ,then by using ,we can easily conclude the axiom of choice.$\Box$