Here $Ord\setminus S$ denotes the class we get by remove a set of ordinals from $Ord$ class. I am asking for a formal proof of any proper initial segment of $Ord\setminus S$ is a set.
I have tried as follows:
For any proper initial segment $I_S(\alpha)=\lbrace x|x<\alpha, x\in Ord\setminus S\rbrace$ of $Ord\setminus S$ for some $\alpha\in Ord \setminus S$, Note that $I_S(\alpha)=\lbrace x|x<\alpha, x\in Ord\setminus S\rbrace \subseteq I(\alpha)=\lbrace x|x<\alpha, x\in Ord \rbrace$ , and $I(\alpha)=\lbrace x|x<\alpha, x,\alpha\in Ord \rbrace$ is a initial segment of Ord and hence a set we conclude that the subclass $I_S(\alpha)$ of $I(\alpha)$ is a set.
But latter I noticed that not all the initial segments are in this form. So how should I fix it? Could someone tell me how to deal with that? Thanks!
EDIT:
The definition of initial segment in my text is:
Now it is my attempt:
For any proper initial segment $X$ of $Ord\setminus S$ , by definition of initial segment, for some $x\in Ord$, it is a collection such that $x ∈X$ and $\forall a\in Ord\setminus S, a ≤ x ⇒ a∈X$. As $Ord\setminus S \subseteq Ord$, for that fixed $x$, it is a subcollection of the collection $Y$ such that $x\in Y$ and $\forall a\in Ord, a ≤ x ⇒ a∈Y$. That is, for any initial segment $X$ of $Ord\setminus S, X\subseteq Y$ for some initial segement $Y$ of $Ord$. Thus any proper initial segment of $Ord\setminus S$ is a subset of a set. Thus it is a set.
But now I realize that I have not prove that the proper initial segment must be a subclass of a proper initial segment other then the whole $Ord$ class.Could someone please help here?