In Munkres' Topology the proof for the Well-Ordering Property is stated as follows:
- I'm having confusion with the first and second underlined part: "Let A be the set of all positive integers n for which the statement holds." Isn't this the very statement that we are trying to show is true in this subproof ? Can you please restate this in a different way because I feel as though it might be ambigious??
- When he says "this set" in the last underlined part does he mean the set $C \cap \{1,..., n \}$ or the set $A$?
- And I'm sorry to put it so generally but what is the key idea behind the subproof; is it to show that intersection between the set of the first $n$ integers and any subset of it has a least element? I don't see how such an intersection should imply the existence of a least element.
Thanks in advance guys, I hate asking to clarify so many little points but this one just ain't clicking for me.
No, this statement is merely a definition. The symbol $A$ is being defined to refer to the set of $n$ such that the statement holds. We don't yet know that the statement holds for all $n$, and we're just collecting all the $n$ such that it does hold into a set and calling that set $A$. We're not asserting that the statement ever actually does hold; maybe $A$ is empty.
The role of $A$ is that we now want to prove $A=\mathbb{Z}_+$. That means that every positive integer is an element of $A$, or in other words that the statement holds for every positive integer $n$.
He means $C \cap \{1,..., n \}$. The statement $n\in A$ by definition means that any nonempty subset of $\{1,\dots,n\}$ has a least element. Since $C\cap \{1,\dots,n\}$ is a nonempty subset of $\{1,\dots,n\}$, it therefore has a least element.
I'm not quite sure what you're asking here. The overall goal is to prove that if $D$ is any nonempty subset of $\mathbb{Z}_+$, then it has a least element. We do this using the following observation: if $n\in D$, then the least element of $D$ must be less than or equal to $n$. So if we find the least element $k$ of $D\cap \{1,\dots,n\}$, then $k$ is actually the least element of all of $D$. Indeed, for any $d\in D$, if $d\leq n$, then $k\leq d$ since $d\in D\cap\{1,\dots,n\}$ and $k$ is the least element of $D\cap\{1,\dots,n\}$. And if $d>n$, then obviously $k\leq d$ since $k\leq n$.
So using this idea, we only need to prove that $D\cap \{1,\dots,n\}$ has a least element. Now we rename $D\cap\{1,\dots,n\}$ to $C$ and think of $C$ as just some arbitrary nonempty subset of $\{1,\dots,n\}$. So what we want to prove now is that any nonempty subset of $\{1,\dots,n\}$ has a least element. This is what the subproof is proving, and it does so by induction on $n$.