# Munkres’ Proof of Well Ordering Property Code Answer

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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.

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??

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\$.

When he says “this set” in the last underlined part does he mean the set \$C cap {1,…, n }\$ or the set \$A\$?

He means \$C cap {1,…, n }\$. The statement \$nin A\$ by definition means that any nonempty subset of \${1,dots,n}\$ has a least element. Since \$Ccap {1,dots,n}\$ is a nonempty subset of \${1,dots,n}\$, it therefore has a least element.

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.

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 \$nin D\$, then the least element of \$D\$ must be less than or equal to \$n\$. So if we find the least element \$k\$ of \$Dcap {1,dots,n}\$, then \$k\$ is actually the least element of all of \$D\$. Indeed, for any \$din D\$, if \$dleq n\$, then \$kleq d\$ since \$din Dcap{1,dots,n}\$ and \$k\$ is the least element of \$Dcap{1,dots,n}\$. And if \$d>n\$, then obviously \$kleq d\$ since \$kleq n\$.

So using this idea, we only need to prove that \$Dcap {1,dots,n}\$ has a least element. Now we rename \$Dcap{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\$.