# Prove ...is a well-order

• Oct 14th 2009, 10:56 AM
tcRom
Prove ...is a well-order
Given P={1,2,3,...} with usual < ordering
Let f(n)= the number of distinct prime divisors of n
(ie: f(1)=0, f(2)=1, f(3)=1, f(4)=1, f(5)=1, f(6)=2,...)

Now, define \$\displaystyle \triangleleft\$ on P where:
a \$\displaystyle \triangleleft\$ b iff f(a) < f(b) OR [f(a) = f(b) and a < b]

So, I need to prove that P with \$\displaystyle \triangleleft\$ is a well-ordering; prove transitive, irreflexive, comparable, and that every non-empty subset has a least element. The last part here is my problem, everything else is fine but I need help proving the least element part.

I know that if you break up P into sets so that all numbers with 1 unique prime divisor goes into a subset, then all the numbers with 2 unique prime divisors go into a disjoint subset, and so on, where each subset is ordered as usual by <. Then, take the union of these subsets and it can be shown that it still holds the usual ordering of <, but we are now dealing with \$\displaystyle \triangleleft\$ ordering so it is a well-ordering (since < is a well ordering). As a restatement of my issue, I need help writing this out and actually making it solid. I've been working on it for a while...

Cheers and thank you!
• Oct 14th 2009, 11:13 AM
clic-clac
You can prove \$\displaystyle \triangleleft\$ is a well ordering using that \$\displaystyle <\$ is a well ordering.

Let \$\displaystyle X\$ be a non empty subset of \$\displaystyle P.\$ Then \$\displaystyle f(X)\$ is a non empty subset of \$\displaystyle P,\$ which has a least element \$\displaystyle b\$ for \$\displaystyle <.\$

Consider \$\displaystyle X\cap f^{-1}(b)\subseteq P,\$ which is also a non empty subset (why?) of \$\displaystyle P.\$ Once again, it has a least element for \$\displaystyle <,\$ let say \$\displaystyle a.\$

Prove that \$\displaystyle a\$ is the least element for \$\displaystyle \triangleleft\$ in \$\displaystyle X.\$