The problem depends on what your previous definition of 'finite' is.
And it would help to know what previous theorems have been established.
What book are you working from? Suppes maybe? What exercise number?
I decided to go for .Show that the following are equivalent:
a). is finite
b). admits a well order <whose inverse> is also a well-order.
c). The structure is well-founded: that is every has a -minimal element.
Assign each element of an ordinal number. Since is finite, and there are an infinite number of ordinals, we will always be able to do this. We can then order the ordinals (and the elements of they represent) in the following way:
ie. , ,
This is a partial order and since every element contains , every non-empty subset contains a least element.
If we take any , contains sets of elements of .
Since is well-ordered, each of the subsets of has a least element.
We can take all these least elements and make a new set.
We can repeat this argument until we get a singleton set (the number of steps required to do this is equal to ).
This singleton set is the -minimal element.
For I run into some trouble.
I figured the best way was to use a proof by contradiction. So suppose is infinite.
Let be any subset of and let be the -minimal element of .
I think I need to show that is finite which is equivalent to a finite number of . Are there any tips for this?