Consider a set . ( can be finite or infinite)
Let be subsets of , such that
1.
2. is finite
Prove that there exits , such that for any ,
I want to prove this in a rigorous way. Here is my attempt
1. Claim there exists an , such = . If this claim is true proving above is trivial.
2. So to prove the claim. Consider any . Thus must belong to some . Let be minimum such . Iterate over all the elements of (say ' ' elements, as is finite). We will have a set . Now claim that . (This claim is easy to prove)
I have two questions here:
1. Is the above proof rigorous enough !
2. How do I prove that exists? More generally if is a finite subset of , prove that exits and is unique?
Please help !
If you consider then the subset of of elements greater than all is non-empty ( belongs to it)
Therefore it has a minimum, and
Indeed, if by definition of so i.e. contradiction.
Hence and forall by definition of i.e.
There are various ways I guess to show that, we can proceed by induction on :
If nothing to show Assume any subset of with elements has a max, consider then has a max, let's say two cases (because the order is total):
either then
or and
Actually the second proof may be better than the first: it only uses the totality of the order, which is a sufficient (and necessary) condition to be able to say that any finite non-empty subset has a maximum.
Thanks very much. I follow it now. I had similar idea but was not sure how to put it. I was trying to repeatedly construct a sub-set by removing the min element (existence of which is guaranteed by well order property) repeatedly till only one element is left and I would then claim that this is the max. But was not able to put it formally. Thanks very much.
So with this - does my proof to the original problem looks convincing enough?