This is a textbook problem.
Here's my proof
Proof: By W.O.P
let C be a set that contains real numbers and doesn't contain any minimum number.
Assume C to be Non Empty for purpose of contradiction.$-(a)$
Assume m $\in$ C, be the smallest number that belong in C. $-(b)$
Therefore there exists following 2 cases:
(Since C doesn't have any smallest number.)
$i)$other number smaller than m exists in C.
Since from $(b)$ m is the smallest number that $\in$ C. Thus this case is False.
$ii)$there exists other element as m.
As per the definition of sets a set contains only non duplicate value.Thus this case is also false
Since the result of above cases result in False.we can conclude that our assumption ($a$) is proven false since assumption ($b$) is proven true from above results.
Thus there exists no non empty set which doesn't contain a minimum element.
I believe that the problem is wrongly proved but having proved it myself i am having trouble in finding the mistake.
Help me find the mistake which i have committed in this proof.
PS. (by cases I mean the cases required to prove C to be finite and not containing a minimum number or cases to prove assumption b false)
Best Answer
Let's start with the same assumption you made " there is no minimum element in $C$ " . But , because of well ordered property in $C$ ,every element is comparable in your non empty subset $C$ .
so take any two arbitrary element from $C$ and then they must be comparable. But because of finiteness property of $C$ , you will find this process should end at a particular point. That will be contradiction