Thread: Set proofs

I am struggling to start these two proofs, I hope you can point me in the right direction.

1. Prove by induction on the size of the set that every finite set has a minimum element.

2. S is a set defined by S= { a + b(2)^0.5 } where a and b are rational
If S has an upper bound does a least upper bound exist? Prove the answer.

"Minimum" would indicate the elements of the set have a total ordering.

1. Let A be a set with a total order and be a finite set. If |B| = 1, then the minimum element is the only element in B. Assume for |B| = n contains a minimum element p. Then if an element j and , has a minimum element, since or , so either p or j is the minimum element. Since | | = n+1, the induction on the size of the set is complete.

Originally Posted by Shakarri

2. S is a set defined by S= { a + b(2)^0.5 } where a and b are rational
If S has an upper bound does a least upper bound exist? Prove the answer.

I am confused by why this question is being asked in this forum.

It is well known that any set of real numbers that has an upper bound has a least upper bound.

Now the set(sequence) of axioms/definitions may vary, but the result is the same.

Please 'flesh out' the setting of this question.

For 2, how about we prove that any subset of reals having an upperbound has a LUB.
Here is a proof by contradiction

Let A be a non empty subset of with an upperbound and no LUB

Let U be an upperbound for A. Then pick an . The interval = (x, U) contains other upper bounds, either in = or . for if it does not then by contradiction there exists a LUB. If both the intervals contain atleast one upper bounds p,q take the interval which contains the smaller value of p and q. so for the sake of brevity, say = contains the smaller upperbound, so either or contains a upper bound, again pick the follow the same rules to pick the interval.Let denote the sequence obtained from this construction.

Now the sequence of lengths of intervals converges as . It follows that there exists a sequence which is monotonously decreasing and since it is bounded, it converges to a limit L in If L is LUB, by contradiction proof is complete. If L is not LUB, then if some R is LUB, it means at some point in the interval sequence picking procedure, we picked the interval containing L instead of R, which would mean L < R. Thus L is LUB.