Given Sup S = inf S, prove that the set S consists of exactly one number?
Suppose that S is a nonempty set of real numbers and that inf S = sup S. Prove that S consists of exactly one number.
Shouldn't this be obvious? I mean, if the least upper bound is the greatest lower bound (meaning two bounds are the same) then stating it is obvious? So isn't this like saying if a set has an upperbound of 1 and a lowerbound of 1 then the set only consists of 1 since no number exists between 1 and 1?
I could really use help on this proof.
- 9 years agoFavorite Answer
Let S be nonempty and inf S = sup S. Then there exists at least one element of S. Assume by contradiction that there exists x, y in S where x =/= y. Then inf S <= x,y <= sup S is clear for any elements in S. Now since inf S = sup S this means inf S <= x,y <= inf S and sup S <= x,y = sup S. This forces x = inf S = sup S and y = inf S = sup S. This forces x = y, a contradiction. Thus, there is only one element of S.
- Anonymous4 years ago
Sup And InfSource(s): https://shorte.im/a9G45
- 9 years ago
Let x belong to S. We know that x<=sup S and we know that x>=inf S. This means that inf S <= x <= sup S = inf S, or inf S <= x <= inf S, or x = inf S. Since inf S can only be one value, x can only be one value therefore S only consists of one number, x = inf S = sup S.