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Let x be an upper bound for A subset Real #'s.
Prove that if x element A, then x = sup(A).
Proof:
Let x be an upper bound for A subset Real #'s.
Assume x element A.
a <= x for all a element A by definition of upperbound.
Since x element A, x <= x.
Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A).
I was wondering if this was a valid proof.
If not could you help correct it please.
Thanks
I would do it this way:Quote:
Originally Posted by tbyou87
Suppose x in A is an upper bound for A, but that x is not the supremum of A.
Then there exists another upper bound y for A, such that y<x. But this conradicts y being an upper bound for A, as there is an element x of A such that x>y.
Hence if x in A, is an upper bound for A a subset of R, then x = sup(A).
RonL
well it sounded great up to the " Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A)." line.Quote:
Originally Posted by tbyou87
remember the definition of supremeum--it is the least upper bound. that is you have to show that two conditions hold:
1) if x is the supremum for a set A, then a <= x for all a in A (you proved this nicely)
2) if y is any other upperbound for A, then y>=x (you have yet to prove this.)
i'd end with something like, "let y be another upperbound for A distinct from x. since x is the upperbound for A and is also an element of A, then x is also max(A), and therefore is y >= x. (since y has to be an upperbound outside the set). thus, since x >= a for all a in A, and x <= y for all y upperbounds of A, x is sup(A)."
well, the language needs to be cleared up a bit, but i'd say something to that effect. no matter how you phrase it though, you have to show those two things.