Proving on the completeness theorem of real number

[Lily@pie]

Homework Statement

Let A be a non empty subset of R that is bounded above

Set D:={2a|a (belongs to) A}
Is it necessarily true that the sup D = 2 sup A? Either prove or provide a counterexample.

Homework Equations

The completeness axiom

The Attempt at a Solution

I am seriously clueless on how to approach... but I still tried something
Let sup D = y and sup A = x
d=2a; d<=y; a<=x

or can I say choose a as the largest value in A, so 2a=d is the largest value in D. Since both are the upper bound for each set and for all upper bound and real number, they are the smallest. so, d is sup D and a is sup A.
Therefore, d=2a => sup D = 2 sup A


But these method seems a bit weird...

 

[HallsofIvy]

No, you can't say "choose a as the largest value in A". The whole point of "sup" is that the set may not have a largest value. And you can't say "Let sup D = y and sup A = x d=2a; d<=y; a<=x" because you didn't say what "a" was.

"sup" has two properties- it is an upper bound (so there are no member of the set larger than the sup) and there is no smaller upper bound (so there are members of the set within any "epsilon" distance of the sup.

Let S be the sup of set A. We can prove that 2S is an upper bound for 2A by contradiction:
suppose there exist b in 2A such that b> 2S. But b= 2a for some a in A. That says 2a> 2S so that a> S which contradicts the fact that S is an upper bound for A.

Similarly, you can prove by contradiction that there is no upper bound less than 2S.

 

[Lily@pie]

But how do we relate this to the prove of sup D = 2 sup A?

Does 2S means D?

Sorry, I'm a bit confused now