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First of all, the sup of c+A exists because A is bounded above; call any such bound B. Then B + c is a bound for c+A. Hence by the Completeness Axiom of the Reals, c+A has a sup.
Now, there are few ways to skin this cat. One way is the show that \[ \sup(c+A) \leq c+\sup(A) \] and \[ c+\sup(A) \leq \sup(c+A) \] Another is use some argument about neighborhoods of the sup.
What have you tried so far?
I honestly haven't had any idea how to approach it, proofs have never been my strong suit and if I can't see a clear line there I usually just stare at it until I do, here nothing was inherently obvious to me, so I just stared for a long time...
by an argument for the neighbourhoods of the sup do you mean show that if you subtract an arbitrary small constant from the sup that it is no longer a sup?
I recommend the two inequalities I wrote down. To make that argument work, show first that c + sup(A) is an upper bound for c+A then by the definition of sup(c+A) it must be that sup(c+A) =< c + sup(A) == Now the other way around ... show that sup(c+A)-c is an upper bound for A then by definition of sup(A), it must be that sup(A) =< sup(c+A) - c ==== Make sense?
yes, that's just step one right? there needs to be more to that doesnt there?
I've laid out the strategy for you. You'll need to fill in the details. But if you do this then you have the inequality both ways so the two terms c+supA and sup(c+A) must be equal. Just try and write it out. I think you'll find it's not too bad. Remember to just use the definition of what an upper bound must be and the fact that the sup is itself an upper bound.
(btw, great profile pic)
Thanks man, I think I see how to finish this off. Very helpful.
If you want, write out what you get and I'll check it later. And if not, no foul.
For what it's worth, questions like these are some of my favorites, some keep them coming.