Denote
(1) True. By contradiction:
If choose .
(2) False. Choose .
(3) False. Choose .
Fernando Revilla
So there are a few questions that go along with this but I've hopefully worked them out correctly so I am only posting the ones I definitely have no clue about.
Let A be a bounded set of reals and z a real number such that the following holds: For every ε>0 there is an a∈A such that a<z+ε (if it is true, prove it. If not, then show a counterexample.
(1) inf A z
I said it was TRUE but I don't know how I would prove it.
(2) inf A z
This one I have no clue about. I'm assuming its FALSE because I said the above statement is true.
(3) sup A z
I said True by trichotomy..but I know that isn't a proof so how would I prove this one as well.
Thanks!
Denote
(1) True. By contradiction:
If choose .
(2) False. Choose .
(3) False. Choose .
Fernando Revilla
Thanks! Now if the same conditions hold For every >0 there is an a A such that a<z+ : what about..
1. A has an element a z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)
2. Inf A=z
I said false: (0,1)=A, z=1
3. Sup A=z
I said false: [-2,-1] z=0 ( would this counterexample also work for sup A z as well?)
Now looking at my paper, the reason I had posted the question A has an element a z and said it was false was because I had another question with the same conditions stated in the first post that said:
A has an element a z and I said True because if a=z then a<z+ holds for all >0 and if a <z then a<z+ holds for all >0.
Do I have them backwards? Which one is true and which one is false and what would be a counterexample for the false one?
Thanks so much! I'm sorry if it got confusing because I did post three additional questions in reply #3 which I guess I should have started a new thread for. However, are the counterexamples in reply #3 for 2 and 3 by any chance correct? Again thanks you've been a big help! This topic isn't my strongest