# Let A be abounded set of reals and z a real number with the following conditions...

• Jan 30th 2011, 03:59 PM
alice8675309
Let A be abounded set of reals and z a real number with the following conditions...
So I had asked this question within another question and it seems it got confusing, so I've decided to create a new post all together with the new questions to clear things up because everything got really confusing(for myself and anyone else if they got thrown off by my last post, sorry about that). But here are the conditions:

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.

(These are new questions with the same conditions, I'm not reposting to repost)

1a) A has an element a $\geq$z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)

1b) A has an element a $\leq$ z

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 $\geq$z as well?)

Thanks!
• Jan 30th 2011, 06:32 PM
tonio
Quote:

Originally Posted by alice8675309
So I had asked this question within another question and it seems it got confusing, so I've decided to create a new post all together with the new questions to clear things up because everything got really confusing(for myself and anyone else if they got thrown off by my last post, sorry about that). But here are the conditions:

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.

This is already confusing: you write above "(if it is true, prove it. If not, then show a counterexample.)" What does this mean? The above can't be

proved, it is given! , unless you meant that we prove, or not, the FOLLOWING...??

(These are new questions with the same conditions, I'm not reposting to repost)

1a) A has an element a $\geq$z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)

This is wrong: if you choose $z=1$ , then ANY $a\in [-1,1]$ works...

1b) A has an element a $\leq$ z

This is true unless $A=\phi$

2. Inf A=z
I said false: (0,1)=A, z=1

Correct

3. Sup A=z
I said false: [-2,-1] z=0 ( would this counterexample also work for sup A $\geq$z as well?)

Correct.

Tonio

Thanks!

.