# For every ε>0 there is an a∈A such that a<z+ε

• Jan 27th 2011, 07:46 PM
alice8675309
For every ε>0 there is an a∈A such that a<z+ε
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 $\leq$ z

I said it was TRUE but I don't know how I would prove it.

(2) inf A $\geq$ z

This one I have no clue about. I'm assuming its FALSE because I said the above statement is true.

(3) sup A $\leq$ z

I said True by trichotomy..but I know that isn't a proof so how would I prove this one as well.

Thanks!
• Jan 28th 2011, 01:03 AM
FernandoRevilla
Denote $m=\inf A\;,\;M=\sup A$

If $m>z$ choose $\epsilon=m-z$ .

(2) False. Choose $A=[0,1)\;,\;z=1$ .

(3) False. Choose $A=[0,1)\;,\;z=1/2$ .

Fernando Revilla
• Jan 30th 2011, 04:38 AM
alice8675309
Thanks! Now if the same conditions hold For every $\epsilon$>0 there is an a $\in$A such that a<z+ $\epsilon$: what about..

1. A has an element a $\geq$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 $\geq$z as well?)
• Jan 30th 2011, 04:50 AM
Plato
Quote:

Originally Posted by alice8675309
Thanks! Now if the same conditions hold For every $\epsilon$>0 there is an a $\in$A such that a<z+ $\epsilon$: what about..
1. A has an element a $\geq$z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)

The above, 1), is true. In #2 is a proof of that.
• Jan 30th 2011, 05:03 AM
alice8675309
Quote:

Originally Posted by Plato
The above, 1), is true. In #2 is a proof of that.

Sorry, I did read reply #2 but I was honestly checking the answers I had. I think I'll have to draw a picture to connect everything.
• Jan 30th 2011, 05:23 AM
alice8675309
Quote:

Originally Posted by Plato
The above, 1), is true. In #2 is a proof of that.

Now looking at my paper, the reason I had posted the question A has an element a $\geq$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 $\leq$ z and I said True because if a=z then a<z+ $\epsilon$ holds for all $\epsilon$>0 and if a <z then a<z+ $^epsilon$ holds for all $\epsilon$>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?
• Jan 30th 2011, 05:39 AM
Plato
Given that $\left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.

If $m=\text{inf}(A)$ suppose that it is true that $z. Then let $\varepsilon =(m-z)>0$.

So $\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + (m - z)} \right]$.
But that says $a. What is wrong with that?

Thus we have that $m\le z.$
• Jan 30th 2011, 05:51 AM
alice8675309
Quote:

Originally Posted by Plato
Given that $\left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.

If $m=\text{inf}(A)$ suppose that it is true that $z. Then let $\varepsilon =(m-z)>0$.

So $\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + (m - z)} \right]$.
But that says $a. What is wrong with that?

Thus we have that $m\le z.$

Ohhh I see it now! So for the one that states A has an element a $\leq$z does this work as a counterexample? A=[0,1) z=1?
• Jan 30th 2011, 06:07 AM
Plato
Quote:

Originally Posted by alice8675309
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 $\leq$ z
(2) inf A $\geq$ z
(3) sup A $\leq$ z

Above I have quoted the OP.
I answered (1), proved it is true.
Both (2) & (3) are false.
To show that we need to find an example of both the set A and the value z.
Let $A=[0,1]$ then $0=\inf(A)~\&~1=\sup(A)$.
If we choose $z=\frac{1}{2}$ both are false.
• Jan 30th 2011, 06:17 AM
alice8675309
Quote:

Originally Posted by Plato
Above I have quoted the OP.
I answered (1), proved it is true.
Both (2) & (3) are false.
To show that we need to find an example of both the set A and the value z.
Let $A=[0,1]$ then $0=\inf(A)~\&~1=\sup(A)$.
If we choose $z=\frac{1}{2}$ both are false.

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 :(