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

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    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!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Denote m=\inf A\;,\;M=\sup A

    (1) True. By contradiction:

    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
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    Thanks! Now if the same conditions hold For every \epsilon>0 there is an a \inA such that a<z+ \epsilon: what about..

    1. A has an element a \geqz
    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 \geqz as well?)
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  4. #4
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    Quote Originally Posted by alice8675309 View Post
    Thanks! Now if the same conditions hold For every \epsilon>0 there is an a \inA such that a<z+ \epsilon: what about..
    1. A has an element a \geqz
    I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)
    Did you read reply #2?
    The above, 1), is true. In #2 is a proof of that.
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    Quote Originally Posted by Plato View Post
    Did you read reply #2?
    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.
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    Unhappy

    Quote Originally Posted by Plato View Post
    Did you read reply #2?
    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 \geqz 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?
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  7. #7
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    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<m. 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<m. What is wrong with that?

    Thus we have that m\le z.
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  8. #8
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    Quote Originally Posted by Plato View Post
    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<m. 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<m. 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 \leqz does this work as a counterexample? A=[0,1) z=1?
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  9. #9
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    Quote Originally Posted by alice8675309 View Post
    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.
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  10. #10
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    Quote Originally Posted by Plato View Post
    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
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