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Thread: 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 $\displaystyle \leq$ z

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

    (2) inf A $\displaystyle \geq$ z

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

    (3) sup A $\displaystyle \leq$ 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 $\displaystyle m=\inf A\;,\;M=\sup A$

    (1) True. By contradiction:

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

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

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


    Fernando Revilla
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    Thanks! Now if the same conditions hold For every $\displaystyle \epsilon$>0 there is an a$\displaystyle \in$A such that a<z+$\displaystyle \epsilon$: what about..

    1. A has an element a$\displaystyle \geq$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 $\displaystyle \geq$z 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 $\displaystyle \epsilon$>0 there is an a$\displaystyle \in$A such that a<z+$\displaystyle \epsilon$: what about..
    1. A has an element a$\displaystyle \geq$z
    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$\displaystyle \geq$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$\displaystyle \leq$ z and I said True because if a=z then a<z+$\displaystyle \epsilon$ holds for all $\displaystyle \epsilon$>0 and if a <z then a<z+$\displaystyle ^epsilon$ holds for all$\displaystyle \epsilon$>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 $\displaystyle \left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.

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

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

    Thus we have that $\displaystyle m\le z.$
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  8. #8
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    Quote Originally Posted by Plato View Post
    Given that $\displaystyle \left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.

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

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

    Thus we have that $\displaystyle m\le z.$
    Ohhh I see it now! So for the one that states A has an element a$\displaystyle \leq$z 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 $\displaystyle \leq$ z
    (2) inf A $\displaystyle \geq$ z
    (3) sup A $\displaystyle \leq$ 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 $\displaystyle A=[0,1]$ then $\displaystyle 0=\inf(A)~\&~1=\sup(A)$.
    If we choose $\displaystyle z=\frac{1}{2}$ 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 $\displaystyle A=[0,1]$ then $\displaystyle 0=\inf(A)~\&~1=\sup(A)$.
    If we choose $\displaystyle z=\frac{1}{2}$ 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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