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Infimum axiom

  1. Apr 26, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    So I have started my long journey through N.L. Carothers Real Analysis and my intention is to work through every single exercise along the way.

    The first problem : http://gyazo.com/ddef0387f04d789c660548c08796585d

    2. Relevant equations



    3. The attempt at a solution

    Suppose A is bounded below, we want to show A has an infimum.

    That is, we want to show ##\exists m \in ℝ## such that :

    (i) ##m## is a lower bound for A.
    (ii) If ##x## is any lower bound for A, then ##x ≤ m##.

    So suppose (i) is satisfied ( Since A is bounded below ), we want to show (ii) holds, so we consider the set ##-A = \{ -a \space | \space a \in A \}##.

    Since A is bounded below by m, -A is bounded above by -m. Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space -m ≤ s##.

    Hence sup(-A) = -m ( i.e, -m is the least upper bound of -A ) so that -sup(-A) = m ( i.e, m is the greatest lower bound of A ), but this is precisely what we desire. Thus -sup(-A) = inf(A) = m as desired.

    That was my go at it. Any thoughts to make this better? I feel as if my argument wasn't as strong as it should be near the end.
     
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  3. Apr 26, 2013 #2

    LCKurtz

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    A is bounded below so it has a lower bound m. That doesn't mean m is the greatest lower bound nor that it satisfies ii.

    No. Nothing says -m is a least upper bound for -A. The least upper bound s will satisfy ##s\le -m##.

    Even if you did have that -m is the l.u.b. of -A, you would still have to prove i and ii hold for it. Back to the drawing board. You might think about showing - sup(-A) works for the g.l.b of A.
     
  4. Apr 26, 2013 #3

    Zondrina

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    Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.

    Since m is a lower bound for A, -m is an upper bound for -A. Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space s ≤ -m##.

    So sup(-A) = s so that 's' is the least upper bound for -A. Hence -sup(-A) = -s is a lower bound for -A.

    Since -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A and hence inf(A) = -sup(-A) = -s.

    This seems to make sense ( hopefully ).
     
    Last edited: Apr 26, 2013
  5. Apr 26, 2013 #4

    LCKurtz

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    It's easy, but you should show this, not just state it.

    Sure, but there may be lots of such s's. What do you really want for s?

    Why? See above.

    Again, you should show this.

    No it doesn't. You have to show that if n is any lower bound for A then ##n\le m##.
    You have a ways to go yet.
     
  6. Apr 26, 2013 #5

    Zondrina

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    Since ##m ≤ a, \space \forall a \in A, \space -m ≥ -a, \space \forall (-a) \in (-A)##. So that -m is an upper bound for -A.

    I want this s so I can show that -A has a supremum. Not just an upper bound ( sup axiom ).

    Above.

    Since -s ≥ m and m is a lower bound for A, would that not show that inf(A) = -s = -sup(-A) ( Even stronger than a lower bound as -s is the greatest lower bound )?

    The point of the question is to show A has an infimum using the fact that A is bounded below ( by m ).

    So I assumed A had a lower bound m. Then I considered -m as the upper bound of -A. Then using the sup axiom I found sup(-A) = s so that inf(A) = -s so A has an infimum. I don't seem to see where this goes wrong?
     
  7. Apr 26, 2013 #6

    LCKurtz

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    That is not what your argument said. You said "Since -A is a nonempty subset of real numbers with an upper bound, ∃s∈R | s≤−m". You said nothing about sup(-A). You haven't written what you mean.

    That is the idea of the argument. But your writeup is not very good. And the copy function on PF makes it very difficult to respond appropriately.
     
    Last edited: Apr 26, 2013
  8. Apr 26, 2013 #7

    Zondrina

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    I would have assumed you had knowledge of the upper axiom, which is perhaps the reason my argument didn't seem too precise.

    Is it okay afterwards though, considering I'm allowed to at least use the upper axiom without proof ( heavily implied by the author ).
     
  9. Apr 26, 2013 #8

    LCKurtz

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    I am not the student here, you are. I have been trying to get you to understand what's wrong with your argument. You don't seem to know how to write up a good argument. I could show you how but you don't seem to actually address the issues I raise.
     
  10. Apr 26, 2013 #9

    Zondrina

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    I apologize for not including the axiom in relevant equations. I've addressed some of your concerns; if it's clear that I have the right idea, then feel free to expand on things.
     
  11. Apr 26, 2013 #10

    LCKurtz

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    This thread has gotten too disjointed to follow. At this point, if you would like to post your careful, most complete, best solution, with all the steps, I will be happy to respond to it. Write it up as if you were going to hand it in for grading. It will be later this evening before I can be back, so take your time and do it right.
     
  12. Apr 26, 2013 #11

    Zondrina

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    The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

    Useable info : Sup Axiom.

    Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n##, we want to show ##A## has an infimum.

    Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##

    We want to show that ##n## which is a lower bound for ##A##, is an upper bound for ##-A##. So, since ##n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)##. So since ##(-a) \in A##, we conclude ##n ≥ -a## and so ##n## is an upper bound for ##-A##.

    Now, since ##-A## is bounded above by ##n## and ##-A## is a nonempty subset of the reals, we know by the supremum axiom that ##\exists m \in ℝ \space | \space sup(-A) = m##

    Now, we want to show ##-m## is such that :

    (i) ##-m## is a lower bound for ##A##.
    (ii) If ##n## is any lower bound for ##A##, then ##n≤m##.

    To prove (i) lets assume the contrary that ##-m## is not a lower bound for ##A##. Then ##\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a##. Since ##(−a) \in (-A)##, this contradicts the fact that ##sup(-A) = m## from earlier and hence ##-m## must be a lower bound for ##A##.

    I'm having some trouble with (ii) for some reason. I assumed the contrary that ##n≥m##, but I'm having a hard time continuing with it.
     
  13. Apr 26, 2013 #12

    Dick

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    You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?
     
  14. Apr 27, 2013 #13

    LCKurtz

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    That isn't even true. The same n that is a lower bound for A is not an upper bound for -A.

    No. Since each element of -A is -a for some a in A, and ##n\le a## you have ##-n\ge -a## which tells you that -n is an upper bound for -A

    Like Dick has observed, you don't need any indirect argument. Just like the argument above where a lower bound of n for A gives an upper bound of -n for -A you should be able to show that an upper bound m for -A gives rise to a lower bound -m for A and you should be able use the fact that you have a least upper bound for -A gives rise to a greatest lower bound for A.
     
  15. Apr 27, 2013 #14

    Zondrina

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    For (ii), suppose that A is bounded below by n. We want to show n ≤ -m. Consider the set ##L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}## which is the set of all lower bounds of A.

    Clearly ##n, -m \in L## by hypothesis and (i) so that ##L## is not empty.
    ---------

    EDIT : I realized after reading LC's comment that there is a very easy way to do this without really putting effort into (i) or (ii).

    Since m is the least upper bound of -A, -m is the greatest lower bound for A. That is ##m ≥ -a \Rightarrow -m ≤ a## so that -m is the greatest lower bound of A.
     
    Last edited: Apr 27, 2013
  16. Apr 27, 2013 #15

    CAF123

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    I think the statement you are trying to prove is the Completeness property for infima. If you know the Completeness Property for suprema and the Reflection principles, then I believe this can be done in a couple of steps.
     
  17. Apr 27, 2013 #16

    LCKurtz

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    But you can't just declare that and call it a proof. You have to get your hands dirty with both i and ii. To prove them for A you have to use the corresponding properties for sup and -A and use them to get the results you want for A.
     
  18. Apr 27, 2013 #17

    Zondrina

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    Back from a long work shift, lets do math. I did (i) by contradiction did I not? Anyway, i'll try them both without any contradiction.

    Since m is the least upper bound of -A, -m is a lower bound for A. That is ##m≥−a \Rightarrow −m≤a##.

    Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?
     
  19. Apr 27, 2013 #18

    LCKurtz

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    That is part of the reason it is so difficult to help you. What is m and where is it defined? That's why I asked you to give a complete argument from the beginning, which you attempted in post #11. In post #13 I indicated some problems with your argument. You have not acknowledged that you understood my corrections so I don't know if you even read them or whether you understood them if you did read them. Now we are back to you being confused about m or -m.

    I'm sorry, but I am tiring of this thread and I think I have said all I want to say.
     
  20. Apr 27, 2013 #19

    Dick

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    The proof is even easier if you don't use the hint. L, the set of lower bounds, has an upper bound, any element of A is an upper bound. That means L has a sup, call it n. Doesn't n have all of the properties you need to be an inf of A?
     
    Last edited: Apr 27, 2013
  21. Apr 27, 2013 #20

    Zondrina

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    The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

    Useable info : Sup Axiom.

    Suppose ##A## is a nonempty subset of ##ℝ## bounded below, we want to show ##A## has an infimum.

    Consider the set of lower bounds of A : ##L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}##. Notice L is not empty because A is bounded below. Since ##x ≤ a##, L is bounded above for any ## a \in A##.

    Now, L is a nonempty subset of real numbers, which is bounded above by at least one element ( Since A is nonempty ), so we know sup(L) exists by the sup axiom, call it n.

    n is in the set of lower bounds of a, so it satisfies (i). Now since ##x ≤ n ≤ a## ( From each respective inequality throughout the proof ) (ii) is satisfied. So we can write inf(A) = n.
     
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