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Nonempty subset homework

  1. Apr 27, 2013 #1

    Zondrina

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

    http://gyazo.com/27a104e05c409c75611ee4250a89c790

    2. Relevant equations

    Sup/Inf axioms as well as the ε-N definition.

    3. The attempt at a solution

    Suppose A is a nonempty subset of ℝ bounded above by ##M##. We want to show that ##lim_{n→∞} x_n = sup(A)## where ##x_n## is a sequence of elements of A.

    That is, ##\forall ε > 0, \exists N \space | \space n ≥ N \Rightarrow |x_n - sup(A)| < ε##

    Note that since A is bounded above by M, we know that ##M > a, \space \forall a \in A## including ##sup(A)##.

    So :

    ##|x_n - sup(A)| ≤ |x_n| + |sup(A)| < |x_n| + M##

    So ##|x_n|## is bounded above by ##ε - M## and below by ##-(ε+M)##

    I'm not quite sure how to continue here.
     
  2. jcsd
  3. Apr 27, 2013 #2

    micromass

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    You need to define the sequence ##(x_n)_n## first. Of course not every sequence will do.
     
  4. Apr 27, 2013 #3

    Zondrina

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    Ohhh I thought this was a general question, so I need a concrete sequence then?

    How about ##A = \{2 - \frac{1}{n} \space | \space n = 1,2,... \}## and ##x_n = 2 - \frac{1}{n}##.

    Then ##sup(A) = 2## and ##x_n → 2## as ##n→∞##

    ##|x_n - sup(A)| = |2 - \frac{1}{n} - 2| = \frac{1}{n}##

    So choosing ##n ≥ \frac{1}{ε} \Rightarrow |x_n - sup(A)| < ε##

    Didn't even need the upper bound except to establish sup(A) existed :).
     
  5. Apr 27, 2013 #4

    micromass

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    You'll need a concrete sequence. But the set ##A## is completely general. You can't pick ##A=\{2-1/n~\vert~n\in \mathbb{N}\}##. It needs to work for every possible set ##A## (that is bounded above).

    So for a set ##A## that is bounded above, you need to find a concrete sequence that converges to ##sup(A)##.
     
  6. Apr 27, 2013 #5

    Zondrina

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    I'm having a bit of trouble actually producing the sequence. I understand why I cannot pre-define A ( I did it mostly to get an idea of whats going on ), but can I choose ANY convergent sequence or is there a method to doing this?

    I mean I can come up with enough convergent sequences like ##x_n = 1/n##, but it's not like I can put a finite number on sup(A).

    All i really have is the fact that A is bounded above by M so that sup(A) exists.
     
  7. Apr 27, 2013 #6

    micromass

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    Here's a hint. Prove the following "lemma":

    Given a positive integer ##n##. There is always an element ##x_n## such that ##sup(A)-\frac{1}{n}< x_n## and such that ##x_n\in A##.
     
  8. Apr 27, 2013 #7

    Zondrina

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    I don't see how that helps me? I've been staring at it for awhile and I'm unclear about your intentions.

    sup(A) < xn + 1/n

    Then i was thinking induction.
     
  9. Apr 27, 2013 #8

    micromass

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    Induction won't help here.

    The intention of my lemma is that it would define a nice sequence ##(x_n)_n##. The sequence as defined in the lemma can be shown to converge to ##sup(A)##.
     
  10. Apr 27, 2013 #9

    Zondrina

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    Waaaait I see what you're getting at now. Since ##x_n + 1/n## will be a sequence as well.

    I unfortunately just got called into work 2 mins ago, I will continue this later today when I get home.
     
  11. Apr 27, 2013 #10

    micromass

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    It will be a sequence as well, but I'm not sure how it helps...

    Have fun!
     
  12. Apr 28, 2013 #11

    Zondrina

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    Let ##A## be a nonempty subset of ##ℝ## which is bounded above.

    We must show ##\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)##

    We must somehow construct ##x_n## from the elements of A so that it converges to sup(A).

    Suppose n is a positive integer and ##x_n## is a sequence of elements of A. Since ##x_n \in A##, we know that ##x_n ≤ sup(A)## so that ##x_n## is bounded.

    I think I have to use the fact it's bounded somehow, I'm having some thoughts, but I'd rather hear some input before going off in potentially the wrong direction.
     
  13. Apr 28, 2013 #12

    Zondrina

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    Sorry for the double, but I just woke up so I want to put a fresh brain effort into this.

    Let ##A## be a nonempty subset of ##ℝ## which is bounded above.

    We must show ##\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)##

    We must somehow construct ##x_n## from the elements of A so that it converges to sup(A). Now because ##x_n \in A##, we know that ##x_n ≤ sup(A)##.

    Using the hint given, I want to show the inequality ##x_n ≥ sup(A) - \frac{1}{n}## holds for any positive integer n because ##sup(A) - \frac{1}{n}## is not an upper bound for A. Inductively, take n=1. Then ##x_1 ≥ sup(A) - 1## and now assume this holds for any n. We want to show this holds for n+1, so ##x_{n+1} ≥ sup(A) - \frac{1}{n+1}##.

    So we can inductively build our sequence ##x_n## with following the relation ##x_n ≥ sup(A) - \frac{1}{n}##.

    Now notice that :
    ##x_n ≥ sup(A) - \frac{1}{n}##
    ##\Rightarrow sup(A) - x_n ≤ \frac{1}{n}##
    ##\Rightarrow |x_n - sup(A)| ≤ \frac{1}{n}##

    So, ##\forall ε>0, \exists N \space | \space n ≥ \frac{1}{ε} \Rightarrow |x_n - sup(A)| < ε##

    Thus making our choice of ##n≥\frac{1}{ε}## we observe ##|x_n - sup(A)| ≤ \frac{1}{n} < ε##.

    Therefore we have found a way to construct a sequence of elements of A which converges to sup(A) as desired.

    I know you said not to use induction micro, but I didn't seem to see any other way. Hopefully this makes sense.
     
  14. Apr 28, 2013 #13
    In the above, you haven't actually shown that [itex]x_n\in A[/itex]; you seem to be assuming this. Specifically, you need to show there exists a sequence that satisfies the inequality you want. Specifically, you need that for every positive integer, there exists an [itex]x_n \in A[/itex] such that [itex]sup(A)<x_n+1/n.[/itex] I would refer you to another problem you are working on that shows you that such an [itex]x_n[/itex] always exists..

    ie. show that s=sup(A) if and only if a≤s for all a in A, and for all [itex]\epsilon>0[/itex], there exists [itex] a\in A[/itex] such that [itex]s-\epsilon<a[/itex].

    Spoiler below (try working it out on your own first).

    To get [itex]x_n=a[/itex], try taking [itex]1/n=\epsilon[/itex].
     
    Last edited: Apr 28, 2013
  15. Apr 29, 2013 #14

    Zondrina

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    Yes my wording was wrong there I meant to say because sup(A) exists, let me try this proof again.

    Let ##A## be a nonempty subset of ##ℝ## which is bounded above.

    We must show ##\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)##

    We must somehow construct ##x_n## from the elements of A so that it converges to sup(A), but because sup(A) exists we know that ##x_n ≤ sup(A)##

    So for any positive integer n, we must show ##sup(A) - \frac{1}{n} < x_n ≤ sup(A)##.

    This comment made me blank for awhile :

    ie. show that s=sup(A) if and only if a≤s for all a in A, and for all ϵ>0, there exists a∈A such that s−ϵ<a.

    I don't see how that relates to this problem?
     
  16. Apr 29, 2013 #15
    Did you read my hint?

    For n=1, you have [itex]\epsilon=1/1=1[/itex], so there exists an [itex]x\in A[/itex] such that
    [itex]sup(A)-1<x[/itex]. Denote this x by [itex]x_1[/itex].

    For n=2, you have [itex]\epsilon=1/2[/itex], so there exists an [itex]x\in A[/itex] such that [itex]sup(A)-1/2<x[/itex]. Now denote THIS x by [itex]x_2[/itex].

    For n=3, [itex]\epsilon=1/3[/itex]. Construct [itex]x_3[/itex] the same way.

    The existence of each [itex]x_n[/itex] at each step is guaranteed by the other problem I cited, which you just finished in another thread. In particular, such an [itex]x_n[/itex] exists for each n, and satisfies... which inequality?

    As for why it's a sequence... to each n, you can associate an [itex]x_n[/itex], and each of these is in [itex]A[/itex]..
     
    Last edited: Apr 29, 2013
  17. Apr 29, 2013 #16

    Zondrina

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    Ohh I see. Let me try again then.

    Let ##A## be a nonempty subset of ##ℝ## which is bounded above.

    We must show ##\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)##

    So for any positive integer n, we must show ##sup(A) - \frac{1}{n} < x_n## so we can construct a sequence which is bounded by ##sup(A)##.

    So take n=1, then we can find ##x_1 \in A## such that ##sup(A) - 1 < x_1## because sup(A)-1 is not an upper bound for A. We can follow this construction for any positive n yielding our desired inequality.

    So we have successfully constructed our sequence ##x_n## by choosing the elements of A which satisfy ##sup(A) - \frac{1}{n} < x_n##. Now because sup(A) exists, we know that ##a ≤ sup(A), \space \forall a \in A##. Since each ##x_n \in A## for each positive integer n, we know that ##x_n ≤ sup(A)## for all n so that the sequence ##x_n## is bounded above by sup(A).

    Hmm I don't have access to the monotone convergence theorem until literally one question after this. How would I finish this up without it? The same way I did before a few posts ago?
     
    Last edited: Apr 29, 2013
  18. Apr 29, 2013 #17

    CAF123

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    I think we are working on the same (or similar question). I just said since you have sup A -1/n < x ≤ sup A and then when you take the limit the left most inequality necessarily becomes a less or equal. So x is less or equal to sup A on both sides which means it is equal to sup A.
     
  19. Apr 29, 2013 #18

    Zondrina

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    Yes I was thinking about this, but I wasn't sure if I could take limits for granted. That would indeed show that ##lim(sup(A) - \frac{1}{n}) < lim(x_n) ≤ lim(sup(A))##, but because we can't have sup(A) < sup(A), it must be the case that ##lim(x_n) = sup(A)##.
     
  20. Apr 29, 2013 #19
    How to bypass this thing with the inequalities;
    Note that if [itex]a<b[/itex], the we also have [itex]a\leq b[/itex]. So whenever you have a less-than sign, you can always instead write a less-than-or-equal-to sign.
    Example: we have 1<2. Also, 1≤2.
    On the other hand, the opposite is false; you can't necessarily turn a ≤ into a <.
    Example: we have 2≤2. We don't have 2<2.

    As for the aspect of "taking the limits for granted", here's a way to do it rigorously:

    For all n>0, we have [itex]x_n-sup(A)\leq 0[/itex]. On the other hand, we have [itex]-\frac{1}{n}\leq x_n-sup(A)[/itex] for all n. Putting these two together, we have [itex]-\frac{1}{n}\leq x_n-sup(A)\leq 0 \leq \frac{1}{n}[/itex]. Hence, [itex]|x_n-\sup(A)|\leq\frac{1}{n}<\frac{2}{n}[/itex].

    Now let [itex]\epsilon>0[/itex]. Then there exists an [itex]N>0[/itex] such that [itex]\frac{2}{N}<\epsilon[/itex] (just take N large enough; [itex]N>\frac{2}{\epsilon}[/itex]). Then for all [itex]n>N[/itex], we have [itex]|x_n-\sup(A)|<\frac{2}{n}<\frac{2}{N}=\epsilon[/itex]. Hence, [itex]x_n[/itex] converges to [itex]sup(A)[/itex] (by definition of convergence; epsilon definition).
     
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