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Show there's a sequence whose limit is its infimum

  1. Oct 22, 2013 #1
    question:

    Suppose ## S \subset \mathbb{R} ## is a nonempty subset of the real numbers that is bounded below.

    Show that there exists a sequence ## <x_n> ## such that ## x_n \in S ## for all n and ## \lim (x_n) = inf(S) ##

    attempt:

    consider an element ## x \in S ## suppose ## x \geq inf(S) + 1/n ## then this would mean that inf(S) is not the greatest lower bound and inf(s) + 1/n is so ## x < inf S + 1/n ## ## \forall n \in \mathbb{N} ## take ## n_1 < n_2 < n_3 < n_4 ... ## then ## x_1 < inf(S) + 1/n_{1} ## and ## x_2 < inf(S) + 1/n_{2} ## and so on so we get a sequence ## inf(S) \leq x_n ... < x_3 < x_2 < x_1 ## hence as n increases 1/n approaches zero (proved already) and x_n approaches inf(S).

    Is this all correct as in the solutions all they have stated is:

    ## inf(S) \leq x_n < inf(S) + 1/n ## and inf(s) + 1/n -> inf(S) so such a sequence exists, I don't understand that
     
  2. jcsd
  3. Oct 22, 2013 #2
    alternatively couldn't I say as ## inf(s) \leq x_n < inf(S) + 1/n ## ##lim inf(S) = inf(S)## and##lim (inf(S) + 1/n) = inf(S) ## therefore by the sandwich theorem x_n approaches inf(S) also?
     
  4. Oct 22, 2013 #3

    LCKurtz

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    You sort of have the idea but that is a very hard to read and over-complicated explanation. To clean it up, you might start by calling ##I = \inf(S)## so you don't have to keep writing it. Then by the definition of ##\inf(S)## you can say that for each natural number ##n## there is a point ##x_n\in S## in the interval ##(I,I+\frac 1 n)##. Then argue that ##x_n\rightarrow I##.
     
  5. Oct 22, 2013 #4
    ok thank you. Is my argument that x_n -> inf(S) valid though? Or could I use the sandwich theorem?
     
  6. Oct 22, 2013 #5

    LCKurtz

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    Probably so, and yes, you could. Just to make it clear, show me below how you would finish the above argument.
     
  7. Oct 22, 2013 #6
    Make sure your proof works for ##\mathbb{N}##; i.e. take into consideration that ##\inf S## may not be a limit point of ##S\setminus\{\inf S\}##.
     
  8. Oct 23, 2013 #7
    the limit of I is I and the limit of (I + 1/n) (as 1/n -> 0 (which I've proved previously)) and seeing as x_n is 'sandwiched' by I and I + 1/n the limit of x_n is I, hence a sequence <x_n> which has the infimum as it's limit exists
     
  9. Oct 23, 2013 #8
    My other argument would be that if we take ## n_1 < n_2 <n_3 ## and## x_1 < I + 1/n_1## ## x_2 < I + 1/n_2 ## etc then we end up with the sequence ## I \leq x_n <...<x_3<x_2<x_1 < I + 1/n_1 ## and as n approaches infinity then x_n approaches I
     
  10. Oct 23, 2013 #9

    LCKurtz

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    Yes. You should probably use the half closed interval ##[I,I+\frac 1 n)## to select your points to address gopher_p's observation.
     
  11. Oct 23, 2013 #10

    LCKurtz

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    While you certainly can make a decreasing sequence ##\{x_n\}##, just picking ##x_k < I+\frac 1 {n_k}## doesn't necessarily do it. But it doesn't need to be decreasing anyway.
     
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