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Metric Spaces

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

    20glgu9.jpg

    2. Relevant equations



    3. The attempt at a solution

    I've done the first 3 parts. I've come to the bit on Cauchy sequences at the end. How do I show [itex]x_n = n[/itex] is/isn't a Cauchy sequence in the 2 metrics?

    [itex](x_n)[/itex] is a Cauchy sequence in a metric space [itex](X,d)[/itex] if for any [itex]\varepsilon >0[/itex] there exists [itex]N\in \mathbb{N}[/itex] such that if [itex]m,n > N[/itex] then [itex]d(x_m , x_n ) < \varepsilon[/itex].

    The metric space [itex](X,d)[/itex] is complete if every Cauchy sequence in [itex](X,d)[/itex] converges to a limit in [itex]X[/itex].
     
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  3. Dec 15, 2011 #2

    micromass

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    So let's prove that it isn't a Cauchy sequence in the standard metric.

    So, we need to find an epsilon (let's take [itex]\epsilon=1[/itex] for a moment), such that for all N, there exists n,m>N such that

    [tex]d(x_n,x_m)\geq 1[/tex]

    So we must find n,m>N such that

    [tex]|n-m|\geq 1[/tex]

    Can you find such a n and m??
     
  4. Dec 15, 2011 #3

    radou

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    Well, let ε be some number in <0, 1>. Does there exist some positive integer N such that for all m, n >= N you have |xm - xn| < ε?

    Edit: sorry, micromass seems to have answered first.
     
  5. Dec 15, 2011 #4
    Setting [itex]m=N+1[/itex] and [itex]n=N+2[/itex] we have that [itex]m,n > N[/itex] and [tex]|m-n|=|N+1-(N+2)|=|1-2|=1 \geqslant 1\;,\;\text{for all}\;N\in \mathbb{N}[/tex]
     
  6. Dec 15, 2011 #5

    micromass

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    Good!! So that proves that it isn't a Cauchy sequence.

    Now, to prove that it is a Cauchy sequence in the other metric, you must make

    [tex]|\tan^{-1}(n)-\tan^{-1}(m)|[/tex]

    smaller than [itex]\varepsilon[/itex]
     
  7. Dec 15, 2011 #6
    How do I make [itex]|\tan^{-1}(m) - \tan^{-1}(n)|<\varepsilon[/itex] ? I'm guessing that the fact given in the question that [itex]d(x,y)< \pi[/itex] might help?

    However, for the last part on whether [itex](\mathbb{R} ,d)[/itex] is complete, can I say: if it is complete then every Cauchy sequence in [itex](\mathbb{R} ,d)[/itex] must converge to a limit in [itex]\mathbb{R}[/itex].

    Suppose the Cauchy sequence [itex]x_n=n[/itex] converges in [itex](\mathbb{R} ,d)[/itex]. Then by part (c), [itex]x_n[/itex] must converge in the standard metric. Every convergent sequence is Cauchy, but [itex]x_n[/itex] is not Cauchy in the standard metric - contradiction. So [itex]x_n[/itex] does not converge in [itex](\mathbb{R} , d)[/itex]; therefore not every Cacuhy sequence in [itex](\mathbb{R} , d)[/itex] converges so it is not complete.
     
    Last edited: Dec 15, 2011
  8. Dec 15, 2011 #7

    micromass

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    Well, this showing that [itex]x_n=n[/itex] is Cauchy in [itex](\mathbb{R},d)[/itex] is equivalent to showing that [itex]y_n=\tan^{-1}(n)[/itex] is Cauchy in the standard metric.

    That is good.
     
  9. Dec 16, 2011 #8
    In examples I've seen on showing a sequence is Cauchy they often involve getting [itex]d(x_m , x_n) \leqslant X \to 0[/itex] where [itex]X[/itex] is some upper bound. Why does this show the sequence is Cauchy?

    For my sequence [itex]x_n = n[/itex] I could use the triangle inequality [itex]d(x_m , x_n) \leqslant d(x_m , \frac{\pi}{2} ) + d(\frac{\pi}{2} , x_n)[/itex]: [tex]d(x_m ,x_n) = | \tan^{-1} (x_m) - \tan^{-1}(x_n) |[/tex]
    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leqslant | \tan^{-1}(m) - \frac{\pi}{2} | + | \tan^{-1}(n) - \frac{\pi}{2} | \to 0 + 0=0[/itex] as [itex]m,n \to \infty[/itex]

    since [itex]\displaystyle \lim_{r\to\infty} \tan^{-1}(r) = \frac{\pi}{2}[/itex]
     
    Last edited: Dec 16, 2011
  10. Dec 16, 2011 #9

    micromass

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    Well, write out what it means that [itex]d(x_m,x_n)\rightarrow 0[/itex]...

    That's good.
     
  11. Dec 16, 2011 #10

    Office_Shredder

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    Be careful when you're writing this up! You're using the triangle inequality for absolute value here, but you're not using the triangle inequality for the metric d here - if you were actually using d(xn,pi/2) you would end up with |arctan(n)-arctan(pi/2)|+|arctan(m)-arctan(pi/2)| which doesn't go to zero as n and m go to infinity
     
  12. Dec 18, 2011 #11
    I'm not getting any credit for this; it's all past exam paper stuff.

    But, yes - it's the scalar triangle inequality I should be using!
     
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