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Show that a metric space is complete

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Given (R+, d), R-Real #
    d= | ln(x/y) |

    Show that this metric space is complete

    2. Relevant equations



    3. The attempt at a solution

    Firstly, I know that to show it is complete I need to have that all Cauchy sequences in that space converge...

    So I'm not 100% sure, but if I know I have to generalize so that it works for every Cauchy sequence, so can I find subsequences that converge, and then say that each sequence converges??? :S if so, how do I start this without picking specific cases, or can I pick a specific sequence in that space?
     
  2. jcsd
  3. Sep 26, 2011 #2
    Hello.

    Trying to find a convergent subsequence of a cauchy sequence is indeed enough, but in a lot of cases you don't need to specifically focus on subsequences. I will tell you that in this problem you won't need to bother with subsequences, it'll be more direct to directly proof that a cauchy sequence has a limit.

    Specific cases are a good way to get a feel for the problem. Problems where specific cases are really useful are those where the question is more open-ended: "Take this metric space ... Is it complete? Yes/no and prove this" but here you already know that it will be complete, so you can jump in:
    [tex]\textrm{Take a cauchy sequence $(x_n)_n$ in the metric space $M := (\mathbb R^+,d)$ with $d(x,y) = |\ln(\frac{x}{y})|$. We will now prove that there is an $x \in M$ such that $x_n \to x$ in $M$:}[/tex]
    And then it's up to you ;)

    Of course, first we need to get an idea of what that limit would look like, before we can prove that it exists. As a hint, rewrite |ln(x/y)| as |ln(x) - ln(y)|. Hint #2: literally write down what it means for x_n to be a cauchy sequence, using the rewritten form of the metric in hint #1. Let this inspire you.
     
  4. Sep 26, 2011 #3
    Hm, ok.. I noticed the rewrite you mentioned, and attempted to do something with it to somehow get a lovely conclusion but I'm not sure what it is.

    I have | ln(xj) - ln(xi) |, what i need is something greater than or equal to that.. i first thought of the triangle inequality but i don't know what i would use...

    Maybe there is another trick I am not seeing.

    Thanks for your help!
     
  5. Sep 26, 2011 #4
    When you have to prove that a metric space is complete, the only space that we know that is complete, is R, so...
     
  6. Sep 26, 2011 #5
    um, can i just say since R is complete, i can find for any sequence xn in R+, an x s/t lim xn = x... and th rest follows?
     
  7. Sep 27, 2011 #6
    Given a cauchy sequence x_n in M (with M defined as in my previous post),
    what can you say about the sequence ln(x_n) in R+ with the normal metric?
     
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