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Question about metric spaces and convergence.

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\left (X,d \right)[/itex] be a metric space, and let [itex]\left\{ x_n \right\}[/itex] and [itex]\left\{ y_n \right\}[/itex] be sequences that converge to x and y. Let[itex] \left\{ z_n \right\}[/itex] be a secuence defined as [itex]z_n = d(x_n, y_n). [/itex] Show that [itex]\left\{ z_n \right\}[/itex] is convergent with the limit [itex]d(x,y)[/itex]


    2. Relevant equations



    3. The attempt at a solution
    This is as far (short) as I've got.

    We know that [itex] \left\{ x_n \right\}[/itex] and [itex] \left\{ y_n \right\}[/itex] are convergents with limits x and y.

    By definition, [itex] \left\{ z_n \right\}[/itex] is convergent to z if [itex] d(z_n, z)<\epsilon[/itex] for any [itex]\epsilon[/itex] when [itex]n>N, N>0[/itex].

    That is to say [itex] d(z_n, z)<\epsilon = d( d(x_n, y_n) , d(x,y))<\epsilon[/itex] when n>N.

    I feel clueless on where to go from here. It seems resonabl that we could get [itex] d(x_n, y_n) < d(x,y) + \epsilon [/itex] with a suitable n, since they are both convergent with the limits x and y. But then again, the whole problem seems intuitive, but i have no good idea about how to formalize this. Could anybody give me a hint or two?

    Thanks! :)
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2

    lanedance

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    Homework Helper

    how about something like

    xn, ym converge so pick e>0, then there exists N, M such that |xn-x|<e and |ym-y|<e for all n>N, m>N

    now consider d(xn,ym)... can you show d(x,y)-2e<d(xn,ym)<d(x,y)+2e
     
  4. Sep 7, 2011 #3
    Thanks. I think i almost could show it.

    Since [itex]x_n, y_n[/itex] converge, there is a N so that [itex]d(x_n, x) <\epsilon[/itex] and [itex]d(y_n, y)<\epsilon[/itex] for n>N. (We pick the largest N of the two)

    Now, since [itex]x_n, y_n[/itex] lay within less than [itex]\epsilon[/itex] distance to x, y,
    [itex]-2\epsilon + x + y < x_n +y_n < x+y+ 2\epsilon [/itex]

    How can i make the transition form the interval above to the general case of the metrics?

    [itex]-2\epsilon +d(x,y) < d(x_n, y_n) < d(x,y) +2 \epsilon [/itex] also seems arguable, since the distance from x_n to y_n will differ less than [itex]2\epsilon[/itex] from d(x,y)

    Now, even so I don't fully see how I formally make the transition to d( d(x_n, y_n), d(x,y)) After all, I don't know how d is defined.

    Thanks again! :)
     
  5. Sep 7, 2011 #4
    I think what you're missing is a metric is a function into R.

    So d(x,y) is a scalar and so is the other one. So in that case you can represent it like this:

    [itex]|d(x_{n},y_{n})-d(x,y)|[/itex]

    I'm not sure about this part, but someone can elaborate. If you treat the metric as a continuous function, we know for continuous functions lim n->inf f(x_n)=x if x_n--> x

    So then you have as you've defined the metric as a continuous function

    lim d(x_n,y_n)-d(x,y)=d(x,y)-d(x,y) |= 0 (as n-> inf)

    its clear intuitively but it is a pain to formalize and it also depends how strict your professor is..

    to formalize it, i think it would be alright to say:
    Choosing the n to ensure both x_n and y_n are within epsilon away from x and y we have:
    [itex]|d(x_{n},y_{n})-d(x,y)|=|d(x,y)-d(x,y)=0<\epsilon[/itex]

    I am pretty sure using the fact that metrics are continuous, then you can simply replace x_n and y_n with x and y if n is sufficiently large. Any opinions?
     
    Last edited: Sep 7, 2011
  6. Sep 7, 2011 #5
    Thanks once again!

    I was wondering about just that: [itex] d(x_n,y_n), d(x,y) =|d(x_{n},y_{n})-d(x,y)| [/itex] Are you certain about this? It would definitely make it simpler, but is not |A-B| the definet metric in the eucledian plane?

    Edit:

    If we allow limits, i agree that once we know that [itex] lim_{n-> \inf} x_n, y_n = x, y [/itex] one could argue that [itex] lim_{n->\inf} d(x_n, y_n) = d(x,y) [/itex] , and problem solved.

    But this problem was given in a problem set where the theme was to use the definition of metrics.

    Edit: Sorry, maybe i misunderstood. So since the two are scalars, it would be ok to define the "distance" between thos two scalars as |A-B|, independantly of the metrics definition? That would make alot of sense. Thanks :)
     
    Last edited: Sep 7, 2011
  7. Sep 7, 2011 #6
    I added more to my post.

    All metrics map into [itex]\mathbb{R}[/itex]. So therefore d(x,y) is just a real number. So you can use |d(x_n,y_n)-d(x,y)|. Yes, |A-B| is the standard metric on R, and this fits into that form. In this case |A-B| simple means absolute value. Ex: |3-5|=2.


    EDIT:

    Absolute value in this context is still a metric, but to look more abstract and fancy you could call one [itex]d_{(X,d)}(x,y)[/itex] for the distance in (X,d) and when you convert it to R use

    [itex]d_{R}(d_{(X,d)}(x_{n},y_{n}),d_{(X,d)}(x,y))[/itex]
     
    Last edited: Sep 7, 2011
  8. Sep 7, 2011 #7
    Great! Could a solution to the problem be something along these lines?:

    Since [itex]x_n \rightarrow x , y_n \rightarrow y[/itex], we know that [itex]d(x_n, x)<\epsilon, d(y_n, y)<\epsilon [/itex] for any n>N.

    And since [itex]x_n, y_n[/itex] lay within an [itex]\epsilon[/itex] interval from x, y for n>N,
    [itex]-2\epsilon +d(x,y) <d(x_n, y_n)< d(x,y) +2\epsilon[/itex]
    [itex]-2\epsilon <d(x_n, y_n) -d(x,y) < 2\epsilon \Rightarrow |d(x_n, y_n) -d(x,y)|<2\epsilon[/itex]
    Now, this must be true for any [itex] \epsilon[/itex]: We can always get [itex] |d(x_n, y_n) -d(x,y)|<\epsilon[/itex] by choosing a number M so that [itex]d(x_n, x)<\frac{\epsilon}{2}, d(y_n, y)<\frac{\epsilon}{2}[/itex] when n>M.
     
  9. Sep 7, 2011 #8
    I think there is a better way to formalize it, which I've been thinking about. The only thing you know about metrics is they must follow the triangle inequality

    So you know by the triangle ineq:
    [itex]d(x_{n},y_{n})\leq d(x{n},x)+d(x,y_{n})\leq d(x{n},x)+d(x,y)+d(y,y_{n})[/itex]

    So from there assume n is large enough, you get what you're trying to say.

    If you start from
    [itex]|d(x_{n},y_{n})-d(x,y)|[/itex]
    you'll be golden. Just make a note that absolute value is R's metric as we said above, just to be thorough.
     
  10. Sep 9, 2011 #9
    Hi, thanks for all the good help!
    I've been thinking about what you said, and I still feel a bit unsure.

    When x_n is a sequence, x is a limit if [itex]d(x_n, x)<\epsilon[/itex] when n>N, by definition.

    So let's say that d is defined as 2|x-y|.
    If we let [itex]z_n=d(x_n, y_n)[/itex] where [itex]z_n, x_n, y_n[/itex] are sequences, and [itex]x_n, y_n[/itex] have limits x,y.

    Then, for [itex]z_n[/itex] to converge to d(x,y), [itex]d( d(x_n, y_n), d(x,y) ) <\epsilon[/itex] must be true.
    But [itex]d( d(x_n, y_n),d(x,y) ) = 2| d(x_n, y_n) - d(x,y) | [/itex] by our definition of d.
    Is this contrary to what we stated above; that [itex]d( d(a,b), d(c,d)) = |d(a,b) - d(c,d)| [/itex]??

     
    Last edited: Sep 9, 2011
  11. Sep 9, 2011 #10

    Where did that plus sign come from?
     
  12. Sep 9, 2011 #11
    Sorry. It was a minus sign in my head. I'll edit it.
     
  13. Sep 9, 2011 #12
    It looks like your confusing is coming from defining it as 2|x-y| and then the 2 not being on what we did. I was thinking of d(x,y)=|x-y| and there was no 2. Either way even if there was a 2 you could choose epsilon such that it doesn't matter.
     
  14. Sep 9, 2011 #13
    Ok. But |x-y| is one metric. The general case d(x,y) could be anything as long as it follows the four conditions of a metric.
     
  15. Sep 9, 2011 #14
    We only used properties available from the metric. For it to work with every metric you could say this:

    by the triangle inequality on (X,d) (from my earlier post) we have:
    [itex]d_{R}(d_{(X,d)}(x_{n},y_{n}),d(x,y)) \leq d_{R}(d_{(X,d)}(x_{n},x)+d_{(X,d)}(x,y)+d_{(X,d)}(y,y_{n}),d(x,y))[\itex]
    Now remember every d_{(X,d)}(?,?) are real numbers so this still makes sense and assuming the correct n is chosen and we have:
    [itex]=d_{R}(d_{(X,d)}(x,y)+2\epsilon),d(x,y))[\itex]
    which if you think about it, clearly equals:
    [itex]=d_{R}(2\epsilon,0)=2\epsilon[\itex]

    If you chose [itex]\epsilon=\frac{\Epsilon}{2}\ \text{for some}\ \Epsilon>0[\itex], then it's finished.
     
  16. Sep 10, 2011 #15
    I think dillinger is right. If you can convince yourself that the limit of z[itex]_{n}[/itex] is going to be the d(x,y) then you just use the triangle inequality he gives ^ there. Basically you're going to prove that |[itex]d(x_{n},y_{n})-d(x,y)| < \epsilon[/itex]. Since you know x[itex]_{n}[/itex] and y[itex]_{n}[/itex] converge, you can say d(x[itex]_{n}[/itex],x) < [itex]\epsilon[/itex]/2 and d(y[itex]_{n}[/itex],y) < [itex]\epsilon[/itex]/2.

    So using the triangle inequality dillinger said, we can say that:

    |[itex]d(x_{n},y_{n})-d(x,y)| \leq [/itex][itex]d(x_{n},y_{n})-d(x,y) \leq d(x{n},x)+d(x,y_{n})-d(x,y) \leq d(x{n},x)+d(x,y)+d(y,y_{n})-d(x,y) = d(x{n},x)+d(y,y_{n})[/itex] < [itex]\epsilon[/itex]/2 + [itex]\epsilon[/itex]/2 = [itex]\epsilon[/itex]
     
  17. Sep 13, 2011 #16
    Yes! I agree guys. Thanks alot for the help and patience :)
     
  18. Apr 7, 2012 #17
    Unfortunately, |[itex]d(x_{n},y_{n})-d(x,y)| \leq [/itex][itex]d(x_{n},y_{n})-d(x,y)[/itex] is an incorrect use of the triangle inequality. There is no guarantee that the value on the right is positive, so the inequality sign should be switched, which invalidates the result.
     
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