Results 1 to 6 of 6

Math Help - Metric Space

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    394

    Metric Space

    Show that the limit of a convergent sequence of distinct points in a metric space is a limit point of the range of the sequence. Give an example to show that this is not true if the word "distinct" is omitted.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by aliceinwonderland View Post
    Show that the limit of a convergent sequence of distinct points in a metric space is a limit point of the range of the sequence. Give an example to show that this is not true if the word "distinct" is omitted.
    Let \{x_n\} is a sequence. And X be the range of the sequence. Now a point x\in X is a limit point if for any r > 0 we have that B(x,r)\cap X has infinitely many points. Thus, we see that if x = \lim x_n is a limit point because for any r > 0 pick \epsilon = r and so d(x_n,y) < \epsilon for n\geq N where N is a natural number. Thus, \{x_N,x_{N+1},x_{N+2}, ... \} is contained in B(x,r)\cap X. However, this set is infinite since \{x_n\} is a sequence of distinct points.

    For a counter example think of konstant sequences.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2008
    Posts
    71
    Mr perfect Hacker and aliceinwonderland YOU are both mistaken in the definition of x being an accumulation point of the sequence { \ x_{n}}

    SERGE LANG in his book analysis I ,and on page 32 ,comes to justify me on that >

    He writes and i quote:


    "LET { \ x_{n}} be a sequence and x a No.We shall say that x is apoint of accumulation (or limit pt) of the sequence if given ε there exists infinitely many integers n such that:
    .............................................| \ x_{n}-x|<ε............................................... ..........................................

    AND then he goes on to give couple of examples:


    1) The sequence {1,1,1,.............} has one point of accumulation namely 1


    2) The sequence {1,1/2,1,1/3,1,1/4............}has two points of accumulations,namely 1 and 0.

    3) The sequence { 1,2,3,.................} has no point of accumulation."


    So the word "distinct" is of no importance as the above definition and examples clearly show

    Another working definition , since the phrase infinitely many is of no substance in a proof, is the following:


    x is a limit point of \ x_{n} iff for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No m\geq k such that:


    ...................................| \ x_{m}-x|<ε............................................... ...............................................


    Now based on that definition i will try to give a proof that:


    If lim\ x_{n} = x ,then, for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No m\geq k such that:


    ...................................| \ x_{m}-x|<ε............................................... ...............................................

    Let ε>0 and kεΝ,now since lim\ x_{n} = x there exists a natural No r ,rεΝ such that.

    ..............................| \ x_{n}-x|<ε...for all n \geq r.................................................. .....................

    And now we distinguish two cases :


    CASE 1  k\geq r.

    CASE 2 r>k.


    IN case 1 we choose  m\geq k and in case we choose  m\geq r

    Considering now case 1 ,let  m\geq k then we have  m\geq r and therefor .


    ................................| \ x_{m}-x|<ε............................................... ..............................................


    NOTE the above proof was done in the metric space of real Nos .It can be generalized in any metric space ( X,d) by substituting the absolute value in real Nos by the general metric ,d,and the word, No by the word ,point

    For e,g | \ x_{n}-x| by d(\ x_{n},x)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by archidi View Post
    1) The sequence {1,1,1,.............} has one point of accumulation namely 1
    I think if the sequence A = {1,1,1,.............} (I assume no element except 1 is in A) is in a metric space, 1 is not a limit point of A. Since any open set containing 1 should intesect A distinct from 1, which is impossible.
    Last edited by aliceinwonderland; January 10th 2009 at 11:12 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    well, the two definitions you gave are indeed equivalent. but did you read the question carefully? what do you think they wanted to prove?

    next,

    Quote Originally Posted by archidi View Post
    Now based on that definition i will try to give a proof that:


    If lim\ x_{n} = x ,then, for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No m\geq k such that:


    ...................................| \ x_{m}-x|<ε............................................... ...............................................
    why do you have to prove this? this is just a part of the definition. i mean, there is nothing to prove in here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,569
    Thanks
    1409
    Quote Originally Posted by archidi View Post
    Mr perfect Hacker and aliceinwonderland YOU are both mistaken in the definition of x being an accumulation point of the sequence { \ x_{n}}
    But the problem was about limit points, not accumulation points. Neither thePerfectHacker nor aliceinwonderland said anything about "accumulation points".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 8th 2011, 02:16 PM
  2. Is it a metric space?
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 3rd 2011, 06:44 PM
  3. metric space
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: October 8th 2010, 06:13 AM
  4. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2010, 02:04 PM
  5. Sets > Metric Space > Euclidean Space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 25th 2010, 10:17 PM

Search Tags


/mathhelpforum @mathhelpforum