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Math Help - Proving a metric space with a particularly strange distance is bounded and separable.

  1. #1
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    Proving a metric space with a particularly strange distance is bounded and separable.

    Hi everyone! I'm trying to solve a problem related to metric spaces and I'm having a really hard time. Here's the given statement:
    Let (A,d) be a metric space of sequences defined as follows:
    1) Any sequence in A is a sequence of natural numbers
    2) For all (a_n)_n, a_n<a_(n+1) (strictly increasing sequence)
    3) For every sequence a in A, the limit as n→∞ of (#{j:a_j<n})/n exists. We call that limit l(a)
    Prove that (A,d) is bounded and separable.



    Now, if a and b are two sequences in A, d is the distance defined as d(a,b)= k^(-1)+|l(a)-l(b)| where k={min j : a_j≠b_j}. By the way, I've already checked this indeed defines a distance.

    The sequence (a_n)_n=(1,2,3,4,...,...) where a_n=n for all n in N is a sequence that exists in A.
    I've tried to prove that (A,d) is bounded by assuming A is unbounded and trying to arrive to an absurd. I supposed that there exists a in A such that for all M>0, d(a,x)>M for all x in A. I couldn't get to anything.

    For the part of separability I have no idea although I've tried to find a dense countable subset in (A,d). I welcome any suggestions or ideas.
    Last edited by andrewkoch; October 8th 2013 at 10:03 PM.
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    Re: Proving a metric space with a particularly strange distance is bounded and separa

    Quote Originally Posted by andrewkoch View Post
    Now, if a and b are two sequences in A, d is the distance defined as d(a,b)= k^(-1)+|l(a)-l(b)| where k={min j : a_j≠b_j}. By the way, I've already checked this indeed defines a distance.
    I do not understand the notation you are using. What is |l(a)-l(b)|? Is that supposed to mean \left( \lim_{n\to \infty} \dfrac{\left|\{j \mid a_j<n \} \right|}{n} \right) - \left( \lim_{n\to \infty} \dfrac{\left|\{j \mid b_j<n \} \right|}{n} \right)?

    Quote Originally Posted by andrewkoch View Post
    The sequence (a_n)_n=(1,2,3,4,...,...) where a_n=n for all n in N is a sequence that exists in A.
    The definition of A that you provided does not even guarantee that A is nonempty. Suppose A = \emptyset. The definition you provided for A is trivially satisfied. You would need some phrase like A is the set of all sequences of natural numbers such that...

    Quote Originally Posted by andrewkoch View Post
    I've tried to prove that (A,d) is bounded by assuming A is unbounded and trying to arrive to an absurd. I supposed that there exists a in A such that for all M>0, d(a,x)>M for all x in A. I couldn't get to anything.
    To answer this, I will assume that I am correct in my first assumption about your notation. For any strictly increasing sequence of natural numbers a, it must be that a_n \ge n for all n. So \left| \{j \mid a_j < n \} \right| \le n-1. This implies 0 < \dfrac{\left| \{j \mid a_j<n \} \right|}{n} < 1 for all n>1 (for n=1, if a_1=1, then you have 0). So, 0 \le |(a) - (b)| < |1 - 0| = 1. Then 0 < k^{-1} \le 1. So, for any two sequences, 0 \le d(a,b) < 1 + 1 = 2. So pick any sequence a_n \in A. Then \left\{b_n \in A \mid d(a_n,b_n)<2 \right\} = A.

    Quote Originally Posted by andrewkoch View Post
    For the part of separability I have no idea although I've tried to find a dense countable subset in (A,d). I welcome any suggestions or ideas.
    For each x \in (0,1), let {a_x}_n = \left\lfloor \dfrac{n}{x} \right\rfloor. Prove that {a_x}_n \in A and \lim_{n \to \infty} \dfrac{\left|\{j \mid {a_x}_j<n \} \right|}{n} = x. Next, for each x\in (0,1), let C_x = \left\{a_n \in A \mid \lim_{n \to \infty} \dfrac{\left| \{j \mid a_j<n \} \right|}{n} = x \right\}. You just proved that each C_x is nonempty. You now have a partition for A. (Edit: Almost... you need a C_0. To show that is nonempty, consider the sequence a_n = n^n. Show that sequence is in A and |(a)| = 0)

    Question: For each r \in (0,1) \cap \mathbb{Q}, is C_r countable or uncountable? I don't know the answer to this. If it is countable, then a countable union of countable sets is countable. If it is uncountable, then you need to find a countable subset of each, let's call it D_r such that if a \in A and |(a)-(d)|<\dfrac{\varepsilon}{2} for all d \in D_r, then there exists d' \in D_r such that [k(a,d')]^{-1} < \dfrac{\varepsilon}{2} where k:A\times A \to \mathbb{N} is defined by k(a,b) = \min \{j \mid a_j \neq b_j \}.
    Last edited by SlipEternal; October 9th 2013 at 05:39 AM.
    Thanks from topsquark and andrewkoch
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    Re: Proving a metric space with a particularly strange distance is bounded and separa

    Quote Originally Posted by SlipEternal View Post
    For each x \in (0,1)
    Oops, that should have been a right-half closed interval: x \in (0,1]. As you showed, the sequence a_n = n is in A, and the limit \lim_{n\to \infty}\dfrac{|\{j \mid a_j<n \}|}{n} = 1.
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    Re: Proving a metric space with a particularly strange distance is bounded and separa

    My notation is horrible, sorry about that. |l(a)-l(b)| it's what you've said but with absolute value. I put something wrong in my post, it's actually limit as n→∞ of (#{j:a_jn})/n. Thanks for your ideas, I'll think about it and try to put a complete answer of the exercise. The set C_r is a subset of the rational numbers. The set of all rational numbers is numerable, so C_r is countable.
    Last edited by andrewkoch; October 9th 2013 at 09:26 AM.
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    Re: Proving a metric space with a particularly strange distance is bounded and separa

    Quote Originally Posted by andrewkoch View Post
    Hi everyone! I'm trying to solve a problem related to metric spaces and I'm having a really hard time. Here's the given statement:
    Let (A,d) be a metric space of sequences defined as follows:
    1) Any sequence in A is a sequence of natural numbers
    2) For all (a_n)_n, a_n<a_(n+1) (strictly increasing sequence)
    3) For every sequence a in A, the limit as n→∞ of (#{j:a_j<n})/n exists. We call that limit l(a)
    Prove that (A,d) is bounded and separable.



    Now, if a and b are two sequences in A, d is the distance defined as d(a,b)= k^(-1)+|l(a)-l(b)| where k={min j : a_j≠b_j}. By the way, I've already checked this indeed defines a distance.

    The sequence (a_n)_n=(1,2,3,4,...,...) where a_n=n for all n in N is a sequence that exists in A.
    I've tried to prove that (A,d) is bounded by assuming A is unbounded and trying to arrive to an absurd. I supposed that there exists a in A such that for all M>0, d(a,x)>M for all x in A. I couldn't get to anything.

    For the part of separability I have no idea although I've tried to find a dense countable subset in (A,d). I welcome any suggestions or ideas.
    I have to correct something: It's limit n→∞ of (#{j:a_jn})/n
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    Re: Proving a metric space with a particularly strange distance is bounded and separa

    Quote Originally Posted by andrewkoch View Post
    My notation is horrible, sorry about that. |l(a)-l(b)| it's what you've said but with absolute value. I put something wrong in my post, it's actually limit as n→∞ of (#{j:a_jn})/n. Thanks for your ideas, I'll think about it and try to put a complete answer of the exercise. The set C_r is a subset of the rational numbers. The set of all rational numbers is numerable, so C_r is countable.
    That just says there are countably many sets C_r. It does not tell us if the set C_r is countable. If you look at the set of all sequences that converge to r, that set is clearly uncountable. So you would need to show that only countably many sequences in A converge to a specific point.
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