Homework 6

Exercise: Give an example of each of the following, or prove that it cannot exist.

  1. A sequence that contains subsequences which converge to every point in the infinite set \{1/n: n\in\mathbb{N}\}
  2. A sequence that contains subsequences converging to every point in the interval [0,1]
  3. A sequence which has a bounded subsequence, but no convergent subsequence.


  1. You can use the example for (b). Also there is a cheap way to get the example. For instance, 1/1, 1/1, 1/2, 1/1, 1/2, 1/3, 1/1, 1/2, 1/3, 1/4, 1/1, 1/2, 1/3, 1/4, 1/5, \ldots. Each time, you go a bit further and rewind.
  2. Listing the rationals in [0,1] suffices. For instance, 0, 1, 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, \ldots.
  3. It is not possible in \mathbb{R}^k. Because the bounded subsequence would have a convergent sub-subsequence. However, in a more general setting, it is possible. Think of \mathbb{N} equipped with the discrete metric and a_n = n. Then \{a_n\} itself is bounded and it doesn’t have any convergent subsequences. Also you may use the examples in Problem 3.

Exercise: Find all subsequential limits of the sequence \{s_n\} defined by
s_1=0; s_{2n}=\frac{s_{2n-1}}{2}; s_{2n+1}=\frac{1}{2}+s_{2n}.

Proof. The odd-numbered terms converge to 1 and the even-numbered terms converge to 1/2.

Exercise: For each of the following metric spaces, given an example of a Cauchy sequence which does not converge. In each case, assume d(x,y)=|x-y|.

  1. X=\mathbb{Q}
  2. X=\mathbb{R}\backslash\mathbb{Q}
  3. X=(-1,1)


  1. Listing rationals approaching \pi suffices. For instance, 3, 3.1, 3.14, 3.141, 3.1415, \ldots.
  2. Listing irrationals approaching 0 suffices. For instance, \frac{\pi}{1}, \frac{\pi}{2}, \frac{\pi}{3}, \ldots.
  3. Any sequence converging to -1 suffices. For instance, -1+\frac{1}{1}, -1+\frac{1}{2}, -1+\frac{1}{3}, \ldots.

Exercise: Suppose that \{p_n\} is a Cauchy sequence in a metric space X, and some subsequence of \{p_n\} converges to p\in X. Prove that then the full sequence converges to p. That is, \lim_{n\rightarrow\infty}p_n=p.

Proof. For any \epsilon>0, there is some ‘good’ subsequence of \{p_n\} in which every element is in (p-\epsilon/2, p+\epsilon/2). This is because some subsequence converges to p, so its tail must lie in the interval above. On the other hand, since \{p_n\} is Cauchy, there is N such that if m,n\geq N then d(p_m, p_n)<\epsilon/2. Let p_n be one of the elements in the ‘good’ subsequence. Then for all m\geq N, d(p_m,p)\leq d(p_m,p_n)+d(p_n,p)<\epsilon/2+\epsilon/2=\epsilon. That is, \lim_{n\rightarrow\infty}p_n=p.

Exercise: Suppose \{a_n\} and \{b_n\} are Cauchy sequences in \mathbb{R} with metric d(x,y)=|x-y|.

  1. Prove that the sequence \{a_n+b_n\} is Cauchy.
  2. Prove that the sequence \{a_n b_n\} is Cauchy.


  1. For every \epsilon > 0 there is a positive integer N_1, N_2\in\mathbb{N} with the following property: If n\geq N_1 and m\geq N_1, then d(a_n, a_m)<\epsilon/2, and if n\geq N_2 and m\geq N_2, then d(b_n, b_m)<\epsilon/2. Now let N=\max(N_1, N_2). Then for any n, m\geq N, we have d(a_n+b_n,a_m+b_m)\leq d(a_n,a_m)+d(b_n,b_m)<\epsilon. This shows that \{a_n+b_n\} is Cauchy.
  2. First we claim that \{a_n\} is bounded. This is because \{a_n\} is Cauchy, so there is N, such that for any n\geq N, d(a_n, a_N)<1. Let A=\max(|a_1|, |a_2|, \ldots, |a_N|)+1. Then A is a bound for \{a_n\}. Also we can find a bound, say B, for \{b_n\}. Then pick any M>A,B. Follow the argument used in (a), for every \epsilon > 0, we can find N such that if n, m\geq N, then d(a_n, a_m)<\frac{\epsilon}{2M} and d(a_n, a_m)<\frac{\epsilon}{2M}. Hence if n,m\geq N, then d(a_nb_n, a_mb_m)\leq d(a_nb_n, a_mb_n)+d(a_mb_n, a_mb_m)=|b_n|d(a_n,a_m)+|a_m|d(b_n.b_m)< B\times\frac{\epsilon}{2M}+A\times\frac{\epsilon}{2M}<\epsilon. This tells us that \{a_nb_n\} is Cauchy.

Leave a comment

Your email address will not be published. Required fields are marked *