I'm working through an analysis text and I've come across this exercise:

Give an example of a complete metric space $X$ and a bounded sequence $\left(x_{n}\right)$ in $X$ such that the sequence $\left(x_{n}\right)$ has no partial limit.

The reason that I'm stuck with this one is because I'm shaky about what it means for a sequence to be bounded. Does this mean that for any positive integers $n$ and $m$, the set of all possible $|x_{n}-x_{m}|$ is a bounded set of real numbers?

If so, I'm not sure how to relate the boundedness of a sequence to the completeness of a metric space.

I know that a metric space $X$ is said to be complete if every Cauchy sequence in $X$ converges to a point in $X$... what does this have to do with bounded sequences?!? Thanks in advance for any help.

EDIT: I'm going to add a couple of definitions.

A sequence is said to be bounded if there exists a constant $C\gt 0$ such that for any positive integers $n$ and $m$ the inequality $|x_{n}-x_{m}|\lt C$ holds.

A point $x$ in a metric space $X$ is said to be a partial limit of a sequence $\left(x_{n}\right)$ if for every $\epsilon\gt 0$ there are infinitely many values of $n$ for which $x_{n}\in B\left(x,\epsilon\right)$.

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    $\begingroup$ Do you want a solution to your exercise or just a definition of boundedness? $\endgroup$ – Samuel May 23 '13 at 5:40
  • $\begingroup$ @user14111: A partial limit of a sequence is the limit of some convergent subsequence of the sequence. $\endgroup$ – Samuel May 23 '13 at 6:23
  • $\begingroup$ I have added my definitions of boundedness and partial limit. I should have clarified what I was after... I don't necessarily want a solution to the exercise, but rather an explanation of why any such solution satisfies the required conditions. Thanks for your answers, folks. $\endgroup$ – Euclid's Compass May 23 '13 at 6:29
  • $\begingroup$ I don't really understand your question then. An answer to "what does the definition of a Cauchy sequence have to do with bounded sequences?" could be "they are not directly related, and therefore it is not unexpected that there exist sequences which is one but not the other". Do you want a solution, or do you want to think about it on your own? $\endgroup$ – Samuel May 23 '13 at 6:45
  • $\begingroup$ @Samuel: I'm going to sleep on it and think about it tomorrow. If I'm still having trouble I'll post here and we'll go from there if that's okay with you. Thanks for the reply, and sorry for the vague post. $\endgroup$ – Euclid's Compass May 23 '13 at 7:02

Fix any point $y_0\in X$, and define a sequence $(x_n)$ to be bounded if there exists a constant $C>0$ such that $d(x_n,y_0)\leq C$ for all $n$. This definition does not depend on $y_0$, since if you choose any other $y_1\in X$, then by the triangle inequality, $$|d(x_n,y_0)-d(x_n,y_1)|\leq d(y_1,y_0),$$ where the right-hand-side is a constant, so the sequence $(x_n)$ is bounded "with respect to $y_0$" if and only if it is bounded "with respect to $y_1$".

Hint for the exercise: Consider a space with no limit points.

  • $\begingroup$ Okay, so this is bringing me back a few sections in my text (good review). Here's my definition of a limit point: A point $x$ is said to be a ${\bf limit point}$ of a set $X$ if for every neighborhood $U$ of $x$ we have $X\cap U\setminus{x}\neq\emptyset.$ $\endgroup$ – Euclid's Compass May 23 '13 at 22:05
  • $\begingroup$ Dang, I ran out of time for my edit... I keep forgetting that hitting enter posts my comment, not goes to a new line. I'm trying to find a set of points that don't touch each other. What if we took ${1}\cup{2}\cup\cdots\cup{k}$ for some positive integer $k$, this would be a subset of the real number line that is a complete metric space, right? Does it contain any Cauchy sequences? Dang, every time I put curly braces in my tex stuff it interprets them as containers. How do you show { and } in tex? $\endgroup$ – Euclid's Compass May 23 '13 at 22:11
  • $\begingroup$ @Euclid'sCompass: That's right, the set $X=\{1,2,\ldots,k\}$ with the usual metric has no limit points. However, any infinite sequence of $X$ must have a constant subsequence, which will be Cauchy and thus convergent. You want a space with infinitely many isolated points. $\mathbb Z$ would work together with the sequence $a_n=n$, except the sequence $a_n=n$ is unbounded. But you can make it bounded by changing the metric... $\endgroup$ – Samuel May 24 '13 at 5:51
  • $\begingroup$ My gut is telling me that you're talking about the discrete metric? So could we do $X=\{1,2,\ldots\}$ with the discrete metric, and the sequence $\left(x_{n}\right)=n$? So this sequence would end up being the constant value of 1, and we're running into the same problem as before. What am I missing? $\endgroup$ – Euclid's Compass May 24 '13 at 8:19
  • $\begingroup$ @Euclid'sCompass: That's exactly the example I'm thinking of. The sequence is not constant; it is $1,2,3,\ldots$; what is constant is the distances between distinct points. It is bounded, and it has no convergent subsequence. $\endgroup$ – Samuel May 24 '13 at 12:09

Let $X=L^{\infty}(\mathbb{R})$, let $f_n \in X$ and define

$$ f_n(x) = \begin{cases} 1 & x \in [n, n+1] \\ 0 & \text{else} \end{cases} $$ The sequence $(f_n)_{n=0}^{\infty}$ is bounded by 1, but has no convergent subsequence.

  • $\begingroup$ What do you mean by $L^{\infty}\left(\mathbb{R}\right)$ ? $\endgroup$ – Euclid's Compass May 23 '13 at 6:31
  • $\begingroup$ $L^{\infty}(\mathbb{R})$ is the space of essentially bounded measurable functions on $\mathbb{R}$ under the identification that two functions are equivalent if they are equal almost everywhere. Essentially you can think of them as bounded functions on $\mathbb{R}$. See here: en.wikipedia.org/wiki/Lp_spaces If this goes beyond your curriculum you should maybe look at the other's answers. $\endgroup$ – N.U. May 23 '13 at 6:46

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