Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that $A\subset \mathbb{R}^{n}$ is bounded iff every sequence in $A$ has a Cauchy sub-sequence? Also in a more general setting, does such a property have a name attached to it or does it have any significance at all? What happens if we drop $\mathbb{R}^{n}$ out and replace it with any metric space $X$?

Here's my thoughts about the first question. I left out the smallest details since it turned out pretty long.

Assume $A\subset \mathbb{R}^{n}$ is bounded and choose a sequence $(x_{n})_{n\in \mathbb{N}}$ of arbitrary elements from $A$. Since $A$ is bounded we can find a large enough closed cube $K\subset \mathbb{R}^{n}$ which covers it (e.g. $K=\{x\in \mathbb{R}^{n}: x_{i}\leq k \forall i\in \{1,...,n\}\}$ with a large enough $k>0$). Splitting $K$ to $2^{n}$ (closed) sub-cubes there must exist a sub-cube $S_{1}\subset K$ which contains members of the sequence with infinitely many indices (since there are only finite amount of sub-cubes in total). Splitting $S_{1}$again to $2^{n}$ sub-cubes, with the same logic we can find $S_{2}\subset S_{1}$. And eventually, for every $k\in \mathbb{N}$ we find $S_{k+1}\subset S_{k}$, and they are all compact subsets of $\mathbb{R}^{n}$ by Heine-Borel. The diameter of the cubes $S_{k}$ goes to zero as $k \rightarrow \infty$, so the intersection $S:=\cap_{k=1}^{\infty}S_{k}$ (which is non-empty and compact) must be a singleton. For every neighborhood $U$ of $S$ there exists $i\in \mathbb{N}$ so that $S_{k}\subset U$ for all $k\geq i$. On the other hand, every $S_{k}$ contained infinitely many members of $(x_{n})_{n\in \mathbb{N}}$ so $S$ is its cluster point, or more precisely the point in $\mathbb{R}^{n}$ which belongs to $S$, and thus there exists a sub-sequence which converges to it. However, there is no guarantee that it would be a member of $A$, so what is left in our hands is that $(x_{n})_{n\in \mathbb{N}}$ has a Cauchy sub-sequence, and that's it?

Now as a second thought, I guess I could have used the compactness of the closure of $A$, find the sub-sequence that way and get the same results with less hassle.

But the other way seems more clear, if $A$ would not bounded we could find a sequence $(x_{n})_{n\in \mathbb{N}}$ in $A$ such that $d(x_{1},x_{k})=k$ for all $k\in \mathbb{N}$, which again could not have a Cauchy sub-sequence and we get a contradiction.

Any improvements to this and answers to my other questions? What happens in a more general setting?

share|improve this question
    
You can choose a subsequence $x_{n_k}$ such that $x_{n_k}\in S_k$. This subsequence will be Cauchy since the diameter of $S_k$ goes to $0$ and the $S_k$ are decreasing. Indeed, it's possible that the limit is not in $A$ if $A$ is not closed (for example the sequence $x_k=\frac 1k$ for $A=(0,1)$, but it doesn't matter. For the second question, look for example at the unit ball of an infinite dimension Hilbert space. –  Davide Giraudo Jan 8 '12 at 14:43
    
Look up the notion of total boundedness. In $\mathbb{R}^n$ boundedness is equivalent to total boundedness. –  t.b. Jan 8 '12 at 14:45
    
@:Davide: Alright, thanks alot. @t.b.: Is this due to the fact that the closure of every bounded set is compact? –  Thomas E. Jan 8 '12 at 14:53
    
Yes. I think Chris's answer addresses this nicely. –  t.b. Jan 8 '12 at 15:02

1 Answer 1

up vote 3 down vote accepted

A set $A$ in a metric space is called totally bounded if every sequence has a Cauchy subsequence (note: this is not the usual definition of total boundedness, but is equivalent to it).

In $\mathbb{R}^n$, as you observe, any bounded set is totally bounded, but this is not true in general. A standard counterexample is $\mathbb{R}$ with the metric $d(x,y)=\min(|x-y|,1)$. Clearly $\mathbb{R}$ is bounded with this metric, but $1, 2, 3, \ldots$ has no Cauchy subsequence. A less artificial example is the unit ball of any infinite-dimensional normed space.

I don't know if there's a standard name for spaces where your result is true; that is, where every bounded set is totally bounded.

share|improve this answer
    
Thanks. Very cool counter-example. Your metric $d$ also seems to generate the same topology as the Euclidean metric. edit: nvm my second question. –  Thomas E. Jan 8 '12 at 14:59
    
Last sentence: spaces in which every bounded set is totally bounded are quite commonly called proper metric spaces (because this is equivalent to the metric function $d(x,\cdot): X \to \mathbb{R}$ being proper). Some people call them metric spaces with the Heine-Borel property. –  t.b. Jan 8 '12 at 15:01
    
Alright, thanks for the info t.b. –  Thomas E. Jan 8 '12 at 15:05
    
@t.b.: The article you link to gives a different definition of "proper". Are both in common use? –  Chris Eagle Jan 8 '12 at 16:01
    
Sorry I was having complete metric spaces in mind for which the two definitions are equivalent, so my comment is somewhat irrelevant. The definition of Wikipedia is the commonly used one. The property "bounded iff totally bounded" is equivalent to the completion being proper and my comment is only correct for complete spaces. –  t.b. Jan 8 '12 at 16:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.