# Something about boundedness in $\mathbb{R}^{n}$ and also in general..

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?

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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? – T. Eskin Jan 8 '12 at 14:53
Yes. I think Chris's answer addresses this nicely. – t.b. Jan 8 '12 at 15:02

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.
Thanks. Very cool counter-example. Your metric $d$ also seems to generate the same topology as the Euclidean metric. edit: nvm my second question. – T. Eskin 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