complete space

A space is complete (or Cauchy complete) if every sequence/net/filter that should converge really does converge. We identify those that should converge as the *Cauchy* sequences/nets/filters. A space that is not complete has gaps that may be filled to form its *completion*; it is rather natural to make the space Hausdorff at the same time.

Forming the completion of a Hausdorff space is an important example of completion in the general sense.

A space (which may be a metric space, a Cauchy space, or anything in between) is **Cauchy complete** if every Cauchy filter converges, equivalently if every Cauchy net converges. A space is **sequentially complete** if every Cauchy sequence converges. Note that a sequentially complete metric space must be complete, but this does not hold for more general spaces.

A compact space is necessarily complete. A space is called **precompact** if its completion (see below) is compact. For metric spaces (or even uniform spaces), there is a natural notion of a totally bounded space, and we have the theorem that a space is totally bounded if and only if it is precompact. Similarly, a space is compact if and only if it is both complete and totally bounded (or more generally, both complete and precompact). Thus the purely topological property of compactness is the conjunction of the nontopological properties of completeness and total boundedness. In some constructive approaches to analysis (particularly Bishop's school), this is taken as the *definition* of compactness (which thereby is no longer a topological property).

I should clarify here exactly which foundational axioms are needed to prove that a complete totally bounded space is compact in the localic sense (the converse is easy) and why some constructivists actually accept this theorem and therefore have no need to redefine βcompactβ. βToby

The set $\mathcal{C}X$ of Cauchy filters on a space $X$ may naturally be given the same sort of structure as $X$ itself has, and this space will be complete. Exactly how to do this depends on what structure $X$ is supposed to have, of course (and one can make the statement false in general by requiring something artificial as the structure in question, most extremely the structure of being a specific non-complete space). But the general idea is this: every point in $X$ generates a principal ultrafilter (consisting of those sets to which the point belongs), so there is a natural map from $X$ to $\mathcal{C}X$. Furthermore, this map is a morphism of the appropriate structure, which in particular makes it Cauchy-continuous (preserving Cauchy filters) and continuous (preserving limits). So all of the limits in $X$ still exist in $\mathcal{C}X$, but now each Cauchy filter in $X$ (having become a Cauchy filter in $\mathcal{C}$) has a limit as well. There is a problem that $\mathcal{C}X$ is rather larger than necessary; for example, all of the filters that converge to a give point in $X$ (not just the free ultrafilter at that point) exist in $\mathcal{C}X$ and converge to one another. But you can take a quotient of $\mathcal{C}X$ to make it Hausdorff or (especially if $X$ was not Hausdorff to begin with) to leave in just as much redundancy as $X$ has but no more.

(details to come, if I get around to it)

We have a picture like this, where $X$ is the original space, $\mathcal{H}$ gives a Hausdorff quotient, and an overline indicates completion:

$\array {
& & \overline{X} \\
& \hookrightarrow & & ↠ \\
X & & & & \mathcal{H}\overline{X} \cong \overline{\mathcal{H}X} \\
& ↠ & & \hookrightarrow \\
& & \mathcal{H}X
}$

(Here the arrows are drawn horizontally to put styles on them; they should all be diagonal in the only possible way.)

When Bill Lawvere interpreted (in tac, 1986) metric spaces as certain enriched categories, he found that a metric space was complete if and only if every adjunction of bimodules over the enriched category is induced by an enriched functor. Accordingly, this becomes the notion of Cauchy complete category. (Note that one *must* say βCauchyβ here, since this is *weaker* than being a complete category, which is based on an incompatible analogy.)

Revised on June 11, 2012 20:40:11
by Stephan Alexander Spahn
(79.227.160.174)