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The impatient reader can skip my attempt at motivation and go straight my "Question formulations for the impatient."

In a failed(?) attempt at discovering something new, some years ago I toyed with the idea of dualizing the notion of a compact topological space. So starting from the "open covers have finite subcovers" formulation, I viewed covers of a space $X$ as surjective maps, to $X$, from a coproduct of subobjects.

Dualizing thus lead me to look at injective maps $X\rightarrow \prod_{i\in I} Q_i$ from $X$ to a product of various quotients $Q_i$ of $X$. So call such a thing a co-cover. Given a co-cover and subset $J\subset I$, projection yields $X\rightarrow \prod_{i\in J} Q_i$, so call that a sub-co-cover; and with $J$ finite, call it a finite sub-co-cover.

Call a co-cover $X\rightarrow \prod_{i\in I} Q_i$ of $X$ open, if for each $x\in X$ there exists some finite $K\subset I$ so that some neighborhood of $x$ maps injectively to $\prod_{i\in K} Q_i$.

Call a space opcompact (since cocompact already has a meaning) if every open co-cover has a finite sub-co-cover.

My disappointment: opcompact turns out equivalent to compact, so nothing new (yet). Proof sketch:

Compact implies opcompact:

For each $x\in X$, pick an open neighborhood $N_x$ of $x$ and a set $K$, so that $N_x$ maps injectively to $\prod_{i\in K} Q_i$. The $N_x$ form a cover with a finite subcover, and the union of the associated $K$'s determines the desired sub-co-cover.

Opcompact implies compact:

Given $X$ with an open cover $\{U_i\}$ that has no finite subcover, get an open co-cover with no finite sub-co-cover from $X\rightarrow \prod X/U_i^c$ where $X/U_i^c$ means the quotient of $X$ where the complement of $U_i$ collapses to a point.

Even with Hausdorff $X$, the "opcompact implies compact" argument may require non-Hausdorff spaces $X/U_i^c$ - we would need $X$ regular to have all these spaces Hausdorff a priori. So call a co-cover $X\rightarrow \prod_{i\in I} Q_i$ Hausdorff if it uses only Hausdorff $Q_i$. Define Hausdorff-opcompact to mean that every Hausdorff co-cover has a finite

Now my question: for Hausdorff spaces, does Hausdorff-opcompact imply compact?

Question formulation for the impatient: Must a noncompact Hausdorff space $X$ admit an infinite family of Hausdorff quotients $Q_i$ so that $X$ maps injectively into $\prod_i Q_i$ but not into any finite projection?

A parallel question arises on replacing Hausdorff by regular (and regular by normal):

Parallel question: for regular spaces, does regular-opcompact (with the obvious meaning) imply compact?

Parallel question formulation for the impatient: Must a noncompact regular space $X$ admit an infinite family of regular quotients $Q_i$ so that $X$ maps injectively into $\prod_i Q_i$ but not into any finite projection?

As always I welcome all pertinent remarks/answers on the general circle of ideas, so not only focused answers to the questions I've actually posed.

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In the language of ultrafilters, a space is compact iff every ultrafilter converges to at least one point. So in some sense the "dual" of compactness is the Hausdorff property: every ultrafilter converges to at most one point. I learned this from Terence Tao's blog. – Qiaochu Yuan Feb 4 '11 at 20:21
@Qiaochu: I'd say that's "a" dual, not "the" dual. But, yes, I agree that that's a very natural dual notion to compactness. Another dual notion might be to develop a theory of "coultrafilters", and decide when they "nverge". – Theo Johnson-Freyd Feb 4 '11 at 21:08
@Theo Years ago a friend and I wrote an unpublished and I fear flawed paper about maximal filters in the partition lattice of a discrete set. If we'd managed to get that right we would have moved on the the topological situation. One has principal co-ultrafilters - a partition with one doubleton and all other cells singletons. And then non-principal co-ultrafilters seem to have something to do with ultrafilters (one gets "big" partition lattice filters from an ultrafilter $\mu$ by making a set in $\mu$ a cell and any every element in the complement a singleton). But not big enough. – David Feldman Feb 4 '11 at 21:56
compact and opcompact may mean the same thing for topological spaces, but they may separate when considered for locales. There is a concept called overtness ( which for spaces holds always, but not always for locales (it depends on the classicality of your foundations, or equivalently, if you are working in a topos or not. Classically all locales are overt). Overtness is in a sense a logical dual to compactness, by swapping quantifiers $\forall \leftrightarrow \exists$. – David Roberts Feb 4 '11 at 22:27
@Theo, @Qiaochu: Yes, it is very often ambiguous to talk about the dual of a concept, since most concepts have many equivalent formulations, which will typically no longer be equivalent when dualised. (The main exceptions, of course, being concepts primarily defined in explicitly categorical terms.) – Peter LeFanu Lumsdaine Feb 5 '11 at 0:40

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