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.

something to dowith 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 21:56overtness(ncatlab.org/nlab/show/overt+space) 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 alogicaldual to compactness, by swapping quantifiers $\forall \leftrightarrow \exists$. – David Roberts Feb 4 '11 at 22:27thedual 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