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.
