Definition 1

A topological space $X$ is said to be quasi-compact if it satisfies the following axiom:

(C) Every filter on $X$ has at least one cluster point.

A topological space is said to be compact if it is quasi-compact and Hausdorff.

It follows immediately from this axiom that if $f$ is a mapping of a set $Z$ into a quasi-compact space $X$, and $\mathfrak{F}$ is any filter on $Z$, then $f$ has at least one cluster point with respect to $\mathfrak{F}$. In particular, every sequence of points of a quasi-compact space has at least one cluster point; but this condition is not equivalent to (C) (Exercise 11).

We give three axioms each of which is equivalent to axiom (C):

(C') Every ultrafilter on $X$ is convergent.

(C') $\Longrightarrow$ (C): If $\mathfrak{F}$ is a filter on $X$ then there is an ultrafilter finer than $\mathfrak{F}$ (§ 6, no. 4, Theorem 1). Since this ultrafilter converges to a point $x$, $x$ is a cluster point of $\mathfrak{F}$.

(C) $\Longrightarrow$ (C'): For if an ultrafilter has a cluster point then it converges to this point (§ 7, no. 2, Corollary to Proposition 4).

If $f$ is a mapping of a set $Z$ into a quasi-compact space $X$, and $\mathfrak{U}$ is an ultrafilter on $Z$, $f$ has at least one limit point with respect to $\mathfrak{U}$ (§ 6, no. 6, Proposition 10).

(C'') Every family of closed subsets of $X$ whose intersection is empty contains a finite subfamily whose intersection is empty.

(C) $\Longrightarrow$ (C''): Let $\mathfrak{G}$ be a family of closed subsets of $X$ with empty intersection. If every finite subfamily of $\mathfrak{G}$ has a non-empty intersection, then $\mathfrak{G}$ generates a filter (§ 6, no. 2, Proposition 1) which has a cluster point by hypothesis. This point belongs to all the sets of $\mathfrak{G}$ (since they are closed); so we have a contradiction.

$(C'') \implies (C)$: For if $(C)$ is false then there is a filter $\mathfrak{F}$ on $X$ which has no cluster point; hence the closures of the sets of $\mathfrak{F}$ form a family of closed subsets of $X$ contradicting axiom $(C'')$.

$(C''')$ (Axiom of Borel-Lebesgue) Every open covering of $X$ contains a finite open covering of $X$.

$(C''') \iff (C'')$ by taking complements.

If $X$ is quasi-compact, then every locally finite covering $\mathfrak{R}$ of $X$ is finite. For each point of $X$ has an open neighbourhood which meets only a finite number of sets of $\mathfrak{R}$, and by $(C''')$ a finite number of these neighbourhoods covers $X$.