In A Concise Course in Algebraic Topology by May, a proposition is stated that any open cover of a paracompact space has a numerable refinement, where the space is assumed to be compactly generated due to his standing hypothesis. Here's some notation from the book:
- A $k-$space $X$ is a space in which a subspace $U\subset X$ is closed if and only if the preimage $t ^{-1}(U)$ under any continuous function $t:C\longrightarrow X$ out of a compact Hausdorff space $C$ is closed.
- A space $X$ is said to be weak Hausdorff if the image of a continuous function $f:C\longrightarrow X$ out of a compact Hausdorff space $C$ is closed. This separation property lies between $T _{1}$ (points are closed) and Hausdorff, but it is not much weaker than the latter.
- A compactly generated space $X$ is a weak Hausdorff $k-$space.
- An open cover $\mathscr{O}$ of a space $X$ is numerable if it's locally finite and for every $U\in \mathscr{O}$ there's a continuous function $\lambda _{U}:X\longrightarrow I=[0,1]$ such that $\lambda _{U}^{-1}(0,1]=U$.
In general topology, a paracompact space is a topological space in which every open cover admits a locally finite refinement. Some authors additionally require a paracompact space to be a Hausdorff space, meaning that every two distinct points have disjoint open neighborhoods. However, without further context, it is unclear if May requires a paracompact space to be Hausdorff.
It’s easy to prove that any open cover of a paracompact Hausdorff space (not necessarily compactly generated) has a numerable refinement: a paracompact Hausdorff space $X$ must have the property ( Theorem 41.7 in Munkres' Topology )that for any indexed open cover $\left \{ U _{\alpha}\right \}$ of $X$, there exists a partition of unity $\left \{ \psi _{\alpha}\right \}$ on $X$ dominated by $\left \{ U _{\alpha}\right \}$. The sets $V_\alpha = \{x\in X : \psi_\alpha(x) > 0\}$ form a numerable refinement of $\{U_\alpha\}$.
But I don't know if the statement that any open cover of a paracompact space has a numerable refinement holds for spaces that are compactly generated, paracompact but not Hausdorff. It’s true if and only if compactly generated and paracompact $\Rightarrow$ Hausdorff: indeed, assume that every open cover of a $T _{1}$ space $X$ has a numerable refinement, and $a\ne b$ are points of $X$, typically $\left \{ X\setminus {\left \{ a\right \}} ,X\setminus {\left \{ b\right \}}\right \}$ has a numerable refinement $\mathscr{O}$. Assume that $a\in U\in \mathscr{O}$, the open set $\left \{ \lambda _{U}(x)>\lambda _{U}(a)/2\right \}$ and the open set $\left \{ \lambda _{U}(x)<\lambda _{U}(a)/2\right \}$ are seperated. So $X$ is Hausdorff.
So the question is :
Must a compactly generated and paracompact space be Hausdorff ?
If the answer is yes, then the statement that any open cover of a paracompact space has a numerable refinement is correct for any interpretation of “paracompact”. If the answer is no, then “a paracompact space” must be understood to mean “a paracompact Hausdorff space” for the statement to hold.