Compactly generated and paracompact $\Rightarrow$ Hausdorff?

Question

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. But I don't know what's the exact meaning of the word "paracompact" here.

I understand that a paracompact space is typically defined as a topological space in which every open cover has a locally finite refinement. However, some authors additionally use the term “paracompact” to imply the space is also Hausdorff. In other words, these authors use “paracompact space” to refer to a space that is both Hausdorff and paracompact in the usual sense. It is unclear if May requires a paracompact space to be Hausdorff.

Here's some notation from the book:

1. 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.
2. 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.
3. A compactly generated space $X$ is a weak Hausdorff $k$-space.
4. 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 $\lambda _{U}^{-1}(0,1]=U$.

If the term “paracompact” in the book also carries the additional meaning of being Hausdorff, then it’s easy to prove the proposition by partition of unity.

But I don't know if the proposition 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 ?

Answer 1

The one-point compactification of the rationals is a Fréchet-Urysohn space, each of whose compact subsets is closed. In particular, it is compactly generated. It is compact, so paracompact by your definition. But it is not Hausdorff.

Answer 2

Questions like these are easily answered with a search of pi-Base:

π-Base, Search for $k_3$-space+Paracompact+~$T_2$

Six counterexamples are listed there today, including Tyrone's example. I'll suggest $\omega_1+1$ with the endpoint doubled (S37) as another. It is compact and therefore paracompact. It is compactly generated because each point has a compact Hausdorff neighborhood.

In fact, this is a general construction: take your favorite paracompact Hausdorff locally compact space, then double a non-isolated point.