Fully normal implies paracompact without a $T_1$ assumption?

Question

It's well-known that a $T_1$ topological space is fully normal if and only if it is $T_2$ and paracompact.

It appears, looking at the proofs from Henno Brandsma's nice exposition here and here, that we can drop the $T_1$ assumption for the implication that fully normal implies paracompactness (without concluding Hausdorff of course).

Concretely, for this direction of the proof, the $T_1$ assumption appears to only be used in the implication $(3)\implies (4)$ of Lemma 1 from the second linked document, and only to assert that by regularity (which follows from $T_1$ and fully normal), the family of all open subsets of a space whose closure lies in some element of an open cover $\mathcal U$, is again an open cover.

But this fact is already true for all fully normal spaces, regardless of regularity, since a star refinement has the property that the closure of each member lies in a member of the original cover.

I want to update $\pi$-base to reflect that $T_1$ is not needed to deduce paracompactness, but I think it would be better, if possible, to cite a source that definitively clarifies this, rather than referring visitors of the site to a collection of notes split across multiple documents with advice to ignore the stated regularity requirements.

Another possibility is to rehash the whole proof here in an answer, but I'd like to avoid that if possible as it is a rather involved proof.

Does anyone have a reference that proves, or at the very least asserts, that fully normal without $T_1$ implies paracompactness?

Answer 1

Here are equivalent conditions for a space $X$ to be fully normal. I assume no separation conditions.

Proposition The following statements about a space $X$ are equivalent.

1. $X$ is fully normal (i.e. every open covering of $X$ has an open star-refinement).
2. Every open covering of $X$ has an open barycentric refinement.
3. Every open cover has a locally-finite closed refinement.
4. Every open covering of $X$ is numerable.
5. Every open covering of $X$ is normal.

Barycentric refinements are defined in the linked $\pi$-base page.

Lemma ([1, Lemma 5.1.15]) Let $\mathcal{U},\mathcal{V},\mathcal{W}$ be open covers of a space $X$ If $\mathcal{U}$ is a barycentric refinement of $\mathcal{V}$ and $\mathcal{V}$ is a barcentric refinement of $\mathcal{W}$, then $\mathcal{U}$ is a star refinement of $\mathcal{W}$. $\quad\blacksquare$

Thus $2.\Rightarrow1.$ That $3.\Rightarrow2.$ is a consequence of the following.

Lemma ([1, Lemma 5.1.13]) If $\mathcal{U}$ is an open cover of a space $X$ and $\mathcal{U}$ has a locally-finite closed refinement, then $\mathcal{U}$ has an open barycentric refinement. $\quad\blacksquare$

A cover $\mathcal{U}$ of a space $X$ is said to be numerable if it has a subordinated partition of unity. It's well-known that a cover is numerable if and only if it is refined by a locally-finite covering of cozero sets. Using this observation, the following is proved in the same way as [1, Theorem 1.5.18].

Lemma Every numerable cover has a shrinking. $\quad\blacksquare$

Here $\mathcal{U}$ is a shrinking of $\mathcal{V}$ if the family of closures $\{\overline U\mid I\in\mathcal{U}\}$ is a precise refinement of $\mathcal{V}$. In any case, this gives $4.\Rightarrow3.$

An open cover $\mathcal{U}$ is normal if it has a sequence of open star-refinement $$\dots\stackrel{\ast}{\prec}\mathcal{U}_3\stackrel{\ast}{\prec}\mathcal{U}_2\stackrel{\ast}{\prec}\mathcal{U}_1\stackrel{\ast}{\prec}\mathcal{U}.$$ Clearly every open cover of a fully normal space is normal, which is the implication $1.\Rightarrow 5.$ Thus it remains to show $5.\Rightarrow4.$ In fact, this is a consequence of a more general statement.

Proposition (Morita [1, Exercise 5.4.H]) An open cover $\mathcal{U}$ of a space $X$ is normal if and only if it numerable. $\quad\blacksquare$

This completes the proof of the opening claim.

Corollary If $X$ is fully normal, then every open cover of $X$ has a locally-finite refinement by cozero sets. In particular, $X$ is paracompact and normal. $\quad\blacksquare$

There is a converse to this which we should also notice. A space $X$ is paracompact if every open cover has a locally-finite refinement. On the other hand, every locally-finite open cover of a normal space has a locally-finite cozero refinement. This can be proved using [1, Theorem 1.5.18]: simply construct a locally-finite shrinking and then choose Urysohn functions for its members. Thus we have shown that in a space which is both normal and paracompact, every open cover is numerable.

Corollary A space is fully normal if and only if it is both paracompact and normal. $\quad\blacksquare$

We have string of implications, proofs of the first two of which are found [1].

paracompact $T_2$ $\Rightarrow$ paracompact regular $\Rightarrow$ paracompact normal $=$ fully normal $\Rightarrow$ paracompact.

Example The line with two origins is paracompact $T_1$, but not fully normal.

Example The Sierpinski dyad is fully normal $T_0$, but not regular.

Example Every pseudometrisable space is paracompact and regular, but there are non-Hausdorff examples of such spaces.

References

1. R. Engelking, General Topology, second edition, Heldermann Verlag Berlin, (1989).