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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?

M W
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2 Answers2

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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).
Tyrone
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    Thank you very much for this extensive answer! Will need to look at it carefully in the next day or two, but this seems extremely clarifying and useful, and much more than I’d hoped for with this question. – M W Feb 14 '24 at 06:39
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    @MW I'm glad it might be useful to you. I'll be happy to try and add more references if you need them, but I think everything is contained in Engelking in some or another. – Tyrone Feb 14 '24 at 07:01
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    Thanks for tackling this Tyrone. My copy of Engelking defines paracompact as requiring Hausdorff, so I appreciate you picking through the details to get a clean result. – Steven Clontz Feb 14 '24 at 15:56
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    For the implication $4.\Rightarrow3.$, I am not sure how [1, thm 1.5.18] can be applied without knowing a priori that $X$ is normal. On the other hand, https://arxiv.org/abs/2203.03120 has good information, in particular Prop. 3.8 (argument of Mather, essentially the same as [1, Lemma 5.1.8] with an extra paragraph at the end to directly give a locally finite closed refinement. (And if desired, it easily follows that every numerable cover has a shrinking.) – PatrickR Feb 16 '24 at 18:43
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    Hi @PatrickR. Suppose that ${\xi_i^{-1}(0,1]}{i\in\mathcal{I}}$ is a locally-finite cozero covering. Put $\xi=\max_i\xi_i$ and let $\varphi_i(x)=\max{2\xi_i(x)/\xi(x)-1,0}$. Then ${\varphi_i^{-1}(0,1]}{i\in\mathcal{I}}$ is a locally-finite cozero covering and $\varphi_i^{-1}(0,1]\subseteq\overline{\varphi_i^{-1}(0,1]}\subseteq\xi_i^{-1}(0,1]$. For more details see Derwent. – Tyrone Feb 17 '24 at 00:52
  • Nice and simple! – PatrickR Feb 18 '24 at 00:17
  • You say that every numerable cover has a shrinking, but isn't it obvious from the definition? A partition of unity $(f_i){i\in I}$ is subordinate to cover $\mathcal{U} = (U_i){i\in I}$ when $\text{supp}(f_i)\subseteq U_i$. – Jakobian Sep 05 '24 at 11:36
  • @Jakobian various authors adopt different definitions of what `subordinated' means. If your definition is that the supports refine the cover, then the nontrivial fact becomes that any cover refined by a locally-finite covering of cozero sets is numerable. Conversely, if the definition of subordinated is that the cozero sets refine the cover, then shrinkability becomes the nontrivial fact (and the bonus fact is that this shrinking can be done by the supports of a locally-finite parition of unity). – Tyrone Sep 05 '24 at 12:03
  • Ah yes. I've already shown that by referring to your above comment. It's just that the lemma felt odd to me, as I thought that while locally finite partitions of unity are somewhat of a convention, that the supports and not just cozero sets of the partition of unity need to refine the cover, was somewhat uniformly accepted. – Jakobian Sep 05 '24 at 12:54
  • A way to verify that a normal open cover $\mathcal{U}$ of $X$ is numerable is to construct a continuous pseudometric $d$ on $X$ such that the balls $B_d(x, 1)$ refine $\mathcal{U}$ (see e.g. Nagami's and Kodama's Dimension theory), and then a partition of unity subordinated to the family of unit balls (this can be done since open covers of metric spaces admit subordinated partitions of unity, and hence so do pseudometric spaces). Although I don't see the converse, this suffices to prove that fully normal and paracompact + normal are equivalent. – Jakobian Nov 26 '24 at 23:49
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While answer of Tyrone is excellent, after studying this subject for a while I feel like there is something missing from it. So I'm going to complement it here. The cited proposition to an exercise by Morita doesn't exactly explain why are normal and numerable covers the same thing. I believe that the raison d'être for this is the following metrization theorem:

Theorem. Suppose that $(\mathcal{U}_n)$ is a sequence of open covers of $X$ such that if $U_1, U_2\in \mathcal{U}_{n+1}$ and $U_1\cap U_2\neq \emptyset$, then there exists $U_3\in\mathcal{U}_n$ with $U_1\cup U_2\subseteq U_3$. Then there exists a continuous pseudometric $d$ on $X$ such that $ \text{St}(x, \mathcal{U}_{n+2})\subseteq B_d(x, 2^{-n})\subseteq \text{St}(x, \mathcal{U}_n)$.

The proof is technical, but quite standard. You first define a function $\delta:X\times X\to \mathbb{R}$, where $\delta(x, y)$ corresponds to biggest $n$ for which there exists $U\in\mathcal{U}_n$ with $x, y\in U$ (the covers $\mathcal{U}_n$ shrink down, we want to measure the exact step for which the family would consider $x, y$ to be distinct). This function is not necessarily a pseudometric, but by standard procedure of considering finite sequences $x_0 = x, x_1, ..., x_n = y$ we can define $d(x, y)$ to be infimum of $\delta(x_0, x_1)+...+\delta(x_{n-1}, x_n)$ over all of such sequences, and this gives the desired pseudometric.

It follows that for any normal cover $\mathcal{U}$ there exists a continuous pseudometric $d$ such that $\{B_d(x, 1) : x\in X\}$ refines $\mathcal{U}$. Say that a set $U\subseteq X$ is $d$-open if its open with respect to the pseudometric $d$, and note that $d$-open subsets of $X$ are cozero sets of $X$. Because pseudometric spaces are paracompact, it follows that we can take a locally finite $d$-open refinement of $\mathcal{U}$, and so a locally finite cozero refinement.

For the converse one needs to know the following lemma.

Lemma. A locally finite open cover of a normal space is normal.

This follows from the following two theorems:

Theorem 1. A point-finite open cover of a normal space has a closed shrinking.

Theorem 2. If $\mathcal{V} = (V_i)$ is a closed cover, $\mathcal{U} = (U_i)$ is a locally finite open cover, then there exists locally finite open cover $\mathcal{W}$ such that $\text{St}(V_i, \mathcal{W})\subseteq U_i$.

Theorem 1 is standard and proven by transfinite induction, for theorem 2 one can take $\mathcal{W}$ to consist of sets of the form $\bigcap_{i\in J} U_i \cap \bigcap_{i\notin J} (X\setminus V_i)$ for finite $J\subseteq I$, then its just a standard verification. To see that those theorems imply the lemma, one observes that they imply that a locally finite open cover of a normal space has a locally finite open barycentric refinement, and so one can iterate the barycentric refinements to obtain a normal sequence.

Having established those, now we can tackle the converse. If $\Phi = (f_i)_{i\in I}$ is a locally finite partition of unity, then $d(x, y) = \sum_{i\in I} |f_i(x)-f_i(y)|$ defines a continuous pseudometric on $X$, and if $f_i(x) > 0$, then $B_d(x, f_i(x)/2)\subseteq \text{coz}(f_i) = \{y\in X : f_i(y) > 0\}$, and so the cozero sets of $f_i$ are $d$-open. Since pseudometric spaces are normal, from the lemma it follows that the family $\{\text{coz}(f_i) : i\in I\}$ is a normal cover of $X$.

Thus we have established:

Theorem. A cover $\mathcal{U}$ is numerable iff it's normal.

Jakobian
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