Use the source, Luke.

Specifically, chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality). You may be particularly interested in:

**Theorem 16.10** If $X$ is Lindelof and $\mathcal{S}$-regular, then $X$ is $\mathcal{S}$-paracompact. In particular, nuclear Frechet spaces are $C^\infty$-paracompact.

For loop spaces (and other mapping spaces with compact source), the simplest argument for Lindelof/paracompactness that I know of goes as follows:

- Embed $M$ as a submanifold of $\mathbb{R}^n$.
- So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$.
- $L\mathbb{R}^n$ is metrisable.
- So $L M$ is metrisable.
- Hence $L M$ is paracompact.

(Paracompactness isn't inheritable by **all** subsets. Of course, if you can embed your manifold as a *closed* subspace then you can inherit the paracompactness directly.)

I use this argument in my paper on *Constructing smooth manifolds of loop spaces*, Proc. London Math. Soc. **99** (2009) 195–216 (doi:10.1112/plms/pdn058, arXiv:math/0612096) to show that most "nice" properties devolve from the model space to the loop space for "nice" model spaces (smooth, continuous, and others). See corollary C in the introduction of the published version.

(I should note that the full statement of Theorem 16.10 (which I did not quote above) is not quite correct (at least in the book version, it may have been corrected online) in that the proof of the claim for strict inductive sequences is not complete. I needed a specific instance of this in my paper *The Smooth Structure of the Space of Piecewise-Smooth Loops*, Glasgow Mathematical Journal, **59**(1) (2017) pp27-59. (arXiv:0803.0611, doi:10.1017/S0017089516000033) (see section 5.4.2) which wasn't covered by 16.10 but fortunately I could hack together bits of 16.6 with 16.10 to get it to work. This, however, is outside the remit of this question as it deals with spaces more general than Frechet spaces.)

On the opposite side of the equation, we have the following after 16.10:

**open problem** ... Is every paracompact $\mathcal{S}$-regular space $\mathcal{S}$-paracompact?

So the general case is not (at time of publishing) known. But for manifolds, the case is somewhat better:

**Ch 27** If a smooth manifold (which is smoothly Hausdorff) is Lindelof, and if all modelling vector spaces are smoothly regular, then it is smoothly paracompact. If a smooth manifold is metrisable and smoothly normal then it is smoothly paracompact.

Since Banach spaces are Frechet spaces, any Banach space that is not $C^\infty$-paracompact provides a counterexample for Frechet spaces as well. The comment after 14.9 provides the examples of $\ell^1$ and $C([0,1])$.

So, putting it all together: **nuclear** Frechet spaces are good, so Lindelof manifolds modelled on them are smoothly paracompact. Smooth mapping spaces (with compact source) are Lindelof manifolds with nuclear model spaces, hence smoothly paracompact.

(Recall that smooth mapping spaces without compact source aren't even close to being manifolds. I know that Konrad knows this, I merely put this here so that others will know it too.)

The chromatic number $\chi(X)$ of a topological space $X$ is related to the separation dimension $t(X)$ introduced and studied by Steinke.

The separation dimension $t(X)$ is defined inductively:

$\bullet$ $t(\emptyset)=-1$

$\bullet$ $t(X)=0$ for any space $X$ of cardinality $|X|=1$;

$\bullet$ if $|X|\ge 2$, then $t(X)\le n$ for some $n\in\mathbb N$ if for each subspace $M\subset X$ with $|M|\ge 2$ there exists a set $A\subset M$ such that $t(A)<n$ and $M\setminus A$ is disconnected;

$\bullet$ $t(X)=n$ for some integer $n\ge 0$ if $t(X)\le n$ and $t(X)\not\le n-1$.

It is easy to see that $t(X)=0$ if and only if the space $X$ is totally disconnected.

In Proposition 3.1 of his paper Steinke proved the following

**Sum Theorem:** For any subspaces $A,B$ of a topological space the union $A\cup B$ has separation dimension $t(A\cup B)\le t(A)+t(B)+1$.

This theorem implies that $t(X)+1\le\chi(X)$ for any topological space $X$.

On the other hand, by the classical Decomposition Theorem of Urysohn (this is Theorem 7.3.9 in Engelking's book "General Topology"), for a metrizable space $X$ of finite dimension $Ind(X)$ the number $Ind(X)+1$ is equal to the smallest cardinality of a partition of $X$ into subsets of large inductive dimension zero.

Since spaces of large inductive dimension zero are totally disconnected, this decomposition theorem implies that
$\chi(X)\le Ind(X)+1$
for any metrizable space $X$.
Therefore, for any metrizable space $X$ of finite large inductive dimension, we obtain the inequalities:

$$t(X)+1\le \chi(X)\le Ind(X)+1.$$

In Corollary on page 279 of his paper, Stainke proves that for each locally compact paracompact space $X$ we have the inequalities

$$dim(X)\le t(X)\le ind(X)\le Ind(X).$$

Since $dim(X)=ind(X)=Ind(X)$ for any separable metrizable space $X$, we finally conclude that

$$dim(X)=t(X)=ind(X)=Ind(X)\quad\mbox{and}\quad\chi(X)=\dim(X)+1$$

for any locally compact separable metrizable space $X$.

In particular, we obtain the following theorem answering the question of N. de Rancourt.

**Theorem 1.** For every $n\in\mathbb N$ the Euclidean space $\mathbb R^n$ has chromatic number $\chi(\mathbb R^n)=n+1.$

For general separable metrizable spaces, we have the following upper bound, which can be interesting for Set Theorists.

**Theorem 2.** Each separable metrizable space $X$ has chromatic number $\chi(X)\le\omega_1$.

*Proof.* Choose a family $(D_\alpha)_{\alpha\in\omega_1}$ of pairwise disjoint dense sets in the real line $\mathbb R$. For every countable ordinal $\alpha$ consider the set $Z_\alpha=\mathbb R\setminus\bigcup_{\alpha\le\beta<\omega_1}D_\beta$ and observe that $(Z_\alpha)_{\alpha\in\omega_1}$ is an increasing transfinite sequence of zero-dimensional subspaces of $\mathbb R$ such that $\bigcup_{\alpha\in\omega_1}Z_\alpha=\mathbb R$.

Taking into account that the cardinal $\omega_1$ has uncountable cofinality, we can show that $\{Z_\alpha^\omega\}_{\alpha<\omega_1}$ is a cover of $\mathbb R^\omega$ by $\omega_1$ many zero-dimensional subspaces, which yields the upper bound $\chi(\mathbb R^\omega)\le\omega_1$.

Since each separable metrizable space embeds into $\mathbb R^\omega$, we finally obtain the desired upper bound $\chi(X)\le\chi(\mathbb R^\omega)\le\omega_1$.

This upper bound is attained for the Hilbert cube.

**Theorem 3.** The Hilbert cube $\mathbb I^\omega=[0,1]^\omega$ has chromatic number $\chi(\mathbb I^\omega)=\omega_1$.

*Proof.* The upper bound $\chi(\mathbb I^\omega)\le\omega_1$ was proved in Theorem 2 and the lower bound $\chi(\mathbb I^\omega)>\omega$ was proved by Krasinkiewicz (see also this paper).

## Best Answer

The proofs rely, in the background, on Urysohn's Lemma, which follows from the Principle of Dependent Choices but is not provable without some Choice. It is false in the ordered Mostowski model, see Geordnete Läuchli Kontinuen by N. Brunner. More information can be found in Versions of normality and some weak forms of the axiom of choice by Howard, Keremedis, Rubin, and Rubin; the link is to the review on zbMath, the paper is behind a paywall.

AddendumIn metric spaces one candefineUrysohn functions, but there one runs into a difficulty with paracompactness: in On Stone’s theorem and the Axiom of Choice Good, Tree, and Watson prove that the paracompactness of metric spaces does not follow from the principle of Dependent Choices.SummaryFor separable metric spaces the existence of partitions of unity can be proven without Choice, because locally finite refinements can be constructed from a countable refinement, and because the continuous functions needed can be defined from the metric. In arbitrary metric spaces the proof of paracompactness needs a certain amount of choice, and in arbitrary paracompact spaces one needs choice twice: first in the proof of Urysohn's Lemma and second in the choice of a set of Urysohn functions that will yield the partition of unity.