Partitions of Unity and Paracompactness: Examples and Counterexamples

For $n\in \mathbb{Z}$, let $\psi (t)=\max \{1-\mathrm{\mid }t\mathrm{\mid },0\}$ and ${\phi }_n(x)=\psi (x-n)$. The functions are continuous, take values in $[0,1]$, and have support $[n-1,n+1]$. They are subordinate to the open cover $U_n=(n-\frac{3}{2},n+\frac{3}{2})$ of $\mathbb{R}$.

If $x\in [n,n+1]$, the only possibly nonzero functions are ${\phi }_n$ and ${\phi }_{n+1}$, and ${\phi }_n(x)+{\phi }_{n+1}(x)=(1-(x-n))+(1-(n+1-x))=1.$ Thus $\{{\phi }_n{\}}_{n\in \mathbb{Z}}$ is a locally finite partition of unity subordinate to $\{U_n\}$. The support intervals show local finiteness directly: every bounded set meets only finitely many support intervals.

On $[0,1]$, take the open-in-the-subspace cover $U=(-1,\frac{3}{4})\cap [0,1]$ and $V=(\frac{1}{4},2)\cap [0,1]$. Define $\phi (x)=\max \{0,\min \{1,2-3x\}\},\psi (x)=1-\phi (x)\mathrm{.}$ Then $\phi$ and $\psi$ are continuous, nonnegative, and sum to one. Their supports are contained respectively in $[0,\frac{2}{3}]\subseteq U$ and $[\frac{1}{3},1]\subseteq V$, so they are a finite subordinate partition of unity.

The explicit pair is an instance of the existence theorem Under choice and dependent choice, every open cover of a compact Hausdorff space admits a finite subordinate partition of unity, under its stated Choice and Dependent Choice hypotheses.

In $\mathbb{R}$, let $I_n=(1\mathrm{/}(n+1),1\mathrm{/}n)$ for positive integers $n$. The intervals are pairwise disjoint, so no point lies in two intervals and the family is point-finite. Every neighbourhood of $0$, however, meets $I_n$ for all sufficiently large $n$. Therefore the family is not locally finite at $0$.

This shows that point-finiteness records membership at a point, whereas local finiteness controls intersections with a whole neighbourhood.

Every pointwise finite sum of continuous real-valued functions is continuous.

Assume the Axiom of Choice. The ordinal space ${\omega }_1$ is Hausdorff, countably compact, and noncompact by Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, ${\omega }_1$ is countably compact and sequentially compact while ${\omega }_1+1$ is compact and Every ordinal with its order topology has a basis of clopen sets, and is $T_1$, Hausdorff and regular. If it were paracompact, then Assuming countable choice, every countably compact paracompact Hausdorff space is compact would make it compact. It is therefore not paracompact.

The use of Choice includes the countable choice hypothesis of the cited ordinal compactness result.

Assuming the Axiom of Choice, every open subspace of a paracompact space is paracompact.

Assuming the Axiom of Choice, a product of paracompact spaces is paracompact.