Why is support of this function compact?

21 Views Asked by At

In these notes under Lemma 4.10, the paper makes the claim:

Lemma 4.10. Suppose $M$ is a manifold. Then there is a basis $\{U_α| α ∈ A\}$ such that:

  1. $\overline{U_{\alpha}}$ is compact and
  2. For each $\alpha \in A$ there is a smooth function $\phi_α : M → R$ such that $\phi_α(x) = 0$ if $x \not\in U_α$ and $\phi_α(x) > 0$ if $x ∈ U_α$.

For example, on $R$ the support of $\phi_{\alpha}$ is compact so the function is zero on $(m, \infty)$ for some $m$.

So, I think they are saying, restrict $\phi_{\alpha} : R \to R$. I thought by assumption the support would be $U_{\alpha}$ (by 2), an open set. But then $R$ is a metric space and compactness would imply that $U_{\alpha}$ is closed.

What am I interpreting incorrectly?