§ Paracompact spaces

• A space is paracompact iff every open cover $\{ U_\alpha \}$has a locally finite refinement $\{ V_\beta \}$.
• That is, every $V_i$ has some $U_i$ such that $V_i \subseteq U_i$, and the set of covers $\{ V_\beta \}$ is locally finite.
• Locally finite cover $V_\beta$: every point $x \in X$ only has finitely many $V_i$ such that $x \in V_i$. Said differently, $|\{ V_i : x \in V_i \}|$ is finite.

§ Hausdorff Paracompact spaces admit partition of unity

• one liner: take locally finite refinement and bash with Urhyson lemma.

§ Compact Implies Paracompact

• Take the open cover $U$, build the finite subcover $V$.
• This is clearly a refinement, because it only has sets from $U$, and it is clearly locally finite, because there is literally only a finite number of sets in $V$.

§ compact space that is not paracompact

• Stone space, or infinite product of $\{0, 1\}$ in the product topology.
• Recall that the open sets here differ from the "full cover" at only a finite number of indexes.
• Suppose that a point $p_i$, which is $1$ at index $i$ and zero everywhere else, does not have infinite cover.
• intuition: so all but a finitely many collection of covers can choose to cover $p_i$.
• But this must happen for each $i$, so there must be soe cover that avoids $p_i$ for all $i$.
• This is impossible?
• Oh god, this proof needs baire category to "push around" an infinite intersection of dense subsets.