see also algebraic topology, functional analysis and homotopy theory
topological space (see also locale)
Kolmogorov space, Hausdorff space, regular space, normal space
compact space (sequentially compact, countably compact, paracompact, countably paracompact, locally compact, strongly compact)
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
compact spaces equivalently have converging subnet of every net
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
Theorems
A cover which is locally a finite cover.
(locally finite cover)
Let $(X,\tau)$ be a topological space.
An open cover $\{U_i \subset X\}_{i \in I}$ of $X$ is called locally finite if for all points $x \in X$, there exists a neighbourhood $U_x \supset \{x\}$ such that it intersects only finitely many elements of the cover, hence such that $U_x \cap U_i \neq \emptyset$ for only a finite number of $i \in I$.
Any open cover defined by a (generalized) partition of unity has a locally finite shrinking. (…details…)
The following says that if there exists a locally finite refinement of a cover, then in fact there exists one with the same index set as the original cover.
(locally finite refinement induces locally finite cover with original index set)
Let $(X,\tau)$ be a topological space, let $\{U_i \subset X\}_{i \in I}$ be an open cover, and let $(\phi \colon J \to I, \{V_j \subset X\}_{j \in J})$, be a refinement to a locally finite cover.
Then $\left\{ W_i \subset X \right\}_{i \in I}$ with
is still a refinement of $\{U_i \subset X\}_{i \in I}$ to a locally finite cover.
It is clear by construction that $W_i \subset U_i$, hence that we have a refinement. We need to show local finiteness.
Hence consider $x \in X$. By the assumption that $\{V_j \subset X\}_{j \in J}$ is locally finite, it follows that there exists an open neighbourhood $U_x \supset \{x\}$ and a finitee subset $K \subset J$ such that
Hence by construction
Since the image $\phi(K) \subset I$ is still a finite set, this shows that $\{W_i \subset X\}_{i \in I}$ is locally finite.
Let $(X,\tau)$ be a normal topological space, and let $\{U_i \subset X\}_{i \in I}$ be a locally finite open cover. Then there exists a shrinking to a locally finite open cover $\{V_i \subset X\}_{i \in I}$ whose closures $Cl(-)$ are still contained in the original cover: