see also algebraic topology, functional analysis and homotopy theory
Basic concepts
topological space (see also locale)
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Basic homotopy theory
A locally finite cover is a cover which in a suitable sense looks locally like a finite cover.
(locally finite cover)
Let $(X,\tau)$ be a topological space.
A cover $\{U_i \subset X\}_{i \in I}$ of $X$ by subsets of $X$ is called locally finite if it is a locally finite set of subsets, hence 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 $i \in I$.
If $\{U_i \subset X\}_{i \in I}$ is an open cover, then it is called a locally finite open cover.
(alternative characterizations of local finiteness)
Let $X$ be a topological space and let $\{U_i \to X\}_{i \in I}$ be a cover by subsets. Then the following are equivalent:
$\{U_i \subset X\}_{i \in I}$ is locally finite (def. 1);
there exist an open cover $\{V_j \subset X\}_{j \in J}$ such that for each $j \in J$ there is a finite number of $i \in I$ that $V_j$ intersects $V_i$.
This is because the various $V_i$ constitute open neighbourhoods for all points $x \in X$.
Moreover, suppose that $\{V_j \subset X\}_{j \in J}$ is a cover by any subsets (not necessarily open), but that it is itself a locally finite set of subsets. Then if for all $j \in J$ there are a finite number of $i \in I$ such that $U_i$ intersects $V_j$, it follows again that also $\{U_i \subset X\}_{i \in I}$ is locally finite.
This is because by the local finiteness of $\{V_j \subset X\}_{j \in J}$ we have for every point $x \in X$ an open neighbourhood $O_x \supset \{x\}$ which intersects only a finite number of the $V_j$, and since each of these intersects only a finite number of the $U_i$, in total also $O_x$ can only intesect a finite number of the $U_i$.
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 $\{V_j \subset X\}_{j \in J}$, be a refinement to a locally finite cover.
By definition of refinement we may choose a function
such that
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 finite 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: