topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
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
continuous metric space valued function on compact metric space is uniformly continuous
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
Analysis Theorems
The image under a continuous function of a compact topological space is itself compact (cor. 1 below.)
This is a generalization of the extreme value theorem in analysis.
In fact the following more general statement holds
Let $f \colon (X,\tau_X) \longrightarrow (Y,\tau_Y)$ be a continuous function between topological spaces such that
$(X,\tau_X)$ is a compact topological space;
$f \colon X \to Y$ is a surjective function.
Then also $(Y,\tau_Y)$ is compact.
Let $\{U_i \subset Y\}_{i \in I}$ be an open cover of $Y$. We need show that this has a finite sub-cover.
By the continuity of $f$ the pre-images $f^{-1}(U_i)$ form an open cover $\{f^{-1}(U_i) \subset X\}_{i \in I}$ of $X$. Hence by compactness of $X$, there exists a finite subset $J \subset I$ such that $\{f^{-1}(U_i) \subset X\}_{i \in J \subset I}$ is still an open cover of $X$. Finally, by surjectivity of $f$ it it follows that
where we used that images of unions are unions of images.
This means that also $\{U_i \subset Y\}_{i \in J \subset I}$ is still an open cover of $Y$, and in particular a finite subcover of the original cover.
If $f \colon X \longrightarrow Y$ is a continuous function out of a compact topological space $X$ which is not necessarily surjective, then we may consider its image factorization
with $im(f)$ regarded as a topological subspace of $Y$. Now by construction $X \to \im(f)$ is surjective, and so lemma 1 implies that $im(f)$ is compact.
maps from compact spaces to Hausdorff spaces are closed and proper
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces