This page is about the concept in topology. For the more general concept see at closed morphism.
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
(open maps and closed maps)
A continuous function $f \colon (X,\tau_X) \to (Y, \tau_Y)$ is called
an open map if the image under $f$ of an open subset of $X$ is an open subset of $Y$;
a closed map if the image under $f$ of a closed subset of $X$ (def. \ref{ClosedSubset}) is a closed subset of $Y$.
(image projections of open/closed maps are themselves open/closed)
If a continuous function $f \colon (X,\tau_X) \to (Y,\tau_Y)$ is an open map or closed map (def. 1) then so its its image projection $X \to f(X) \subset Y$, respectively, for $f(X) \subset Y$ regarded with its subspace topology.
If $f$ is an open map, and $O \subset X$ is an open subset, so that $f(O) \subset Y$ is also open in $Y$, then, since $f(O) = f(O) \cap f(X)$, it is also still open in the subspace topology, hence $X \to f(X)$ is an open map.
If $f$ is a closed map, and $C \subset X$ is a closed subset so that also $f(C) \subset Y$ is a closed subset, then the complement $Y \backslash f(C)$ is open in $Y$ and hence $(Y \backslash f(C)) \cap f(X) = f(X) \backslash f(C)$ is open in the subspace topology, which means that $f(C)$ is closed in the subspace topology.
(maps from compact spaces to Hausdorff spaces are closed and proper)
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;
$(Y,\tau_Y)$ is a Hausdorff topological space.
Then $f$ is
a closed map (def. 1);
a proper map.
(proper maps to locally compact spaces are closed)
Let
$(X,\tau_X)$ be a topological space,
$(Y,\tau_Y)$ a locally compact topological space according to def. \ref{LocallyCompactSpace},
$f \colon X \to Y$ a continuous function.
Then:
If $f$ is a proper map, then it is a closed map.