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 following terminology is used in topology:
(saturated subset)
Let $f \;\colon\; X \longrightarrow Y$ be a function of sets. Then a subset $S \subset X$ is called an $f$-saturated subset (or just saturated subset, if $f$ is understood) if $S$ is the pre-image of its image:
If $S$ is not necessarily $f$-saturated, then $f^{-f}(f(S))$ is called its saturation. Notice that $S \subset f^{-1}(f(S))$.
Let $f \colon X \longrightarrow Y$ be a function. Then a subset $S \subset X$ is $f$-saturated (def. ) precisely if its complement $X \backslash S$ is so.
whose underlying function $f \colon X \longrightarrow Y$ is surjective exhibits $\tau_Y$ as the corresponding quotient topology precisely if $f$ sends open and $f$-saturated subsets in $X$ (def. ) to open subsets of $Y$. By lemma this is the case precisely if it sends closed and $f$-saturated subsets to closed subsets.
Let
$f \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a closed map.
$C \subset X$ be a closed subset which is $f$-saturated;
$U \supset C$ an open subset containing $C$
then there exists a smaller open subset $V$ still containing $C$
and such that $V$ is $f$-saturated.
We claim that
has the desired properties. To see this, observe first that
the complement $X \backslash U$ is closed, since $U$ is assumed to be open;
hence the image $f(X\backslash U)$ is closed, since $f$ is assumed to be a closed map;
hence the pre-image $f^{-1}\left( f\left( X \backslash U \right)\right)$ is closed, since $f$ is continuous, therefore its complement $V$ is indeed open;
this pre-image $f^{-1}\left( f\left( X \backslash U \right) \right)$ is saturated, as all pre-images clearly are, and hence also its complement $V$ is saturated, by lemma .
Therefore it now only remains to see that $U \supset V \supset C$.
The inclusion $U \supset V$ means equivalently that $f^{-1}\left( f\left( X \backslash U \right)\right) \supset X \backslash U$, which is clearly the case.
The inclusion $V \supset C$ meas that $f^{-1}\left( f\left( X \backslash U \right) \right) \,\cap \, C = \emptyset$. Since $C$ is saturated by assumption, this means that $f^{-1}\left( f\left( X \backslash U \right)\right) \,\cap \, f^{-1}(f(C)) = \emptyset$. This in turn holds precisely if $f\left( X \backslash U \right) \,\cap \, f(C) = \emptyset$. Since $C$ is saturated, this holds precisely if $X \backslash U \cap C = \emptyset$, and this is true by the assumption that $U \supset C$.
Last revised on May 21, 2017 at 18:14:24. See the history of this page for a list of all contributions to it.