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
(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;
a proper map.
For the first statement, we need to show that if $C \subset X$ is a closed subset of $X$, then also $f(C) \subset Y$ is a closed subset of $Y$.
Now
since closed subsets of compact spaces are compact it follows that $C \subset C$ is also compact;
since continuous images of compact spaces are compact it then follows that $f(C) \subset Y$ is compact;
since compact subspaces of Hausdorff spaces are closed it finally follow that $f(C)$ is also closed in $Y$.
For the second statement we need to show that if $C \subset Y$ is a compact subset, then also its pre-image $f^{-1}(C)$ is compact.
Now
since compact subspaces of Hausdorff spaces are closed it follows that $C \subse Y$ is closed;
since pre-images under continuous of closed subsets are closed, also $f^{-1}(C) \subset X$ is closed;
since closed subsets of compact spaces are compact, it follows that $f^{-1}(C)$ is compact.
(continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms)
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 homeomorphism, i. e. its inverse function $Y \to X$ is also a continuous function.
In particular then both $(X,\tau_X)$ and $(Y, \tau_Y)$ are compact Hausdorff spaces.
Write $g \colon Y \to X$ for the inverse function of $f$.
We need to show that $g$ is continuous, hence that for $U \subset X$ an open subset, then also its pre-image $g^{-1}(U) \subset Y$ is open in $Y$. By passage to complements, this is equivalent to the statement that for $C \subset X$ a closed subset then the pre-image $g^{-1}(C) \subset Y$ is also closed in $Y$.
But since $g$ is the inverse function to $f$, its pre-images are the images of $f$. Hence the last statement above equivalently says that $f$ sends closed subsets to closed subsets. This is true by prop. .
The idea captured by corollary is that Hausdorffness is about having “enough” open sets whilst compactness is about having “not too many”. Thus a compact Hausdorff space has both “enough” and “not too many”. This theorem says that both conditions are at their limit: if we try to have more open sets, we lose compactness. If we try to have fewer open sets, we lose Hausdorffness.
Last revised on June 21, 2017 at 04:49:51. See the history of this page for a list of all contributions to it.