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
(quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff)
Let
be a continuous function between topological spaces such that
$(X, \tau)$ is a compact Hausdorff topological space;
$\pi$ is a surjection and $\tau_Y$ is the corresponding quotient topology.
Then the following are equivalent
$(Y, \tau_Y)$ is itself a Hausdorff topological space;
$\pi$ is a closed map.
The implicaton $\left( (Y, \tau_Y)\, \text{Hausdorff} \right) \Rightarrow \left( \pi \, \text{closed} \right)$ follows since maps from compact spaces to Hausdorff spaces are closed. We need to show the converse.
Hence assume that $\pi$ is a closed map. We need to show that for every pair of distinct points $y_1 \neq y_2 \in Y$ there exist open neighbourhoods $U_{y_1}, U_{y_2} \in \tau_Y$ which are disjoint, $U_{y_1} \cap U_{y_2} = \emptyset$.
First notice that the singleton subsets $\{x\}, \{y\} \in Y$ are closed. This is because they are images of singleton subsets in $X$, by surjectivity of $f$, and because singletons in a Hausdorff space are closed, and because images under $f$ of closed subsets are closed, by the assumption that $f$ is a closed map.
It follows that the pre-images
are closed subsets of $X$.
Now since compact Hausdorff spaces are normal it follows that we may find disjoint open subset $U_1, U_2 \in \tau_X$ such that
Moreover, by this lemma we may find these $U_i$ such that they are both saturated subsets. Therefore finally this lemma says that the images $\pi(U_i)$ are open in $(Y,\tau_Y)$. These are now clearly disjoint open neighbourhoods of $y_1$ and $y_2$.
Consider the function
from the quotient topological space of the closed interval by the equivalence relation which identifies the two endpoints
to the unit circle $S^1 = S_0(1) \subset \mathbb{R}^2$ regarded as a topological subspace of the 2-dimensional Euclidean space equipped with its metric topology .
This is clearly a continuous function and a bijection on the underlying sets. Moreover, since continuous images of compact spaces are compact and since the closed interval $[0,1]$ is compact we also obtain another proof that the circle is compact.
Hence since continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms the above map is in fact a homeomorphism