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
A G-delta, $G_\delta$, subset of a topological space is a set that can be written as the intersection of a countable family of open sets.
One place where $G_\delta$-subsets occur is when looking at continuous maps from an arbitrary topological space to a metric space (or, more generally, a first countable space). In particular, when considering continuous real-valued functions. Thus we have the following connections to the separation axioms.
A normal space in which every closed set is a $G_\delta$-set is perfectly normal.
In a completely regular space, every singleton set that is a $G_\delta$-set is the unique global maximum of a continuous real-valued function.
One direction is obvious. For the other, let $v$ be a point in a completely regular space $X$ such that $\{v\}$ is a $G_\delta$-set. Let $\{V_n\}$ be a sequence of open sets such that $\bigcap V_n = \{v\}$. We now define a sequence of functions $(f_n)$ recursively with the properties:
Having defined $f_1, \dots, f_{n-1}$, we define $f_n$ as follows. Since $V_n \cap f_{n-1}^{-1}(1)$ is a neighbourhood of $v$ and $X$ is completely regular, there is a continuous function $\tilde{f}_n \colon X \to [0,1]$ with support in this neighbourhood and such that $\tilde{f}_n(v) = 1$. We then compose with a continuous, increasing surjection $[0,1] \to [0,1]$ which maps $[\frac12,1]$ to $1$. The resulting function is the required $f_n$.
We then define a function $f \colon X \to [0,1]$ by
By construction, $f^{-1}(1) = \{v\}$.
We need to prove that this is continuous. First, note that if $f_n(x) \ne 0$ then $f_k(x) = 1$ for $k \lt n$ and if $f_n(x) \ne 1$ then $f_k(x) = 0$ for $k \gt n$. Hence the preimage under $f$ of $(\frac{2^k-1}{2^k}, \frac{2^{k+1}-1}{2^{k+1}})$ is $f_n^{-1}(0,1)$ and $f$ restricted to this preimage is a scaled translate of $f_n$. From this, we deduce that the preimage of any open set not containing $1$ is open. Thus $f$ is continuous everywhere except possibly at $v$. Continuity at $v$ is similarly simple: given a set of the form $(1 -\epsilon,1]$ then there is some $n$ such that $2^{-n} \lt \epsilon$, whence $f^{-1}(1-\epsilon,1]$ contains all points such that $f_k(x) = 1$ for $k \le n$, which by construction is a neighbourhood of $v$. Hence $f$ is continuous and has a single global maximum at $v$.