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
Theorem named after Urysohn.
Every second-countable regular Hausdorff space $X$ is metrizable
The proof can be divided into three parts. Recall that “second-countable” means having a countable base. All spaces considered are assumed to be $T_1$ (points are closed), so that “regular” means regular Hausdorff.
A regular space with a countable base is normal.
A normal space $X$ is a completely regular space, i.e., given a point $x \in X$ and an open set $U \subseteq X$ containing $x$, there exists a continuous function $f: X \to [0, 1]$ such that $f(x) = 1$ and $f(x) = 0$ for $x \notin U$.
A completely regular space $X$ with a countable base can be embedded in the Hilbert cube $[0, 1]^\mathbb{N}$. Since $[0, 1]^\mathbb{N}$ is metrizable, so is $X$.
The second assertion being proved at Urysohn lemma, we prove the first and third assertions.
A regular space $X$ with a countable base is normal.
Let $A, B$ be disjoint closed sets of $X$. The collection of open sets $U$ such that $U \cap A$ is inhabited and $\widebar{U} \cap B = \emptyset$ is, by regularity, an open covering of $A$. By second-countability, we may index it as $U_1, U_2, \ldots$. Similarly, there is a countable open covering $V_1, V_2, \ldots$ of $B$ such that $\widebar{V_k} \cap A = \emptyset$.
Now form open sets $Y_n \coloneqq U_n \cap \bigcap_{k=1}^n \neg \widebar{V_k}$ and $Z_n = V_n \cap \bigcap_{k=1}^n \neg \widebar{U_k}$. It is clear that the $Y_n$ cover $A$ and the $Z_n$ cover $B$. Moreover, $Y_m \cap Z_n = \emptyset$ for all $m, n$. For if $m \geq n$ say, then
It follows that $\bigcup_m Y_m$ and $\bigcup_n Z_n$ are disjoint open sets containing $A$ and $B$, respectively. This completes the proof.
A completely regular space $X$ with countable base can be embedded in $[0, 1]^\mathbb{N}$.
$X$ has a countable base $\mathcal{B}$. The set $S = \{(U, V) \in \mathcal{B} \times \mathcal{B}: \widebar{U} \subseteq V\}$ is countable. For each $s = (U, V) \in S$ there is by the Urysohn lemma a continuous map $g_s: X \to [0, 1]$ such that $g_s$ is identically $1$ on $\widebar{U}$ and identically $0$ on $\neg V$. The map
is a continuous injection, since if $x \neq y$, we can find a pair $s = (U, V) \in S$ with $x \in U$ and $y \notin V$, so that $g_s(x) = 1$ differs from $g_s(y) = 0$, whence $g(x) \neq g(y)$. The subspace topology on $X$ induced from the monomorphism $g$ is contained in the given topology of $X$, simply by continuity of $g$. On the other hand, if $W$ is an open neighborhood of $x$ in $X$, there exists a smaller open neighborhood $V \in \mathcal{B}$ and $s = (U, V) \in S$ such that $g_s(x) = 1$ and $g_s$ is identically $0$ outside $V$. Provided that $g_s(x) - g_s(y) \lt 1$, we see $g_s(y) \neq 0$, so $y \in V$. In other words, letting $\pi_s: [0, 1]^S \to [0, 1]$ be the evident projection, we have $g_s = \pi_s \circ g$, so that the inverse image under $g$ of the subbase element $\pi_s^{-1}((0, 1])$ is $g_s^{-1}((0, 1]) \subseteq V \subseteq W$. This shows that the subspace topology induced by $g$ contains the topology of $X$. It follows that $g: X \to [0, 1]^S \cong [0, 1]^\mathbb{N}$ is an embedding into the Hilbert cube.