see also algebraic topology, functional analysis and homotopy theory
topological space (see also locale)
fiber space, space attachment
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
subsets are closed in a closed subspace precisely if they are closed in the ambient space
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
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
Theorems
A (topological) space whose only connected subspaces are singletons is called totally disconnected.
Discrete spaces are totally disconnected.
(the rational numbers are totally disconnected)
The rational numbers $\mathbb{Q} \subset \mathbb{R}$ equipped with their subspace topology inherited from the Euclidean metric topology on the real numbers, form a totally disconnected space.
By construction, a base for the topology is given by the open subsets which are restrictions of open intervals of real numbers to the rational numbers
for $a \lt b \in \mathbb{R}$.
Now for any such $a \lt b$ there exists an irrational number $r \in \mathbb{R}\backslash \mathbb{Q}$ with $a \lt r \lt b$. This being irrational implies that $(a,r)_{\mathbb{Q}} \subset \mathbb{Q}$ and $(r,b)_{\mathbb{Q}} \subset \mathbb{Q}$ are disjoint subsets. Therefore every basic open subset is the disjoint union of (at least) two open subsets:
Hence no inhabited open subspace of the rational numbers is connected.
A product in $Top$ of totally disconnected spaces is totally disconnected. A subspace of a totally disconnected space is totally disconnected. Hence limits in $Top$ of diagrams of totally disconnected spaces are totally disconnected.
For example, the Baire space of irrational numbers is homeomorphic to a countable product space $\mathbb{N}^\mathbb{N}$ (via continued fractions), so it is totally disconnected. Similarly, Cantor space $2^\mathbb{N}$ is totally disconnected. Another notable special case of the preceding class of examples is the following.
Every profinite group is totally disconnected and in particular the set of p-adic numbers is totally disconnected.
See also Stone space.
The general class of examples in Example 3 may be seen in the following light.
The category of totally disconnected spaces and continuous maps is a reflective subcategory of Top.
The reflector takes a space $X$ to the space of connected components, i.e., the quotient space $X/\sim$ of $X$ where $\sim$-equivalence classes are precisely the connected components of $X$. We check that connected components $C$ of $X/\sim$ are singletons. Let $q: X \to X/\sim$ be the quotient map; it suffices to check that $q^{-1}(C) \subseteq X$ is connected (because then $q^{-1}(C)$ is contained in a single $\sim$-equivalence class, making $C = q q^{-1}(C)$ a single point). So, suppose the (closed) set $q^{-1}(C)$ is a disjoint union $K_1 + K_2$ of closed sets $K_1, K_2$; we are required to show one or the other is empty. For each $c \in C$, the inverse image $q^{-1}(\{c\})$ is connected, hence we must have $q^{-1}(\{c\}) \subseteq K_1$ or $q^{-1}(\{c\}) \subseteq K_2$. Thus we can partition $C$ into sets
and we observe $q^{-1}(C_1) = K_1$ and $q^{-1}(C_2) = K_2$. By definition of quotient topology, since $K_1, K_2$ are closed we infer $C_1, C_2$ are closed. They are also disjoint and $C = C_1 + C_2$, so by connectedness of $C$ either $C_1 = \emptyset$ or $C_2 = \emptyset$, and therefore $K_1 = q^{-1}(C_1)$ or $K_2 = q^{-1}(C_2)$ is empty, as required.
Finally, given a continuous map $f: X \to Y$ with $Y$ totally disconected, each connected component $C$ of $X$ is mapped to a connected set $f(C)$ of $Y$ which is a singleton by total disconnectedness of $Y$, and so we get a (unique) factoring through a map $X/\sim \to Y$, continuous of course by virtue of the quotient topology. This completes the proof.