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In a Hausdorff space, a point $x$ outside a compact subset $K$ can be separated from all of $K$ by disjoint open sets. Choose disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$ for each $k\in K$. Finitely many $U_k$ cover $K$ by compactness. Their union and the matching finite intersection of the $V_k$ give the displayed separation. If $K$ is empty, choose $U=\varnothing$, $V=X$. Thus $X\setminus K$ is open, proving that $K$ is closed.

Two disjoint compact subsets of a Hausdorff space have disjoint open neighbourhoods. For every $p\in P$, separation of a point and a compact set gives disjoint open sets $G_p\ni p$ and $H_p\supseteq Q$. Choose finitely many $G_p$ covering $P$, then take their union for $U$ and the matching finite intersection of the $H_p$ for $V$. The compactness of both subsets is used, once in the point-versus-compact set result and once in the outer finite selection. The ambient Hausdorff space need not be compact or a normal space.

A topological space is sober when every nonempty irreducible closed subset is the closure of a unique point. Every Hausdorff space is sober. In contrast, an infinite set with the cofinite topology has the whole space irreducible and closed, while every singleton is closed, so it is not sober.

A Stone space is a compact Hausdorff space that is totally disconnected, equivalently a compact Hausdorff space with a basis of clopen sets. The topology on a type space gives an important example.

A topological space is T1 when every point is a closed point. Equivalently, for distinct $x,y$ there is an open set containing $x$ but not $y$, and an open set containing $y$ but not $x$. These open sets need not be disjoint. Every Hausdorff space is T1, but an infinite set with its cofinite topology is T1 without being Hausdorff.

A continuous map injective map $f:X\to Y$ that is a homeomorphism onto its image with the subspace topology. A continuous map injective map from a compact space into a Hausdorff space is a topological embedding.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

A topological surface is a Hausdorff space with a countable basis in which every point has a neighborhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

A topological space is both a compact space and a Hausdorff space exactly when every ultrafilter on its underlying set converges to exactly one point. Existence of all ultrafilter limits characterizes compactness. If a space is not Hausdorff, two distinct points have neighbourhood systems whose combined finite intersections are nonempty; the ultrafilter lemma extends these to an ultrafilter converging to both points. Conversely, disjoint neighbourhoods prevent such a common limit.

A continuous dense map into a compact Hausdorff space assigns to each ultrafilter on its domain the unique limit of its pushforward. This limit map is surjective by the ultrafilter characterization of image closure. It is injective exactly when disjoint domain subsets have disjoint image closures.

A point $x$ is the limit of a map $f:I\to X$ along an ultrafilter $\mathcal{U}$ on $I$ when $f^{-1}(V)\in \mathcal{U}$ for every neighbourhood $V$ of $x$. Every ultrafilter on a compact space has a limit, and that limit is unique in a Hausdorff space.

The same Banach space can acquire incomparable weak-star topologies from different isometric preduals. Pair ${\ell }^1$ with $c_0$ by $\sum {\alpha }_n{\beta }_n$ and with the space $c$ of convergent sequences by ${\alpha }_1{\gamma }_{\infty }+{\sum }_{n\ge 2}{\alpha }_n{\gamma }_{n-1}$. The functional $\alpha \mapsto {\alpha }_1$ is continuous only for the first topology: $e_1-e_n\to 0$ for the second but its values remain one. Conversely $\alpha \mapsto \sum {\alpha }_n$ is continuous only for the second: $e_n\to 0$ for the first but its values remain one. More generally, comparable weak-star topologies arising from isometric preduals agree on the common closed unit ball, since its identity map is a continuous bijection from a compact space to a Hausdorff space by the Banach-Alaoglu theorem.