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A family $(A_i)$ of subsets of a topological space is locally finite if every point has a neighbourhood meeting only finitely many of its members. On a compact space such a family has only finitely many nonempty members. For singleton subsets of a Riemann surface, this is the condition permitting prescribed poles of a meromorphic function without accumulation inside the surface.

A loop in a topological space is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=\gamma (1)$. Equivalently it is a map from the circle. A based loop also specifies the common endpoint as a chosen base point.

The lower limit topology on $\mathbb{R}$ has basis of a topology $\{[a,b):a<b\}$. Every such nonempty interval contains a rational, so this topological space is separable. Its topology is finer than the usual Euclidean topology; the one-sided basic neighbourhoods allow its square to have an uncountable discrete antidiagonal.

The manifold boundary of a topological manifold with boundary consists of the points which, in a half-space coordinate chart, have last coordinate zero. These are precisely the points without a neighbourhood homeomorphic to an open set in the full Euclidean space of the same dimension. It is an intrinsic property, independent of the coordinate chart. It differs from the boundary of a set in an ambient topological space: a circle has empty manifold boundary, but its boundary of a set in the Euclidean plane is the entire circle.

A net is a family indexed by a directed ordered set: every pair of indices has a common upper bound. It converges to $x$ when it is eventually in every neighbourhood of $x$. Unlike sequences, nets characterize closure in every topological space: a point is in the closure of a set precisely when some net in that set converges to it.

A non-Hausdorff manifold is a locally Euclidean topological space for which distinct points need not have disjoint neighborhoods. The real line with two origins is a basic example. Nonseparated algebraic schemes exhibit analogous gluing phenomena, but are not generally topological manifolds.

A non-isolated point $x$ of a topological space $X$ is one for which $\{x\}$ is not open, equivalently every neighbourhood of $x$ contains another point of $X$. A homeomorphism preserves this property because it and its inverse preserve open sets, in particular open singletons. In a product space $X\times Y$, a point $(x,y)$ is isolated exactly when both coordinates are isolated: use basic open rectangles and fix the other coordinate. Thus the non-isolated points of $X\times Y$ are $(X^{\prime }\times Y)\cup (X\times Y^{\prime })$, where the primes denote the non-isolated points in the spaces themselves.

An open arc is a topological space homeomorphic to an open interval. Unlike an arc, which is homeomorphic to a closed interval, an open arc does not include its two endpoints.

An open neighbourhood of a point $x$ in a topological space $(X,\mathcal{T})$ is an open set $U$ containing $x$. A general neighbourhood need only contain an open neighbourhood; it need not itself be open. A finite intersection of open neighbourhoods of the same point is again an open neighbourhood, by the topology axioms.

An open set is a member of the topology on a topological space. In a metric space, a subset $U$ is an open set precisely when every $x\in U$ has an open ball $B(x,r)$ contained in $U$ for some $r>0$. For $U=(0,1)$, one can take $r=\frac{1}{2}\min (x,1-x)$. At $x=0.2$, this gives $r=0.1$ and $B(x,r)=(0.1,0.3)\subset U$. The radius depends on the point; no uniform radius is required. ^[ref]

Openness depends on the ambient topological space. In $A=[0,1]$ with its subspace topology, $[0,1/2)=A\cap (-1,1/2)$ is an open set in $A$, although it is not an open set in the real line. Similarly, the whole closed interval $A$ is both an open set and a closed set in itself. These relative open sets are what the continuity test uses for maps involving a subset with its subspace topology.

Arbitrary intersections of open sets need not be open sets: ${\bigcap }_{n\ge 1}(-1/n,1/n)=\{0\}$, a singleton that is not an open set in the usual real line. Any point $x\ne 0$ is excluded once $n>1/|x|$. This is why the topology axiom specifies finite intersections.

The open subsets of a topological space form a frame. Arbitrary joins are unions, finite meets are intersections, and an arbitrary meet is the interior of an intersection. A continuous map induces a frame homomorphism in the opposite direction by inverse image.

A connected space $S$ cannot be expressed as $S=A\cup B$ with $A,B$ disjoint nonempty relatively open sets. Equivalently, the only subsets simultaneously relatively open and closed are $\varnothing$ and $S$. A decomposition into such $A,B$ is a separation of the topological space.

A topological space is Lindelöf if every open cover has a countable subcover. We also prove the requested equivalence for metrizable spaces here, after the two source definition leaves.

Choose a compatible metric $d$. First suppose that $D$ is a countable dense subset. The balls $B(d_0,q)$ with centers $d_0\in D$ and positive rational radii $q$ form a countable topological basis. Indeed, if $x\in U$ and $B(x,ϵ)\subset U$, choose $d_0\in D$ with $d(x,d_0)<ϵ/3$ and a rational $q$ between $d(x,d_0)$ and $ϵ-d(x,d_0)$. The triangle inequality gives $x\in B(d_0,q)\subset U$.

For any open cover, retain the topological basis elements which are contained in some member of the cover, and choose one covering member for each such topological basis element. There are only countably many choices, and the resulting subfamily covers the topological space, since each point has a topological basis neighborhood inside a covering member. Thus a separable metrizable space is Lindelöf.

Conversely, suppose the metric space is Lindelöf. For each positive integer $n$, the balls $B(x,1/n)$ cover it, so select a countable subcover with centers in a countable set $D_n$. The union $D={\bigcup }_{n\ge 1}{D}_n$ is countable and dense: if $B(x,ϵ)$ is a nonempty open ball, choose $n$ with $1/n<ϵ$. One of the selected radius-$1/n$ balls contains $x$, so its center lies in $D\cap B(x,ϵ)$. ThereforeThe metric hypothesis is used in producing uniform small-ball covers and a countable ball topological basis; the equivalence need not hold for general topological spaces.

A topological space is separable if it has a countable dense subset. A subset $D$ is dense when its closure is the whole topological space, equivalently when every nonempty open set intersects $D$. Countable here permits a finite set, so finite topological spaces and the empty topological space fit the same definition.

Choose countable dense sets $D_X\subseteq X$ and $D_Y\subseteq Y$. Their Cartesian product is countable. Every nonempty basic open set of the product topology has the form $U\times V$ with $U,V$ nonempty open in their respective topological spaces. Density gives $d_X\in D_X\cap U$ and $d_Y\in D_Y\cap V$, hence $(d_X,d_Y)\in (D_X\times D_Y)\cap (U\times V)$.

Every nonempty open set in the product contains such a basic open set. Thus $D_X\times D_Y$ is dense in $X\times Y$, proving topological separability of the product. If one factor is empty the product is empty and the conclusion holds directly.

Let $\mathcal{B}$ be a topological basis consisting of clopen sets. For $U\in \mathcal{B}$ its characteristic function ${\chi }_U:X\to \{0,1\}$ is continuous, since the preimages of both singleton values are open. DefineThe coordinate criterion for product topology makes this map continuous. If $x\ne y$, the $T_0$ property gives an open set containing one and not the other. A topological basis element contained in that open set still distinguishes them, so their corresponding characteristic coordinates differ. Therefore $e$ is injective.

Finally, for each topological basis element $U$,The right-hand side is the preimage of an open cylinder in the product of discrete two-point topological spaces. These cylinders recover a topological basis for the topology of $X$, proving continuity of the inverse onto the image. Hence

A Hausdorff compactification of $X$ is a pair $(h,Y)$ in which $Y$ is compact Hausdorff and $h:X\to Y$ is a topological embedding with dense image.

For a locally compact Hausdorff noncompact topological space, form $X^{\ast }=X\cup \{\infty \}$ with one new point. Keep all open subsets of $X$, and take the neighborhoods of infinity to contain setsCompact subsets of a Hausdorff space are closed. Finite unions of compact sets are compact, so finite intersections of these basic neighborhoods are again of the same form; their intersections with ordinary open sets are open in $X$. This specifies a topology restricting to the original one on $X$.

It is compact: from an open cover choose a member containing infinity. Its complement is contained in some compact $K$, and finitely many more members cover that compact remainder. It is Hausdorff: two points of $X$ are separated as before; to separate $x\in X$ from infinity, choose an open neighborhood $U$ of $x$ lying inside a compact neighborhood $K$. The open set $U$ and the infinity neighborhood $X^{\ast }\setminus K$ are disjoint.

The inclusion of $X$ is an embedding. Every basic neighborhood of infinity meets $X$, because otherwise its compact complement $K$ would equal all of $X$, contrary to noncompactness. Thus $X$ is dense in $X^{\ast }$. This proves that $X^{\ast }$ is the one-point compactification.

A proper filter $\mathcal{F}$ on $X$ is a nonempty family of subsets, closed under finite intersection and passage to supersets, with $\varnothing \notin \mathcal{F}$. An ultrafilter is a maximal proper filter, equivalently one which contains, for each $A\subseteq X$, exactly one of $A$ and $X\setminus A$. The pushforward filter isInverse images preserve intersections and complements. Thus it is a proper filter, and the either-a-set-or-its-complement criterion shows it is an ultrafilter whenever $\mathcal{F}$ is one.

Convergence of a filter $\mathcal{F}\to x$ means that every neighborhood of $x$ belongs to $\mathcal{F}$. A mapping is continuous exactly when $\mathcal{F}\to x$ implies $f_{\ast }\mathcal{F}\to f(x)$ for all proper filters and points. The forward implication follows from inverse images of neighborhoods. For the converse apply the criterion to the neighborhood proper filter at $x$: each inverse image of a neighborhood of $f(x)$ must be a neighborhood of $x$, which is continuity at $x$.

A topological space is compact exactly when every ultrafilter converges to at least one point; equivalently every proper filter has a cluster point. For completeness, in a compact space the closed sets $\overline{A}$, $A\in \mathcal{U}$, have the finite intersection property, hence a common point $x$. Every neighborhood $N$ of $x$ belongs to the ultrafilter. Indeed, if $N$ did not, choose an open $V$ with $x\in V\subseteq N$. Then $V$ would not belong either, so $X\setminus V$ would belong. That set is closed and excludes $x$, contradicting the choice of $x$ in every closure. Conversely, if an open cover has no finite subcover, its closed complements generate a proper filter. An ultrafilter extension cannot converge anywhere, because each point has a covering neighborhood whose complement belongs to that ultrafilter. This proves the compactness criterion using the permitted extension principle.

Now suppose $b\in \overline{f[A]}$. The familyis a filter base of nonempty sets: closure guarantees each intersection is nonempty, and finite intersections correspond to intersections of neighborhoods. Extend its generated proper filter to an ultrafilter $\mathcal{H}$. It contains $A$ and every $f^{-1}[V]$, so $f_{\ast }\mathcal{H}\to b$.

Conversely, if such an ultrafilter exists, each neighborhood $V$ of $b$ has $f^{-1}[V]\in \mathcal{H}$. Its intersection with $A\in \mathcal{H}$ is nonempty. Thus $V$ meets $f[A]$, provingThis part includes the unnumbered proper filter definitions and criteria preceding the printed subparts.

Use the Stone-Čech extension property: a discrete topological space $X$ embeds densely into a compact Hausdorff space $\beta X$, and every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Apply it to the characteristic function ${\chi }_A:X\to \{0,1\}$ of an arbitrary subset $A$; discreteness makes it continuous. Its extension ${\tilde{\chi }}_A$ has clopen fibers, andIndeed that fiber is closed and contains $A$, while its openness and density of $X$ show that every point in the fiber is approached by points of $A$.

If $A$ and $B$ are disjoint, the extension of ${\chi }_A$ is one on $\overline{A}$ and zero on $\overline{B}$, so these closures are disjoint. The inclusion $X\to \beta X$ therefore satisfies the criterion from part (b), and its ultrafilter-limit map is both surjective and injective. ThusThis is a bijection of underlying sets; identifying the compatible topology on the ultrafilter topological space would give the familiar stronger homeomorphism statement, but is not needed for the requested conclusion.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.