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Cantor space is the binary sequence space with the product topology of discrete two-point factors. Here the Bernoulli-space synonym refers to this topology, without selecting a particular probability measure. The metric $d(x,y)={\sum }_{j\ge 1}{2}^{-j}|x_j-y_j|$ induces the topology. A diagonal subsequence argument gives compactness. Fixing a finite prefix gives a clopen cylinder set. The map $x\mapsto {\sum }_{j}2x_j{3}^{-j}$ is a homeomorphism to the usual Cantor set.

The smallest closed convex set containing a set $C$, equivalently the closure of its convex hull in the specified topology. The topology must be stated: the norm topology, weak topology and weak-star topology need not give the same closure for an arbitrary set.

Even a bounded infinite input need not have a closed convex hull. In the real line, let $S=\{1/n:n\ge 1\}$. Every finite convex combination of these positive points is positive, so zero is absent. Conversely, any $x\in (0,1]$ lies between some $1/n$ and $1$, and is their convex combination. HenceThe closed convex hull adds the missing limit point zero. For the finite eight-point example in convex hull, the convex hull is already closed, so taking the closed convex hull adds nothing.

A subset $F$ of a topological space $(X,\tau )$ is closed when its complement in that same ambient space is open: $X\setminus F\in \tau$. For example, the closed interval $[0,1]$ is closed in the usual real line because its complement is the open set $(-\infty ,0)\cup (1,\infty )$. Always specify the ambient topological space: a subset can be closed in a subspace topology without being closed in the surrounding space. ^[ref]

Being closed is not the logical negation of being open. In the usual real line, all four combinations occur: $(0,1)$ is open but not closed; $[0,1]$ is closed but not open; the half-open interval $[0,1)$ is neither open nor closed; and both the empty set and the whole real line are open and closed, hence clopen. Nor does being closed imply being a bounded set: $[0,\infty )$ is closed because its complement $(-\infty ,0)$ is open, but it is not a bounded set. The symbol $\infty$ is not an extra point that this interval must contain.

The coloured strokes mark membership; filled endpoint circles mean included points, and hollow endpoint circles mean excluded points. The empty set has no coloured points, while arrows at both ends of the coloured real line indicate that membership continues indefinitely. These examples use the usual topology of the real line.

In a metric space $X$, a subset $F$ is closed exactly when every convergent sequence $(x_n)$ in $F$ whose limit of a sequence $x$ lies in $X$ has $x\in F$. For example, $x_n=1-1/n$ lies in $[0,1)$ for $n\ge 1$ and has limit of a sequence $1$, which is missing; thus this half-open interval is not closed in the real line. This criterion is useful for keeping limiting constructions inside a prescribed set. ^[ref]

In every topological space, the general criterion is $F=\overline{F}$, using closure; equivalently, $F$ contains all its limit points. A closed set is always a sequentially closed set, but the converse can fail outside metric spaces. Arbitrary intersections and finite unions of closed sets are closed, by taking complements in the topology axioms.

A closed subgroup is a subgroup which is a closed set in the ambient topology. It is a topological group in its subspace topology, but need not be an open subgroup.

The additive subgroup $\mathbb{Z}\subset \mathbb{R}$ of integers is a closed subgroup: every noninteger lies in an open interval $(n,n+1)$ avoiding $\mathbb{Z}$. It is also a discrete subgroup, since $(n-1/2,n+1/2)\cap \mathbb{Z}=\{n\}$. Nevertheless it is not an open subgroup of $(\mathbb{R},+)$: every open interval containing an integer contains nonintegers too. Thus having a discrete topology internally says nothing about being an open set in the ambient topology.

A subset $A\subseteq S$ is closed under a binary operation on $S$ when the result of combining any two elements of $A$ is again in $A$. The even integers are closed under addition, but the odd integers are not: $1+3=4$ is even. Closure permits restricting the binary operation to $A\times A\to A$. It is different from being a closed set in topology.

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

A Hausdorff space is completely regular if a point $x$ outside a closed set $F$ can be separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. These functions determine the original topology: for any neighborhood $O$ of $x$, choose $F=X\setminus O$ to obtain $x\in \{u>1/2\}\subseteq O$. This is exactly the point-separation property needed for the evaluation embedding into the weak-star dual of bounded continuous functions.

For an infinite coinfinite set $A\subseteq \mathbb{N}$, let $E_A=[A{]}^{\omega }$ in the space of infinite subsets of the natural numbers. This cone is a nonempty clopen set in the Ellentuck topology, because it is the basic neighbourhood $[\varnothing ,A]$, and any $X\notin E_A$ has a finite initial segment containing a point outside $A$ and hence a basic neighbourhood disjoint from $E_A$.

The cone is a closed set with empty interior in the ordinary topology on infinite subsets: membership of a point outside $A$ witnesses its complement as an open set, while every ordinary finite-stem neighbourhood can be extended by an arbitrarily large point outside $A$. Thus $E_A$ is an ordinary nowhere dense set, but it is neither ordinary open nor Ellentuck nowhere dense. It provides both distinctions between these two topologies.

A subset $A$ of a topological space $X$ is a connected subset when $A$, with the subspace topology inherited from $X$, is a connected space. Thus connectedness is tested inside $A$: a forbidden separation of a topological space consists of nonempty subsetswhere $O_1,O_2$ are open sets in $X$. The ambient open sets need not be disjoint outside $A$.

For example, in the real line with its usual topology, $[0,2]$ is a connected subset, whereas $\{0,2\}$ is not: its two singleton sets are open in its subspace topology. A subset of a connected space need not be connected. Conversely, $[-2,-1]$ is a connected subset of the disconnected topological space $[-2,-1]\cup [1,2]$. The ambient topological space need not itself be connected.

A finite set of distinct points in the real line inherits the discrete topology: a sufficiently small open interval around each point isolates that point. Its only nonempty connected subsets are singleton sets. This depends on the topology: a nonempty finite set with the indiscrete topology is a connected space, since its only open sets are itself and the empty set. Here, as in the definition of connected space, the empty set is also connected because it admits no separation of a topological space.^[ref]

The distinction matters when finding connected components: we search among connected subsets of the specified topological space, rather than judging whether the surrounding drawing looks like one piece.

A continuous extension of a function $f:D\to Y$ to a larger domain $E\supseteq D$ is a continuous function $F:E\to Y$ satisfying $F(x)=f(x)$ for $x\in D$. The domain and codomain carry specified topologies. Prescribing values on a dense subset does not by itself prove the existence of a continuous extension.

Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

Give $\mathbb{N}$ the topology whose nonempty open sets are $\mathbb{N}$ and the finite initial segments. Every nonempty open set contains $1$, so any two nonempty open sets intersect. A continuous map from this topological space into a Hausdorff space is constant: distinct image points would have disjoint open neighbourhoods, whose inverse images would be disjoint nonempty open sets. This includes continuous real-valued maps.

For arbitrary topological spaces $X_i$, an index set $I$, a sequence $x^{(n)}\in {\prod }_i{X}_i$, and a point $x$ in that Cartesian product, convergence in a product space to $x$ in the product topology is equivalent to coordinatewise convergence in every factor. No countability or Hausdorff assumption on the factors is needed. For a product space of copies of the real line, regarding each point as a function $I\to \mathbb{R}$ identifies this with pointwise convergence.

Proof outline: the forward implication follows because each projection map is continuous. Conversely, given a basic neighborhood $B(F,U)$ of $x$, each of its finitely many coordinate tests holds for all $n\ge N_i$. Taking $N={\max }_{i\in F}{N}_i$ makes all tests hold simultaneously; if $F$ is empty, any threshold works. This proves eventual membership in every neighborhood, rather than only in a sampled collection of cylinders. The same equivalence holds for nets: a directed set supplies a common upper bound for finitely many thresholds. Sequences alone need not characterize the whole topology of an arbitrary product space.

A filter on a set equipped with a topology converges to $x$ if it contains every neighbourhood of $x$. A Hausdorff space gives every convergent filter at most one limit: disjoint neighbourhoods of distinct purported limits would have empty intersection in the filter.

On ${\mathbb{R}}^2$, the coordinate seminorms $p_x$ and $p_y$ measure horizontal and vertical components separately. Their kernels of seminorms are the vertical and horizontal axes. Neither is a norm, but together they form a separating family of seminorms. A basic zero neighbourhood isEach condition alone is an unbounded strip; their intersection is an open rectangle. For $a=b=1$, this is the unit open ball of the maximum norm $\Vert (x,y){\Vert }_{\infty }=\max \{|x|,|y|\}$. For unequal tolerances it is the unit open ball of the weighted norm $\max \{|x|/a,|y|/b\}$. Thus the two coordinate measurements generate the usual topology of the plane.

For a numerical check, take $a=1$, $b=1/2$: $(0.8,0.4)$ passes both tests, $(0.8,0.7)$ passes only the horizontal test, and $(1.2,0.4)$ passes only the vertical test. A direction missed by one seminorm is detected by the other.

This concerns arbitrary compact spaces, with no Hausdorff restriction. For nonempty sets $A_j$, adjoin a distinguished point ${\infty }_j$ and use the four-open-set topology $\{\varnothing ,A_j,\{{\infty }_j\},A_j\cup \{{\infty }_j\}\}$. Each resulting space is compact without choice. The product has the canonical all-infinity point. The closed coordinate conditions $x_j\in A_j$ have the finite intersection property, since finite choice is provable without the axiom of choice. If the countable product is compact, their intersection is nonempty, and any point in it supplies a choice function.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

A subgroup $G$ of a topological group is discrete when the identity has a neighbourhood containing no other element of $G$. Translation then isolates every group element. In a Hausdorff topological group, such a subgroup is closed: any accumulation at a point would make differences of distinct nearby group elements approach the identity. Consequently its intersection with a compact set is finite. For isometries of hyperbolic space, this converts compactness of the set of isometries moving one point a bounded distance into a properly discontinuous group action.

For example, $(-1/2,1/2)\cap \mathbb{Z}=\{0\}$ isolates the identity element of the additive discrete subgroup $\mathbb{Z}\subset \mathbb{R}$. Its induced subspace topology is the discrete topology, while the surrounding real numbers retain their usual topology; see closed subgroup for why this subgroup is closed but not open.

An edge path in a simplicial complex is a finite sequence of vertices of a simplicial complex such that $\{v_i,v_{i+1}\}$ is a simplex for each consecutive pair. Repeated consecutive vertices are allowed. Its geometric realization is a continuous path formed by traversing the corresponding line segments, with constant pieces for repetitions.

A connected geometric realization has an edge path between any two vertices. For a fixed vertex, the reachable vertices form a union of whole simplexes: all vertices of a simplex are pairwise joined by edges. The reachable and unreachable geometric parts therefore intersect each simplex in either the whole simplex or the empty set. Both parts are closed in the realization topology and are complements, so both are open. Connectedness forces the unreachable part to be empty. Conversely, edge paths between vertices and straight segments inside simplexes join arbitrary points, so the realization is path-connected.