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Bounded completeness requires the defining implication for a complete statistic only for bounded measurable functions. Completeness implies it, but the converse need not hold. If a sufficient statistic $T$ gives a parameter-independent event probability $c$ and the conditional probability $S(T)$ of that event is nonconstant, then $S-c$ is a bounded zero-mean obstruction to bounded completeness.

A metric space is complete if and only if every decreasing sequence of nonempty closed sets whose diameters tend to zero has nonempty intersection. In that case the intersection contains exactly one point.

For two Cauchy sequences, $|d(x_n,y_n)-d(x_m,y_m)|\le d(x_n,x_m)+d(y_n,y_m)$, so the displayed limit exists. It defines a pseudometric on the Cauchy sequences, and its metric quotient identifies sequences whose mutual distance tends to zero. Constant sequences embed the original metric space isometrically. Their image is dense, because the class of $(x_n)$ is approached by the constant classes of $x_n$. The quotient is complete: from a Cauchy sequence of its points select a subsequence whose successive distances are below $2^{-j}$ and choose original-space points within $2^{-j}$ of these subsequence points. Those original-space points form a Cauchy sequence and its class is the limit of the subsequence, hence of the whole sequence. The original embedding is onto exactly when the original space is complete.

In a one-period complete market with a strictly positive numéraire, at most one positive state-price density prices every traded asset. Suppose $\rho ,{\rho }^{\prime }$ do so. For each event $A$, replicate the contingent claim $S_1^{(0)}{1}_A$. Linearity of the pricing equations givesBoth weighted densities are integrable because they price the numéraire. An integrable random variable whose integral over every event is zero vanishes almost surely: take the events where it is positive and negative. Therefore $(\rho -{\rho }^{\prime })S_1^{(0)}=0$, and strict positivity of the numéraire gives $\rho ={\rho }^{\prime }$. This argument avoids assuming that the unweighted densities themselves are integrable.

In an arbitrage-free one-period market with finitely many assets and an equivalent martingale measure, market completeness is equivalent to uniqueness of the dominated martingale measure. Completeness prices every indicator function, forcing two measures to agree on every event. Conversely, if the attainable payoff span has dimension $k$ but is not the whole payoff space, choose $k+1$ disjoint positive-probability events. Their moment vectors against a basis of attainable payoffs are linearly dependent. A nonzero bounded linear combination $h$ of their indicators consequently has zero pricing moments. Multiplying an existing equivalent density by $1+\epsilon h$, for sufficiently small positive and negative $\epsilon$, constructs distinct equivalent martingale measures. This proves the converse without assuming that the original space was finite. On a general probability space, completeness here means replication of all bounded claims, or equivalently all square-integrable claims when asset payoffs are square-integrable.

A star graph is the complete bipartite graph with one centre and $m$ leaves. Any pair of its edges meet at the centre.

Fix a $(p+1)$-set $B\subseteq [4p]$. Take every $2p$-set $B\cup X$ with $X$ a $(p-1)$-subset of the remaining $3p-1$ points, together with the complements of all these sets. Within either half, intersections have size at least $p+1$. Across the halves they have size $|X\setminus Y|\le p-1$. Thus no intersection has size $p$, and the two disjoint halves give the displayed family size.

A universal set $U$ fixes the objects under consideration in a particular discussion; each set under consideration is a subset of $U$. For example, $U=\{1,2,3,4,5,6\}$ makes the complement of $\{1,2,3\}$ equal to $\{4,5,6\}$. This use of “universal” is local to the discussion: it does not assert the existence of a set containing every set. Stating $U$ makes exclusion rules precise.

Let eventual outcome $I=j$ have probability ${\pi }_j$, with $\log T\mid I=j\sim N({\mu }_j,{\sigma }_j^2)$. An observed outcome at time $t$ contributes ${\pi }_j{f}_j(t)$ to the likelihood, while right censoring at $c$ contributes ${\sum }_j{\pi }_j{S}_j(c)$. The expectation-maximization algorithm imputes both the censored outcome and its latent log event time. Its E-step uses the displayed conditional weights and the first two moments of a truncated normal distribution above $\log c$. With $z=(\log c-{\mu }_j)/{\sigma }_j$ and inverse Mills ratio $\psi (z)$, these moments are $m={\mu }_j+{\sigma }_j\psi (z)$ and $s=m^2+{\sigma }_j^2[1-\psi (z)(\psi (z)-z)]$. Observed outcomes use deterministic class weights and moments $\log t,(\log t{)}^2$. The M-step updates each class probability to its average weight, its mean to the weighted first moment, and its variance to the weighted second moment minus the squared new mean. Outcome-specific cumulative probabilities are cumulative incidence functions, not the conditional distributions $F(t\mid I=j)$ themselves.

For a connected $n$-dimensional manifold without boundary,Choose a locally finite good cover by coordinate disks, each with a local orientation. Compactly supported classes involve only finitely many of these disks at a time. The compact-support Mayer–Vietoris sequence gives one generator $u_i$ per disk and overlap relations $u_i=\pm u_j$, with the sign determined by the two local orientations. Passing to the direct limit over finite unions handles a noncompact manifold; a finite good cover of the whole manifold need not exist.

The overlap graph is connected, so all generators reduce to one up to sign. If local orientations can be chosen coherently, every relation has positive sign and the result is $\mathbb{Z}$. An orientation-reversing loop imposes $u=-u$, giving $\mathbb{Z}/2$. This is also the top-degree case of Poincare duality with the orientation local system, using compactly supported cohomology on the cohomology side.

The compact set $K=\{0\}\cup \{1/n!:n\ge 1\}$ generates the additive group $\mathbb{Q}$, since every positive integer denominator divides some factorial. With its usual topology, $\mathbb{Q}$ is metrizable and is not locally compact. Nor can it have a nonzero Haar measure finite on compact sets: invariance gives equal masses to all singletons; a positive singleton mass makes $K$ have infinite mass, whereas zero singleton masses make the countable group have zero total mass.

// Target: analysis.bigb

Let finitely many arcs have centers ${\theta }_j$, half-lengths $0<{\delta }_j\le 1$ and $D={\sum }_j{\delta }_j$. Each summand has positive real part on the closed disk, and $\Phi (0)=1$. On the $j$th arc its contribution has real part at least $1/(4D)$. Thus the analytic logarithm $A=\log \Phi$ obeys $A(0)=0$, $|\mathrm{Im}A|<\pi /2$, and $\mathrm{Re}A\ge \log (1/(4D))$ throughout the arc union. Approximate $A$ by an analytic polynomial and modulate its bounded imaginary part; a partial Fourier sum isolates one half of the polynomial, producing a large value from a bounded function.

A metric space is totally bounded when, for every $\epsilon >0$, it is covered by finitely many open balls of radius $\epsilon$.

A metric space is totally bounded when, for every $\epsilon >0$, finitely many open balls of radius $\epsilon$ cover it. Equivalently it has a finite $\epsilon$-net at every positive scale.

Embed $\mathbb{N}$ into the compact product $\{0,1{\}}^{\mathcal{P}(\mathbb{N})}$ by $j(n{)}_S=1_S(n)$ and take its closure $X$. The Tychonoff theorem makes $X$ a compact Hausdorff space, while $j(\mathbb{N})$ is countable, dense and discrete in its subspace topology. Yet $X$ is not metrizable. Any convergent subsequence $j(n_k)$ of distinct points would have a convergent coordinate for $S=\{n_2,n_4,\dots \}$, whereas that coordinate alternates between zero and one. This contradicts sequential compactness of a compact metric space. Thus a dense metrizable subspace does not force a separable compact Hausdorff ambient space to be metrizable.

Let continuous maps $f,g:X\rightrightarrows Y$ between compact Hausdorff spaces have an underlying split coequalizer $q:Y\to Z$ with set maps $s,t$ satisfying $qs=1$, $ft=1$ and $gt=sq$. Its equivalence relation is the image of the closed subset $\{(x,x^{\prime })\in X^2:g(x)=g(x^{\prime })\}$ under $(f,f)$. One inclusion follows from $qf=qg$; for the other use $x=t(y)$ and $x^{\prime }=t(y^{\prime })$ whenever $q(y)=q(y^{\prime })$. This image is compact and hence closed in the Hausdorff product $Y^2$. The quotient is therefore compact Hausdorff, giving the preserved coequalizer required by the Beck monadicity theorem.

On a nonempty compact Hausdorff space, let continuous nonnegative functions decrease pointwise. Their suprema decrease to a finite number $L$. For every $t<L$, the sets $\{f_n\ge t\}$ are nonempty nested closed sets; compactness gives a common point with limit value at least $t$. Thus the supremum of the pointwise limit is $L$. The limit need not be continuous, and this conclusion does not claim uniform convergence as in the Dini theorem.

A nonempty closed subsemigroup $S$ of a compact group is a subgroup. For $x\in S$ and a neighbourhood $U$ of the identity, choose $V$ with $V^{-1}V\subseteq U$ and cover the compact group by finitely many translates of $V$. Two of sufficiently many powers $1,x,x^2,\dots$ lie in the same translate, so some positive power of $x$ belongs to $U$. Hence the identity is a limit of positive powers and lies in $S$. The preceding powers converge to $x^{-1}$, which also lies in $S$. Applied to the closure of a positive-word monoid, this proves that it equals the closed subgroup generated by its generators, without assuming they already include their inverses.

The Peter-Weyl theorem makes finite-dimensional unitary representations separate elements of a compact group. Intersecting their closed kernels gives the identity subgroup. If the descending chain condition holds for closed subgroups, choose finitely many representations whose kernel intersection is already trivial: otherwise repeatedly exclude a surviving nonidentity element to obtain an infinite strict chain. Their direct sum is a faithful representation into a finite-dimensional unitary group. Compactness makes its image closed, and the closed-subgroup theorem gives the original group a compatible Lie group structure.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.

On the circle group $\mathbb{T}$, normalized Haar measure is the pushforward measure of uniform angle on $[0,2\pi )$. For a continuous function $f:\mathbb{T}\to \mathbb{C}$, defineThe factor $1/(2\pi )$ makes the integral of the constant function $1$ equal to $1$. For $g=e^{i\varphi }$, a change of angle givesThe last equality uses the $2\pi$-periodic function $u\mapsto f(e^{iu})$. Since the circle group is an abelian group, this also verifies right invariance. An arc of angular length $L$ has normalized Haar measure $L/(2\pi )$. In particular, the arc $A=\{e^{i\theta }:\pi /6\le \theta \le \pi /2\}$ and its quarter-turn $iA$ both have measure $(\pi /2-\pi /6)/(2\pi )=1/6$.

The continuous unitary characters ${\chi }_n(e^{i\theta })=e^{in\theta }$, indexed by integers, give a fully computed family of integrals:Here $n$ must be an integer, so $z^n$ is well defined and $e^{2\pi in}=1$. Every nonconstant mode cancels exactly over a full turn. For example, the integral of $f(e^{i\theta })=3+2\cos \theta -\sin (2\theta )$ is $3$, and the integral of ${\cos }^2\theta =(1+\cos (2\theta ))/2$ is $1/2$. More generally,This is the orthogonality of complex exponentials underlying the Fourier basis and Fourier series: invariant averaging separates distinct frequencies and retains the constant component.