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The level-two principal congruence subgroup $\Gamma (2)\subset {\mathrm{PSL}}_2(\mathbb{R})$ consists of integer determinant-one matrices congruent to the identity modulo two, modulo the scalar sign. It acts freely and properly discontinuously on the complex upper half-plane. An ideal quadrilateral with vertices $-1,0,1,\infty$ has its paired sides identified by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(1-2\tau )$. The quotient has genus zero and three punctures, so it is conformally the sphere with three points removed. The invariant metric $|d\tau |/\mathrm{Im}\tau$ descends to its curvature-minus-one hyperbolic metric.

// Destination: analysis.bigb

A generating family consists of sets of size at most $r$ whose $r$-element supersets are exactly the prescribed uniform set family. Small-support generators express dependence on a bounded prefix rather than on the full ground set. Taking the entire family itself as generators always works, but extremal intersection problems admit much smaller supporting intervals.

Every nonzero Noetherian module contains a uniform module: otherwise repeatedly split a nonuniform nonzero submodule into two disjoint nonzero submodules, retaining one and collecting the other, to build an infinite direct sum. Repeatedly adding a uniform submodule disjoint from the current finite sum must likewise stop. The final finite sum is an essential submodule, since any disjoint nonzero submodule would allow another addition.

The uniform dimension of a module is the supremum of the numbers of nonzero submodules that can occur in an internal direct sum. Finite uniform dimension is equivalent to having no infinite internal direct sum of nonzero submodules. A finite direct sum of uniform modules contained as an essential submodule gives that number as the dimension. The finite uniform decomposition of a Noetherian module therefore gives finite uniform dimension.

On a finite time horizon, an adapted process is of class D if its values at all stopping times in that horizon form a uniformly integrable family. Deterministic boundedness implies this property. The stopping-time uniformity is stronger than having an integrable value at each fixed time.

An $r$-uniform hypergraph has the extension property when, for every finite vertex set $S$ and every subfamily $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{S}{r-1}\right)$, there is a vertex $v\notin S$ such that $F\cup \{v\}$ is a hyperedge exactly for $F\in \mathcal{A}$, among all $F\in \left(\genfrac{}{}{0ex}{}{S}{r-1}\right)$. Every prescribed pattern has infinitely many witnesses: enlarge $S$ by any finite set of witnesses already found and prescribe the original pattern on its old $(r-1)$-subsets, with arbitrary choices on the new ones. For $r=2$, this says that any finite adjacency and nonadjacency prescription has a new realizing vertex.

Choose each $r$-element subset of the natural numbers as a hyperedge independently with probability $p$, where $0<p<1$. This defines a countable random uniform hypergraph. For any finite pattern in the hypergraph extension property, an outside vertex realizes it with probability $q=p^{|\mathcal{A}|}(1-p{)}^{|(\genfrac{}{}{0ex}{}{S}{r-1})\setminus \mathcal{A}|}>0$. Distinct outside vertices use disjoint edge decisions, so the probability that the first $N$ candidates all fail is $(1-q{)}^N\to 0$. Taking a countable intersection over all finite patterns proves the extension property almost surely. The bounds $0<p<1$ are necessary for all patterns to have positive probability.

The link at a fixed vertex $v$ of an $r$-uniform hypergraph has vertex set $V(H)\setminus \{v\}$ and the displayed $(r-1)$-element hyperedges. For a universal countable uniform hypergraph, its link has the hypergraph extension property: include $v$ in the finite set of vertices and prescribe only those $r$-edges containing $v$, completing the remaining prescription arbitrarily. Thus the link is a universal countable $(r-1)$-uniform hypergraph. In particular the link at a fixed vertex in the three-uniform case is the Rado graph. This operation retains a specified vertex; deleting the maximum from every edge is a different operation.

For a tripartite three-uniform hypergraph on nonempty finite parts $X,Y,Z$, let $h:X\times Y\times Z\to \{0,1\}$ be its hypergraph edge indicator, let $p=\mathbb{E}h$ and put $g=h-p$. In the eighth-power convention, $\alpha$-quasirandomness meansAll averages are uniform and independent, with repeated sampled vertices of a hypergraph allowed. This expression is nonnegative because it equals ${\mathbb{E}}_{x_0,x_1,y_0,y_1}({\mathbb{E}}_z{\prod }_{i,j}g(x_i,y_j,z){)}^2$. It measures correlations across the eight faces of a tripartite octahedral configuration. If instead the norm itself is used as the quasirandomness parameter, the parameter in this definition is its eighth power.

A property of uniform hypergraphs is hereditary if every induced subhypergraph of a hypergraph with that property also has the property. It is understood to be invariant under relabelling the vertex set. For example, excluding a specified induced hypergraph is hereditary. This permits counting restrictions to smaller vertex sets using the same counting function.

A uniform Newtonian gravitational field inside one spherical void can be cancelled at a point by the repulsive field representing another removed sphere. The latter field has a Jacobian matrix proportional to $I-3\hat{r}{\hat{r}}^T$: its axial eigenvalue is negative and both transverse eigenvalues positive. The equilibrium is restoring along the axis but unstable transversely. A zero force is thus not enough for stable rest.

On a finite-measure measurable set, a sequence of measurable functions converging pointwise almost everywhere converges uniformly outside a set of arbitrarily small measure. For tolerance $1/k$, the sets on which every index after $N$ has error at most $1/k$ increase to full measure as $N\to \infty$. Choose $N_k$ with exceptional measure at most $\epsilon 2^{-k-1}$ and intersect the good sets. Countable subadditivity bounds the discarded measure, and the resulting tail estimates prove uniform convergence. Finite measure is essential: $1_{[n,\infty )}$ on $[0,\infty )$ is a counterexample.

Suppose continuous functions $f_n$ converge by uniform convergence to $f$ on a domain of a function $X$, and $x_n\to x$ with all these inputs in $X$. The uniform limit theorem makes $f$ a continuous function, and the triangle inequality givesThe first term handles the changing approximation and the second handles the changing input. For $f_n(x)=x^n$ on $[0,1]$, the inputs $x_n=2^{-1/n}$ instead give $x_n\to 1$, $f_n(x_n)=1/2$, and $f(1)=1$. Pointwise convergence cannot supply the first uniform bound, and this pointwise limit also lacks continuity at one.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

On the half-line, $\sup f_n=1/(en)\to 0$, but ${\int }_0^{\infty }{f}_n=1$. The change of variable $x=ny$ shows why: the decreasing amplitude is spread over an increasing interval. Uniform convergence controls integrals on finite-measure domains; it does not alone control total mass on an infinite-measure domain.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.

Suppose $d$ is nonnegative, integrable and uniformly continuous on the positive half-line. If $d(t_n)\ge ϵ>0$ at times tending to infinity, uniform continuity supplies intervals of one fixed positive length around those times on which $d\ge ϵ/2$. An infinite disjoint subsequence of these intervals contradicts integrability. Thus the displayed implication holds. Mere smoothness at every finite time is insufficient: smooth peaks of shrinking widths can have finite integral and heights bounded away from zero.

A uniform algebra on $K$ is natural if every character of an algebra is evaluation at a point of $K$. The evaluation embedding then identifies its character space homeomorphically with $K$. A natural uniform algebra is characterized by the finite-generator criterion for a natural uniform algebra. In particular, $C(K)$ is natural; the disk algebra is natural when its underlying space is the closed disk.

The disk algebra consists of functions continuous on the closed unit disk and holomorphic in its interior, with the supremum norm. Polynomials are dense: dilations $f_r(z)=f(rz)$ converge uniformly to $f$ as $r\uparrow 1$, and each $f_r$ is uniformly approximated on the closed disk by its Taylor polynomials. A character of an algebra is therefore determined by its value $\lambda$ at the coordinate function; boundedness gives $|\lambda |\le 1$, and polynomial approximation yields $\phi (f)=f(\lambda )$. Consequently the disk algebra is a natural uniform algebra on the closed disk.

Let $C$ consist of continuous functions on the closed unit disk whose boundary values extend to a member of the disk algebra. The extension is unique by the maximum modulus principle; the extension map $T:C\to A(\overline{\mathbb{D}})$ is a surjective unital algebra homomorphism of norm at most one. Its kernel $I$ is the ideal of functions vanishing on the boundary. A character vanishing on $I$ factors through $T$, so it is $f\mapsto T(f)(\lambda )$ for some closed-disk point $\lambda$. For a character $\chi$ not vanishing on $I$, choose $h\in I$ with $\chi (h)\ne 0$. The rule $F\mapsto \chi (hF)/\chi (h)$ extends $\chi$ to a character of all continuous functions on the disk: multiplicativity follows from $\chi (hF)\chi (hG)=\chi (h)\chi (hFG)$. It is therefore evaluation at an interior point, since its value at $h$ is nonzero. These two closed-disk families agree exactly on their boundaries. Their glued compact space maps continuously and bijectively to the Hausdorff character space of an algebra, giving the stated homeomorphism. The algebra is closed because uniform convergence of boundary data gives uniform convergence of their analytic extensions.