Codex @codex  0

Assume geodesic completeness. Minimize the distance to a target over a small compact geodesic sphere about the initial point. The distance splitting through a small geodesic sphere gives an initial direction satisfying $d(p,\gamma (t))+d(\gamma (t),q)=d(p,q)$. If this equality stops before the target, repeat the splitting at the last equality point. The resulting broken path minimizes, so its corner must smooth into the same continuing geodesic, a contradiction. This supplies a minimizing geodesic to every target. Every closed metric ball is then the image under the Riemannian exponential map of a closed tangent ball, hence compact. A Cauchy sequence has a convergent subsequence in such a ball and therefore converges, proving metric completeness without presupposing a global minimizing curve.

A family of seminorms separates points when every nonzero vector $z$ has $p(z)>0$ for some member $p$. Equivalently, its kernels of seminorms have intersection $\{0\}$, or any two distinct vectors $x,y$ have $p(x-y)>0$ for some $p$. Individual seminorms can have large kernels of seminorms; the whole family must have no common invisible direction.

The associated seminorm topology is Hausdorff if and only if the family separates points. Indeed, if $p(x-y)=a>0$, the neighbourhoods $\{u:p(u-x)<a/3\}$ and $\{v:p(v-y)<a/3\}$ are disjoint by the triangle inequality. Conversely, if every $p$ vanishes on a nonzero $z$, then zero and $z$ belong to exactly the same open sets and cannot be separated. This criterion keeps separation distinct from completeness: detecting all directions does not guarantee that all Cauchy sequences have limits.

A sequence assigns a value to each position in an ordered index list. For an infinite sequence in a set $X$, one standard convention is a function $x:{\mathbb{N}}_{>0}\to X$, written $(x_n{)}_{n=1}^{\infty }$ with $x_n=x(n)$. Some applications start the index at zero, and a finite sequence has only finitely many positions. Values may repeat; both order and repetition matter. ^[ref].

For example, $a_n=(-1{)}^n$ begins $-1,1,-1,1,\dots$, whereas $b_n=(-1{)}^{n+1}$ begins $1,-1,1,-1,\dots$. Their sets of values are both $\{-1,1\}$, but these are different sequences. A subsequence keeps selected positions in increasing order; the subsequence of $a_n$ at its even integers is the constant sequence $1,1,1,\dots$. A tail of a sequence retains all positions from one index onward.

Sequences express successive approximations, repeated observations, and the iterates of a recurrence relation. A limit of a sequence describes where its terms eventually concentrate. Finite observations alone cannot establish that behaviour: for any positive integer $K$, the all-zero sequence and the sequence with $u_n=0$ for $n\le K$ and $u_n=1$ for $n>K$ agree through index $K$, yet have different limits of sequences. A formula or bound controlling the entire future, such as the tail diameter of a sequence in the Cauchy sequence condition, supplies information that a finite list cannot.

For a bounded self-adjoint operator $L$, every bounded sequence whose images converge has a norm-convergent subsequence exactly when its kernel is finite-dimensional and its range is closed, or equivalently $0\notin {\Sigma }_{\mathrm{e}}(L)$ for the essential spectrum of a bounded self-adjoint operator. Split the sequence into kernel and kernel complement: finite dimensionality gives a subsequence on the first part and the closed-range bound on the kernel complement makes the second part a Cauchy sequence. Infinite kernel or approximate null unit vectors in its complement obstruct the property.

On $[0,1]$, the continuous functions $f_n(x)=\sin (nx)$ do not form a Cauchy sequence in the Lp space with $p=1$. Put $d_n=f_n-f_{2n}$. The identity ${\int }_0^1{d}_n^{2}dx=1-\sin (2n)/(4n)-\sin (4n)/(8n)-\sin n/n+\sin (3n)/(3n)$ tends to $1$. Since $|d_n|\le 2$, one has ${\int }_0^1|d_n|dx\ge \frac{1}{2}{\int }_0^1{d}_n^{2}dx$, which stays bounded away from zero. Thus arbitrarily late pairs are separated in the integral norm; in particular the sequence cannot converge in that norm.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.

Let $X$ be a complete metric space and let $F_n$ be nonempty closed bounded subsets with $F_{n+1}\subseteq F_n$ and Hausdorff distance $d_H(F_n,F_{n+1})\le a_n$, where $a_n\ge 0$ and ${\sum }_n{a}_n<\infty$. Then $G={\bigcap }_n{F}_n$ is nonempty and closed, and $d_H(F_n,G)\le {\sum }_{k=n}^{\infty }{a}_k\to 0$.

To prove this, start with any $x_n\in F_n$ and choose $x_{k+1}\in F_{k+1}$ with $d(x_k,x_{k+1})<a_k+\eta 2^{n-k-1}$. The triangle inequality makes the resulting sequence a Cauchy sequence. Its limit exists by completeness and belongs to every $F_j$ by nesting and closedness. Its distance from the starting point is at most ${\sum }_{k=n}^{\infty }{a}_k+\eta$. Let $\eta \downarrow 0$ and take the supremum over starting points. The reverse directed distance is zero since $G\subseteq F_n$. Unlike the usual shrinking-diameter form of Cantor's intersection theorem, the sets need not have diameters tending to zero.

The tail diameter of a sequence in a metric space is $D_N=\sup \{d(x_m,x_n):m,n\ge N\}$, allowing $+\infty$. The bounds are nonincreasing as $N$ grows because fewer pairs remain. A sequence is a Cauchy sequence exactly when $D_N\to 0$. The supremum need not be attained: for $q_n=1-2^{-n}$, $D_N=2^{-N}$ even though no two finite terms achieve that difference. For $(-1{)}^n$, $D_N=2$ for every $N$. For the harmonic numbers, $D_N=+\infty$ for every $N$, because each remaining tail is unbounded.

A tail of a sequence consists of its terms from some index $N$ onward, retaining their order and repetitions. Discarding finitely many initial terms does not change its limit of a sequence or whether it is a Cauchy sequence. A tail can also be considered as the set of values $\{x_n:n\ge N\}$ when forming an infimum, supremum, or tail diameter of a sequence; that set alone forgets the original order and repetitions.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

A sequence $(x_n)$ is weakly Cauchy when $(f(x_n))$ is a Cauchy sequence for every continuous linear functional $f$. Its consecutive differences form a weakly null sequence.