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The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

A normed vector space is a Banach space exactly when every absolutely convergent series converges in the space. Completeness implies the assertion because the triangle inequality makes the partial sums a Cauchy sequence. Conversely, select from any Cauchy sequence a subsequence whose consecutive differences have norm at most $2^{-k}$. The series of these differences converges by hypothesis, so the subsequence converges. The original Cauchy sequence then has the same limit. This provides a useful way to demonstrate incompleteness by a summable-norm series whose candidate sum lies outside the space.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$. The partial sums $A_N={\sum }_{n=0}^N{p}^{2n}{\zeta }_{p^{n+1}}$ form a Cauchy sequence. Each $A_N$ generates ${\mathbb{Q}}_p({\zeta }_{p^{N+1}})$: for a nonidentity cyclotomic automorphism with $r=v_p(a-1)$, the first changed term has valuation $2r+1/(p-1)$, strictly below every later changed term. Thus the conjugate distance is at least $p^{-2N-1/(p-1)}$. The tail bound is $p^{-2N-2}$, strictly smaller. If the limit were algebraic, Krasner's lemma would place every one of these unbounded-degree cyclotomic fields in one finite extension, a contradiction.

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

Let $X$ be a normed space, $Y$ a Banach space, and $A_n:X\to Y$ be bounded linear operators with $C={\sup }_n\Vert A_n\Vert <\infty$. Convergence of $A_{n}d$ for every $d$ in a dense subset $D$ implies convergence of $A_{n}x$ for every $x\in X$ to a bounded linear operator $A$, with $\Vert A\Vert \le {\mathrm{liminf}}_n\Vert A_n\Vert$. Indeed, $\Vert A_{n}x-A_{m}x\Vert \le 2C\Vert x-d\Vert +\Vert A_{n}d-A_{m}d\Vert$ makes $(A_{n}x)$ a Cauchy sequence; completeness gives its limit, and linearity and the bound pass to that limit. If $X$ is a Banach space, pointwise convergence conversely implies the uniform bound by the Uniform boundedness principle. If the limit bounded linear operator is specified in advance, completeness of $Y$ is unnecessary.

If $\sum a_n$ converges and $b_n$ is bounded with $\sum |b_{n+1}-b_n|<\infty$, then $\sum a_n{b}_n$ converges. For $T_j={\sum }_{k=m}^j{a}_k$, summation by parts givesThe first factor tends uniformly to zero by the Cauchy sequence criterion, and the second has a uniform bound. A bounded monotone sequence has finite total variation, so monotone multipliers are included. Eventual monotonicity also suffices, since finitely many initial terms do not affect convergence. Absolute convergence of $\sum a_n$ is unnecessary.

For a function $F:(0,1)\to Z$ into a complete metric space, the displayed pairwise condition is equivalent to existence of a limit as $h\to 0^{+}$. Necessity follows from the triangle inequality. For sufficiency, the values $F(1/(n+1))$ form a Cauchy sequence, hence converge to some $z\in Z$. Given $\epsilon >0$, use the pairwise condition with $\epsilon /2$ and then send $n\to \infty$ in $d(F(h),F(1/(n+1)))<\epsilon /2$. This gives $d(F(h),z)\le \epsilon /2<\epsilon$ for every sufficiently small positive $h$.

A series over the real numbers or complex numbers converges if and only if its tails satisfy: for every $\epsilon >0$, some $N$ makes $|{\sum }_{k=n+1}^m{a}_k|<\epsilon$ whenever $m>n\ge N$. This says exactly that its partial sums form a Cauchy sequence; sufficiency uses completeness.

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

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.

A Cauchy subsequence is a subsequence that is a Cauchy sequence. Every subsequence of a Cauchy sequence is a Cauchy sequence, because its increasing indices eventually lie beyond any index required by the original pairwise bound. Every convergent subsequence is also a Cauchy sequence.

The converse about the original sequence fails: $a_n=(-1{)}^n$ has a constant subsequence at its even integers, hence a Cauchy sequence, while $a_n$ itself is not a Cauchy sequence. If the original sequence is already a Cauchy sequence and one subsequence converges to $L$, however, the whole sequence converges to $L$. Choose a late subsequence term close to $L$ and use the triangle inequality to compare it with every sufficiently late original term. ^[ref].

For a bounded linear operator $T$ between Hilbert spaces, its range is closed exactly when there is $b>0$ with $\Vert Tv\Vert \ge b\Vert v\Vert$ for all $v\in (\ker T{)}^{\perp }$. The restriction to this orthogonal complement is a bounded bijection onto the range, so the bounded inverse theorem proves necessity. Conversely, the bound makes preimages of a Cauchy sequence of image points a Cauchy sequence, proving closedness by completeness.

For $2\le p<\infty$, a nonempty closed convex set $K$ in an Lp space has exactly one nearest point to every $f$. A minimizing sequence $g_j$ is a Cauchy sequence: the Clarkson inequality bounds $\Vert g_j-g_k{\Vert }_p^p$ by $2^{p-1}(\Vert g_j-f{\Vert }_p^p+\Vert g_k-f{\Vert }_p^p)-2^p{d}^p$, where $d$ is the infimum of the distances. Completeness of Lp spaces and the closed set property give a limit in $K$. Applying the same inequality to two minimizers proves uniqueness. Nonemptiness is essential.

A metric space is complete when every Cauchy sequence converges to a point of the space.

If a bounded linear operator from a Banach space onto a normed vector space admits the displayed lift for every $y$, then its target is complete. Given a Cauchy sequence, choose a subsequence with summable successive differences. Lift those differences to a summable sequence in the source, and use its convergent sum to produce a limit of the target subsequence. The original Cauchy sequence has the same limit.

Let $K$ be a nonempty compact space and let $\alpha :K\to \mathbb{R}$ be continuous and nowhere zero. On $C(K)$, define $\Vert f{\Vert }_{\alpha }={\sup }_{x\in K}|\alpha (x)f(x)|$. Multiplication by $\alpha$ is an isometry onto $C(K)$ with its ordinary uniform norm, because division by the continuous nonvanishing $\alpha$ provides the inverse. Thus the weighted space is a Banach space. Equivalently, the extreme value theorem gives $0<m=\min |\alpha |\le \max |\alpha |=M<\infty$, and $m\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_{\alpha }\le M\Vert f{\Vert }_{\infty }$. A weighted Cauchy sequence is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.

Let $X$ be a normed space, $Y$ a Banach space, and $T:X\to Y$ a linear map. If its kernel is complete in the original norm and $S:Y\to X$ is a bounded right inverse, then the graph norm $\Vert x{\Vert }_T=\Vert x{\Vert }_X+\Vert Tx{\Vert }_Y$ is complete, even if $X$ with its original norm is incomplete or $T$ is unbounded.

For a Cauchy sequence $(x_n)$ in the graph norm, $y_n=T{x}_n$ converges to $y\in Y$. The elements $w_n=x_n-S{y}_n$ lie in the kernel and satisfy $\Vert w_n-w_m{\Vert }_X\le (1+\Vert S\Vert )\Vert x_n-x_m{\Vert }_T$. Hence $w_n\to w\in \ker T$. The element $x=w+Sy$ satisfies $Tx=y$ and

In a homogeneous Riemannian manifold, choose $r>0$ so that one closed metric ball $\overline{B}(o,r)$ is compact. Such a ball exists from local compactness and the metric topology. Every radius-$r$ ball is isometric to it. A Cauchy sequence eventually lies in one such compact ball, has a convergent subsequence, and therefore converges. This proves metric completeness; the Hopf-Rinow theorem gives geodesic completeness and minimizing geodesics between points.

A Cauchy sequence $(u_n)$ in the BV space converges in $L^1$ to $u$. For fixed sufficiently large $n$, lower semicontinuity gives $|D(u_n-u)|(\Omega )\le {\mathrm{liminf}}_m|D(u_n-u_m)|(\Omega )$. The $L^1$ norm of the same difference converges, so the small full-norm Cauchy bound passes to $u_n-u$. The triangle inequality then puts $u$ in the BV space and proves convergence in its full norm. This argument turns completeness of $L^1$ and lower semicontinuity of variation into completeness of the combined norm.

For any nonempty complete metric space $X$, the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.

For a Cauchy sequence $(F_n)$ of these sets, define $A_n=\overline{{\bigcup }_{m\ge n}{F}_m}$ and ${\delta }_n={\sup }_{p,q\ge n}{d}_H(F_p,F_q)$. A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded $F_N$; hence every $A_n$ is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving $d_H(F_n,A_n)\le {\delta }_n$ and $d_H(A_n,A_m)\le {\delta }_n$ for $m\ge n$. Choose $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. The Summably nested Hausdorff intersection theorem gives $A_{n_j}\to G\ne \varnothing$. The triangle inequality now gives $F_n\to G$, proving completeness.