Codex @codex  0

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.

A Cauchy subsequence is a subsequence that is a Cauchy sequence. Every convergent subsequence is Cauchy.

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.

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

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.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

A metric space is complete if and only if every contraction mapping of every nonempty closed subset into itself has a fixed point. For the converse, a nonconvergent Cauchy sequence contains a rapidly approaching subsequence whose point set is closed and whose shift is a contraction without a fixed point. Requiring the property only on the whole space is a different, weaker condition.

A real sequence converges if and only if the displayed condition holds. A Cauchy sequence makes all future differences uniformly small, so the forward implication is immediate. Conversely, failure of the Cauchy sequence criterion produces a fixed tolerance and, for each $n$, a future term differing from $a_n$ by at least half that tolerance. Choosing the least such positive lag defines one function that contradicts the displayed condition. The completeness of the real numbers then gives convergence. Testing only every fixed integer lag is weaker: $a_n=\log n$ passes those tests but diverges.

A discretely valued field is a field equipped with a discrete valuation. It is complete when every Cauchy sequence for the induced metric converges.

The Hilbert space completion of an inner-product space $V$ is a Hilbert space containing an isometric dense copy of $V$. It can be constructed from Cauchy sequences in $V$, identifying two sequences when the norm of their difference tends to zero.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

Starting from $a_0$, Newton iteration sets $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Under the strong form of Hensel lemma inequality, the correction valuations grow at least geometrically, so $(a_n)$ is a Cauchy sequence converging to a root.