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The necessary formulation uses a real Hilbert space $V$ of functions satisfying the essential boundary conditions, a continuous symmetric bilinear form $a$, and a continuous linear functional $\ell$. Assume constants $C,c>0$ withThe latter is coercivity, the positive-definiteness condition needed in the chosen complete space. On the domain of the strong differential operator, integration by parts and the boundary conditions identify $a(v,w)=⟨\mathcal{L}v,w⟩$; this form extends to $V$. Let $\ell (w)=⟨f,w⟩$ for admissible forcing. DefineWhen $u$ has the regularity to belong to the strong operator domain, this is $\mathcal{L}u=f$ with its boundary conditions. For complex spaces use a Hermitian form and $-2\mathrm{Re}\ell (v)$ to make the functional real.

Here is an existence proof by minimization. Continuity of $\ell$ and coercivity give $I(v)\ge c\Vert v{\Vert }_V^2-2\Vert \ell {\Vert }_{V^{\ast }}\Vert v{\Vert }_V$, so $m={\inf }_{V}I$ is finite and a minimizing sequence $v_n$ is bounded. The quadratic parallelogram identity isSince its left side is at least $m$,Coercivity makes $v_n$ Cauchy in $V$. Completeness gives $v_n\to u\in V$, and continuity of the form and forcing gives $I(u)=m$.

For every $w\in V$, differentiation along $u+tw$ yieldsThus the Euler-Lagrange equation is exactly the stated weak solution equation. Conversely, if $u$ satisfies it, expansion givesThis proves that the weak solution is the unique global minimum and that two weak solutions coincide. It proves the symmetric coercive case underlying the Lax-Milgram theorem directly, rather than taking existence as a theorem name. The framework is symmetry and coercivity in quadratic energy minimization.

The qualifications matter if positive definiteness is read more weakly. For a nonsymmetric operator the derivative of $⟨\mathcal{L}v,v⟩$ uses $(\mathcal{L}+{\mathcal{L}}^{\ast })/2$, so it need not give $\mathcal{L}u=f$. For example, on smooth periodic real functions $\mathcal{L}=1+{\partial }_x$ has $⟨\mathcal{L}v,v⟩=\Vert v{\Vert }_2^2$, but the functional's equation is $u=f$, not $u+u_x=f$. Even a symmetric operator can be strictly positive without being coercive: on l2 sequence space, $(Lv{)}_n=v_n/n$ with $f_n=1/n$ would require a solution $u_n=1$, which is not square summable. The associated functional has values $-{\sum }_{n=1}^{N}1/n$ on vectors with the first $N$ entries equal to one, so it has no minimum in that space. This is strict positivity without coercivity can fail variational solvability. Under the usual symmetric coercive meaning of positive definiteness, the proof above supplies every assertion requested.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

Let $d=[L:K]$, with normalized discrete valuations $v_K,v_L$. Suppose first that $L/K$ is totally ramified, so $e(L/K)=d$. Choose a uniformizer $x$ of $L$. The elements $1,x,\dots ,x^{d-1}$ are linearly independent over $K$: the nonzero terms $c_j{x}^j$ in any proposed relation have valuations $d{v}_K(c_j)+j$, distinct modulo $d$, and a sum with a unique least valuation cannot vanish by the ultrametric inequality. Hence $[K(x):K]\ge d$, proving $L=K(x)$.

Extend $v_K$ to an algebraic closure with values in $\mathbb{Q}$. Every conjugate of $x$ has valuation $1/d$, by uniqueness of the extended valuation on finite extensions of the complete field $K$. In its monic minimal polynomial of an algebraic element $g(T)=T^d+a_{d-1}{T}^{d-1}+\cdots +a_0$, the coefficient $a_{d-j}$ is a sum of products of $j$ conjugates, counted with multiplicity. Thus $v_K(a_{d-j})\ge j/d>0$ for $j\ge 1$, so every nonleading coefficient is in ${\mathfrak{m}}_K$. The constant coefficient is the product of all $d$ conjugates, so has valuation exactly one. This proves that $g$ is an Eisenstein polynomial. This argument also works for inseparable finite extensions, since the conjugates can be counted with their multiplicities.

Conversely let $g$ be an Eisenstein polynomial of degree $d$, with root $x$ and $L=K(x)$. The Eisenstein criterion proves irreducibility. For completeness, monic factors would have integral coefficients; their reductions would be positive-degree powers of $T$, so both constant coefficients would be divisible by the uniformizer of $K$, contradicting $v_K(g(0))=1$. To determine the root valuation, put $t=v_K(x)$ in the extended valuation. If $t<0$, the leading term is the unique term of least valuation; if $t=0$, the leading term again has uniquely least valuation zero. Both are impossible. If $t>0$, every nonconstant lower term has valuation at least $1+jt>1$. Cancellation in $g(x)=0$ therefore requires the leading term and constant term to have equal valuation:Since $t=v_L(x)/e(L/K)$ and $v_L(x)$ is a positive integer, $e(L/K)=d{v}_L(x)\ge d$. The opposite inequality follows from $[L:K]=e(L/K)f(L/K)=d$. Hence $e=d$, $f=1$, and $v_L(x)=1$. We have proved both the equivalence and the uniformizer assertion:

Now take $q=p^a$ with $a\ge 1$ and $\zeta ={\zeta }_{p^a}$. The translated cyclotomic polynomial isIn characteristic $p$, putting $U=T^{p^{a-1}}$ reduces this to ${\sum }_{j=0}^{p-1}(1+U{)}^j=U^{p-1}=T^{p^{a-1}(p-1)}$. Its constant coefficient over the integers is $p$. It is therefore Eisenstein at $p$, of degree $d=p^{a-1}(p-1)$. Consequently ${\mathbb{Q}}_p({\zeta }_{p^a})/{\mathbb{Q}}_p$ is totally ramified of degree $d$, with uniformizer ${\zeta }_{p^a}-1$. If $q=1$, the field is simply ${\mathbb{Q}}_p$, a trivial totally ramified extension.

For a general positive integer $n=p^{a}m$ with $p\nmid m$, the field generated by the prime-to-$p$ root of unity ${\zeta }_m$ is unramified. Indeed $T^m-1$ has distinct roots modulo $p$. In a finite unramified field whose residue field contains those roots, Hensel lemma lifts all of them uniquely. Reduction preserves their orders: a prime-to-$p$ root reducing to one must itself be one, by Hensel uniqueness at the root one of $T^m-1$. The arithmetic Frobenius therefore acts as ${\zeta }_m\mapsto {\zeta }_m^p$, and the extension degree is the multiplicative order of $p$ modulo $m$, with degree one when $m=1$.

The identityfollows by taking powers of a primitive nth root of unity and using the coprimality of $p^a,m$ in the opposite direction. The prime-to-$p$ field is unramified and the $p$-power field is totally ramified; unramified base change preserves total ramification proves that their compositum has the same residue degree as the prime-to-$p$ field. Thus the total ramification criterion for local cyclotomic extensions gives the full classification:This includes trivial extensions when $a=0$. For $p=2$ only $m=1$ is possible.

For a finite totally ramified Galois extension $L/K$ with Galois group $G$, define its lower ramification numbering byHere $G_0$ is the inertia group, which equals $G$, and $G_1$ is the wild inertia group. For a uniformizer $\pi$ the condition is equivalent to $v_L(\sigma (\pi )-\pi )\ge i+1$. To justify this uniformizer criterion for lower ramification groups, expand every $b\in {\mathcal{O}}_L$ as a convergent series ${\sum }_{j\ge 0}{c}_j{\pi }^j$ with the coefficients $c_j$ chosen from representatives of $k_K$ in ${\mathcal{O}}_K$. Such an expansion is obtained by successive subtraction of a residue representative and division by $\pi$, since total ramification gives $k_L=k_K$. The automorphism fixes each $c_j$, and $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$. Thus the required valuation bound for $\pi$ implies it for every $b$, including the limit by completeness.

For $L={\mathbb{Q}}_p({\zeta }_{p^a})$, all primitive conjugates lie in $L$ and the Eisenstein degree is $\phi (p^a)$. HenceIf ${\sigma }_b\ne 1$, put $s=v_p(b-1)$, so $0\le s<a$. The element ${\zeta }^{b-1}$ is a primitive $p^{a-s}$th root of unity. Applying the already proved uniformizer result to its field, and multiplying its valuation by the ramification index of $L$ over that field, givesThe identity automorphism has infinite valuation. Therefore the lower ramification filtration of a prime-power cyclotomic extension is explicitlyThe order of the middle group is $p^{a-j}$. At odd $p$, $G_0/G_1$ has order $p-1$; at $p=2$, the condition $b\equiv 1(\mathrm{mod}2)$ is automatic, so $G_0=G_1$ when $a\ge 2$. The case $p=2,a=1$ has trivial Galois group, and the displayed formula remains valid. For $a=0$ all groups are trivial. The valuation calculation also provides the upper ramification numbering, if desired, by integrating the reciprocal indices with the Herbrand function; no change of numbering is implicit in the displayed answer.

Let $\pi \in K$ be a uniformizer. Because $L/K$ is unramified, the same $\pi$ is a uniformizer of $L$, and $v_L{|}_K=v_K$. The conjugates of $z\in L^{\times }$ all have the same valuation, soThis proves that every field norm has valuation divisible by $n$. Conversely $N_{L/K}({\pi }^r)={\pi }^{nr}$ for every integer $r$, so it remains to prove surjectivity of unit norms in unramified extensions.

Write $|k_K|=Q$, so $|k_L|=Q^n$. Reduction of the field norm on units is the finite-field normThe multiplicative group of a finite field is cyclic. The displayed power map on a group of order $Q^n-1$ has image of order $Q-1$, and is therefore onto $k_K^{\times }$. Given a unit $u\in {\mathcal{O}}_K^{\times }$, first choose $b_1\in {\mathcal{O}}_L^{\times }$ with $N(b_1)\equiv u(\mathrm{mod}\pi )$.

We can correct the norm at each subsequent power of $\pi$. For $r\ge 1$ and $a\in {\mathcal{O}}_L$, expanding the product over the cyclic Galois group givesAll products involving at least two correction terms are divisible by ${\pi }^{2r}$, hence by ${\pi }^{r+1}$. The residue field trace is onto even if the residue characteristic divides $n$. Indeed its polynomial $T(X)=X+X^Q+\cdots +X^{Q^{n-1}}$ is nonzero of degree $Q^{n-1}<Q^n$; it cannot vanish at every element of $k_L$. Its values lie in $k_K$, since raising $T(a)$ to $Q$ cyclically permutes the summands. Thus it is a nonzero $k_K$-linear map to the one-dimensional space $k_K$, and is surjective.

Suppose $N(b_r)\equiv u(\mathrm{mod}{\pi }^r)$. Choose ${\bar{a}}_r$ whose trace is the coefficient of ${\pi }^r$ in $u/N(b_r)-1$, and put $b_{r+1}=b_r(1+{\pi }^r{a}_r)$. Then $N(b_{r+1})\equiv u(\mathrm{mod}{\pi }^{r+1})$, and $b_{r+1}\equiv b_r(\mathrm{mod}{\pi }^r)$. By completeness the units $b_r$ converge to a unit $b$. The field norm is continuous, being a polynomial in coordinates of multiplication, so $N(b)=u$. Multiplying such a unit by a power of $\pi$ provesThe proof uses no local class field theory, and works in either characteristic.

A metric space is complete if every Cauchy sequence in it converges to a point of that space. The stated shrinking-set characterization is the Cantor intersection theorem together with its converse.

First assume completeness, and choose $x_n\in A_n$. If $m,n\ge N$, then both points belong to $A_N$, so $d(x_m,x_n)\le \mathrm{diam}(A_N)$. The vanishing diameters imply that $(x_n)$ is a Cauchy sequence, hence it converges to some $x\in X$. For every fixed $N$, its tail belongs to the closed set $A_N$, so $x\in A_N$. Thus $x\in {\bigcap }_N{A}_N$. In fact the intersection contains only one point, since any two of its points have distance at most every diameter.

Conversely assume the stated intersection property and let $(x_n)$ be a Cauchy sequence. PutThese closures are nonempty, closed and nested. Taking a closure does not increase the diameter: approximate two points of the closure by points of the set and use continuity of distance. The Cauchy sequence therefore makes $\mathrm{diam}(A_N)\to 0$. Choose $x$ in the intersection. For $n\ge N$, both $x_n$ and $x$ belong to $A_N$, and $d(x_n,x)\le \mathrm{diam}(A_N)\to 0$. Every Cauchy sequence converges in $X$, proving completeness.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

For fixed $m$, the reverse triangle inequality givesThe right side tends uniformly to zero as $n,k\to \infty$ because $(x_n)$ is a Cauchy sequence. Hence $(d(x_m,x_n){)}_n$ is a real Cauchy sequence. By completeness of $\mathbb{R}$, it has a finite, nonnegative limit:This conclusion uses completeness only of the real line, without assuming that $X$ is complete.

The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

The Lax-Milgram theorem assumes that $H$ is a real Hilbert space and $B:H\times H\to \mathbb{R}$ is a bounded bilinear form and a coercive bilinear form: there are $M,\alpha >0$ such thatThen for every continuous linear functional $\ell \in H^{\prime }$ there is a unique $u\in H$ satisfyingSymmetry of $B$ is not required.

For the proof, the Riesz representation theorem provides a bounded linear operator $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v{)}_H$ and $\ell (v)=(g,v{)}_H$. Coercivity and the Cauchy-Schwarz inequality implyThus $T$ is injective and its range is closed: if $T{u}_j$ converges, the displayed inequality makes $u_j$ a Cauchy sequence; completeness gives $u_j\to u$ and $T{u}_j\to Tu$. If $w$ lies in the orthogonal complement of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha \Vert w{\Vert }^2\le B[w,w]=0$, so $w=0$. A closed subspace with zero orthogonal complement is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

One strong form of Hensel lemma is this: if $K$ is complete for a discrete valuation $v$, $f\in {\mathcal{O}}_K[X]$, and $a_0\in {\mathcal{O}}_K$ satisfiesthen there is a unique root $\alpha$ satisfying

Define the Newton iteration over a valued fieldTaylor expansion and the ultrametric inequality show inductively that $v(f^{\prime }(a_n))=v(f^{\prime }(a_0))$ andThus the valuations of the corrections $a_{n+1}-a_n$ tend to infinity, so $(a_n)$ is a Cauchy sequence. Completeness gives a limit $\alpha$, and continuity gives $f(\alpha )=0$. If $\beta$ is another root in the stated ball, Taylor expansion of $f(\beta )-f(\alpha )$ shows that the linear term has strictly smaller valuation than all higher terms unless $\beta =\alpha$, proving uniqueness.

Let $(x_n,y_n)$ be a Cauchy sequence in the product. Since each coordinate distance is bounded by $\delta$, both $(x_n)$ and $(y_n)$ are Cauchy in their respective factors. By completeness they have limits $x\in M$, $y\in N$. Part (ii) then gives convergence of the pairs to $(x,y)$.

Thus the product of two complete metric spaces is complete under the maximum product metric. This checks every Cauchy sequence, rather than assuming beforehand that it has a coordinate limit.

A complete metric space is one in which every Cauchy sequence converges to a point of that space. For an incomplete example, take $(0,1)$ with the usual Euclidean metric and $x_n=1/(n+1)$. It is Cauchy because it converges to zero in $\mathbb{R}$. It cannot converge inside $(0,1)$: any such convergence would also hold in $\mathbb{R}$, contradicting uniqueness of the limit, since zero is excluded. Thus the ambient metric and the underlying set both matter to completeness.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.

The space of polynomials is not complete for the supremum norm metric $d$. ConsiderThe geometric series has sum $F(t)=1/(1-t/2)$, and for $0\le t\le 1$ its tail satisfiesThus $(p_n)$ converges uniformly to $F$ and is a Cauchy sequence for $d$.

There is no polynomial $p$ agreeing with $F$ on $[0,1]$: otherwise the polynomial $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero polynomial. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the uniform limit, $(p_n)$ has no limit in $X$. This proves the claimed failure of completeness.

A metric space $(X,d)$ is a bounded set if there is a finite constant $M\ge 0$ such thatFor nonempty $X$, this is equivalent to containment in an open ball of finite radius: the triangle inequality converts a radius bound into a diameter bound, and conversely a diameter bound gives a ball of any slightly larger radius about a chosen point.

It is a complete metric space if every Cauchy sequence in $X$ has a limit of a sequence belonging to $X$. Explicitly, the hypothesis on $(x_n)$ isand completeness requires an $x\in X$ with $d(x_n,x)\to 0$. ThusBoundedness controls distances; completeness ensures that these limiting points remain in the space.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

With $d=\Delta x$, define the semidiscrete Fourier transform byFor finite-support sequences, expand the squared modulus and use orthogonality:ConsequentlyFor a general square-summable sequence, truncate it to $|j|\le N$. The identity makes the transformed truncations Cauchy in $L^2$, and completeness defines the transform as their $L^2$ limit. Passing to the limit proves the Parseval identityThe infinite series need not converge absolutely pointwise; the $L^2$ interpretation is sufficient for the stability analysis.