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An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfyingA Non-Archimedean absolute value satisfies the stronger ultrametric inequality $|x+y|\le \max (|x|,|y|)$. Two equivalent absolute values induce the same topology, or equivalently differ by a positive real power. The trivial absolute value on a field takes value one on every nonzero element. The rational classification and the compactness criterion below concern nontrivial absolute values on a field; the trivial exceptions are given explicitly.

Here the additive valuation has real values. For any $c>1$, the mutually inverse constructions areMultiplicativity becomes $v(xy)=v(x)+v(y)$, and the ultrametric inequality becomes $v(x+y)\ge \min (v(x),v(y))$. Equivalent real-valued valuations differ by positive scaling, so these constructions give the required bijection on equivalence classes. Changing $c$ merely rescales the valuation.

To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to 0$. Thus two nontrivial absolute values on a field with the same topology give the same strict positivity relation on their additive valuations. Fix $t$ with $v_1(t)>0$. Comparing the signs of $m{v}_j(t)-n{v}_j(x)=v_j(t^m{x}^{-n})$, for integers $m$ and positive integers $n$, shows that $v_1(x)/v_1(t)$ and $v_2(x)/v_2(t)$ have identical rational cuts. They are equal, proving $v_2=c{v}_1$ for $c>0$. If one allows valuations in arbitrary ordered groups, the nontrivial classes arising this way are precisely rank-one valuations: higher-rank ordered value groups do not embed order-preservingly in $\mathbb{R}$.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, $|n|\le 1$ for every integer $n$, by repeatedly applying the ultrametric inequality to sums of ones. Some prime $p$ must have $|p|<1$, otherwise prime factorization and multiplicativity would make every nonzero rational have value one. There is at most one such prime: if both $|p|,|q|<1$, a Bezout identity $ap+bq=1$ contradicts the ultrametric inequality. If $p\nmid m$, another Bezout identity gives $|m|=1$. HenceThis proves the non-Archimedean part of the Ostrowski theorem. If the trivial absolute value on a field is admitted, it supplies one additional class and is not equivalent to any p-adic absolute value.

The valuation ring, its maximal ideal, and its residue field areSuppose the absolute value on a field is nontrivial and $R$ is compact. The ideal $\mathfrak{m}$ is an open additive subgroup of $R$, so $k$ is discrete; as a continuous image of a compact space it is finite. The subgroup $\mathfrak{m}$ is also closed, since all its cosets are open, and is therefore compact. The continuous function $|\cdot |$ attains a maximum $\rho$ on $\mathfrak{m}$, with $0<\rho <1$. Choose $\pi$ with $|\pi |=\rho$. Then the positive values of $v=-\log |\cdot |$ have least element $-\log \rho$. Division with remainder in this additive subgroup of $\mathbb{R}$ proves $v(K^{\times })=v(\pi )\mathbb{Z}$. Thus $K$ is a discretely valued field and $\pi$ is a uniformizer.

Conversely, normalize the discrete valuation by $v(\pi )=1$. If $k$ has $q$ elements, $R/{\pi }^{N}R$ has $q^N$ elements. Each quotient therefore supplies a finite cover by balls of radius tending to zero. The ring $R$ is a closed subset of the complete metric field $K$, hence complete and totally bounded, so it is compact. Equivalently,This is the local compactness criterion for a complete non-Archimedean field, with nontriviality understood. For the trivial absolute value on a field, $R=K$ has the discrete topology and is compact exactly when $K$ is a finite field; its value group is zero rather than a nonzero discrete cyclic group.

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.

The weak topology on a normed vector space $X$ is $\sigma (X,X^{\ast })$, the coarsest topology making every bounded linear functional continuous. A neighbourhood base at $x$ is given by finitely many inequalities $|f_j(y-x)|<\epsilon$, with $f_j\in X^{\ast }$. In the complex case, separation uses real parts of bounded linear functionals.

Mazur theorem states that the weak closure of a convex set equals its closure in the norm topology. The weak topology is coarser than the norm topology, so the norm closure is contained in the weak closure. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and $a\in \mathbb{R}$ with $\mathrm{Re}f(x)>a\ge {\sup }_{c\in C}\mathrm{Re}f(c)$. The corresponding weak topology neighbourhood of $x$ misses $C$, so $x$ is outside its weak closure. This proves Mazur theorem, including the empty-set case. In particular, if $x_n$ converges weakly to $x$, then $x$ is in the weak closure of every tail and therefore in the norm closure of its convex hull. Choosing a finite convex combination of the $n$th tail within $1/n$ of $x$ proves the usual Mazur lemma formulation as well.

The weak-star topology on the continuous dual space $X^{\ast }$ is $\sigma (X^{\ast },X)$: convergence means pointwise convergence on $X$, and a neighbourhood base prescribes finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this weak-star topology, even if $X$ is incomplete. Embed this closed unit ball intowhere $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Each factor is compact, so $P$ is compact by the Tychonoff theorem. Inside $P$, the equations $a_{x+y}=a_x+a_y$ and $a_{\lambda x}=\lambda a_x$ define a closed set. Every such point defines a linear functional satisfying $|a_x|\le \Vert x\Vert$, hence belongs to the closed unit ball of $X^{\ast }$. Thus this image is closed in $P$. The product topology on it is exactly the weak-star topology, proving Banach-Alaoglu theorem. Evaluations also separate its points, so the weak-star topology is Hausdorff.

For the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, we have $(J_{X}x)(f)=f(x)$. Restricting all evaluations at $f\in X^{\ast }$ therefore gives exactly $\sigma (X,X^{\ast })$. Thus the induced subspace topology is the weak topology on $X$.

Now identify $X$ with $J_{X}X$. Let $C$ be a bounded convex set, write $D$ for its norm closure in $X$, and let $K$ be its weak-star topology closure in $X^{\ast \ast }$. Boundedness places $K$ in a multiple of the closed unit ball of $X^{\ast \ast }$, so $K$ is compact by Banach-Alaoglu theorem. The preceding subspace topology identification and Mazur theorem giveIf $D$ is a weakly compact set, its image in the Hausdorff weak-star topology is compact and therefore closed. It contains $C$, so $K\subseteq D\subseteq X$. Conversely, if $K\subseteq X$, the displayed identity gives $K=D$, and its compactness is precisely weak compactness in $X$. Hence $\boxed{D\text{ is weakly compact}⟺K\subseteq X}$. This argument also covers $C=\varnothing$.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint is $T^{\ast }:Y^{\ast }\to X^{\ast }$, defined by $(T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$. Evaluation at any fixed $x$ is thus evaluation at $Tx$ after applying $T^{\ast }$. Each is continuous in the relevant weak-star topology, proving that $T^{\ast }$ is weak-star continuous. Applying the same result to $T^{\ast }$ shows that $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star continuous. Direct evaluation gives

Use $B_X$ for the closed unit ball; using the open ball gives the same norm closure of $T(B_X)$. Goldstine theorem says that $J_X{B}_X$ is weak-star dense in $B_{X^{\ast \ast }}$. Put $K=T^{\ast \ast }(B_{X^{\ast \ast }})$. It is compact and closed in the weak-star topology by Banach-Alaoglu theorem and the established continuity. It contains $J_{Y}T(B_X)$. Conversely, Goldstine theorem gives, for every $x^{\ast \ast }\in B_{X^{\ast \ast }}$, a net $(J_X{x}_{\alpha })$ from $J_X{B}_X$ converging weak-star to $x^{\ast \ast }$; its image converges weak-star to $T^{\ast \ast }{x}^{\ast \ast }$. ConsequentlyApply the preceding bounded convex set criterion in $Y$, and then scale the closed unit ball. We obtain the bidual characterization of weakly compact operators:The canonical embedding into the bidual on the right specifies exactly which copy of $Y$ is intended.

The dual space here is the algebraic dual space $X^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(X,\mathbb{R})$, consisting of every linear functional; no topology or boundedness is imposed. For a basis $e_1,\dots ,e_n$, define $e_i^{\ast }({\sum }_j{x}_j{e}_j)=x_i$. These are linear functionals satisfying $e_i^{\ast }(e_j)={\delta }_{ij}$. Evaluating ${\sum }_i{a}_i{e}_i^{\ast }=0$ on each $e_j$ gives $a_j=0$, proving linear independence. Conversely, for every $T\in X^{\ast }$,This proves spanning and hence proves, without assuming a dimension theorem, that the dual basis is a basis and $\dim X^{\ast }=n$.

The canonical bidual evaluation map isIts definition makes no choice of basis and it is linear. If $x\ne 0$, one coordinate functional is nonzero on $x$, so $J$ is injective. For any $B\in X^{\ast \ast }$, choose $x={\sum }_{i}B(e_i^{\ast })e_i$. Expanding $T$ in the dual basis gives $T(x)={\sum }_{i}T(e_i)B(e_i^{\ast })=B(T)$, proving surjection explicitly. Thus $J$ is an isomorphism.

If $T_1,\dots ,T_n$ is any basis of $X^{\ast }$, its dual basis $B_1,\dots ,B_n$ in $X^{\ast \ast }$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a basis of $X$, and $T_i(v_j)=B_j(T_i)={\delta }_{ij}$. Every basis of the finite-dimensional dual is a dual basis, of a uniquely determined basis of $X$.

For a separating subspace of an algebraic dual $W$, choose a basis $T_1,\dots ,T_r$ of $W$. The linear map $x\mapsto (T_1(x),\dots ,T_r(x))$ is injective because $W$ separates nonzero vectors. The rank-nullity theorem implies $n\le r$; but $W\subseteq X^{\ast }$ implies $r\le n$. HenceThe zero-dimensional case also has this conclusion.

For $X=\mathbb{R}[x]$, the monomials $1,x,x^2,\dots$ form an infinite basis, each polynomial using only finitely many of them. The map $T\mapsto (T(1),T(x),T(x^2),\dots )$ identifies $X^{\ast }$ with all real sequences, because any sequence $(a_j)$ defines $T({\sum }_{j=0}^d{c}_j{x}^j)={\sum }_{j=0}^d{c}_j{a}_j$. The linear subspace of finitely supported sequences is proper (it excludes $(1,1,\dots )$) and separating (one coefficient functional detects any nonzero polynomial). Thus

Since $U_m\cap U_n=U_{\gcd (m,n)}$ when the greatest common divisor is at least $2$, and is empty otherwise, finite intersections of the $U_n$ are again among the allowed unions. Arbitrary unions are built into the definition; the empty union is empty, and $X={\bigcup }_{n\in X}{U}_n$. Thus these sets form a topology.

The space is not Hausdorff: every neighborhood of $4$ contains its divisor $2$, and every neighborhood of $2$ contains $2$, so $2$ and $4$ cannot have disjoint neighborhoods.

On $\mathcal{U}=\mathbb{R}$ with its usual topology, the following examples isolate the three failures.

A nonproper functional is $E(x)=+\infty$ for every $x$. Its effective domain is empty, so it has no finite-valued global minimizer. It nevertheless has coercivity and is sequentially lower semicontinuous. If a minimizer is instead defined only by $E(x)=\inf E$ without requiring finiteness, every point formally minimizes this function; under that convention the requested nonproper counterexample is impossible with the given codomain.

A noncoercive functional is $E(x)=e^x$. It is a proper extended-real function and is continuous, but its infimum zero is approached as $x\to -\infty$ and is never attained. Hence it has no global minimizer.

A failure of sequential lower semicontinuity isIt is a proper extended-real function with coercivity, but $E(1/n)\to 0<E(0)$. Its infimum is zero, while every function value is positive. Thus it has no global minimizer.

The epigraph isAssume $E$ is $\tau$-sequentially lower semicontinuous. If $(u_n,t_n)$ lies in the epigraph and converges to $(u,t)$ in the product topology, thenThe limit remains in the epigraph, so it is a sequentially closed set.

Conversely, suppose the epigraph is a sequentially closed set and sequential lower semicontinuity fails along $u_n\stackrel{\tau }{\to }u$. Choose a finite real number $a$ with $\mathrm{liminf}E(u_n)<a<E(u)$. There is a subsequence along which $E(u_{n_j})\le a$. Thus $(u_{n_j},a)$ lies in the epigraph and converges to $(u,a)$, which does not lie there. This contradiction provesThis is a sequential statement; identifying it with ordinary closedness requires an appropriate assumption on the topology.

Write $D(f)=\mathbb{C}\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb{C}$. Also,so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\dots ,a_m\}$ is the zero set of ${\prod }_j(z-a_j)$. Consequently $\tau$ is exactly the cofinite topology: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a topology.

The product topology on $X\times Y$ is the topology with basisThe proposed complement need not be open. TakeThen $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb{C}$, so $U\cap V\ne \varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence

A sublevel set consists of points whose objective value is at most a chosen threshold. Coercivity makes finite sublevel sets bounded, while sequential lower semicontinuity makes them sequentially closed sets in the selected topology.

A topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

The trivial absolute value on a field is zero at zero and one elsewhere. Its metric topology is discrete. Its valuation ring and residue field are both the field itself. This is why rational absolute-value classification and nontrivial discrete-value compactness criteria need a nontriviality hypothesis.