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For a Hilbert space $H$, the linear span of the matrix-coefficient functionals $T\mapsto ⟨Tx,y⟩$ is a predual of $\mathcal{B}(H)$. The unit ball is compact in the resulting weak operator topology by the Tychonoff theorem, and the compact norming dual-pair criterion identifies $\mathcal{B}(H)$ with the dual of that span.

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

Give each nonempty finite set $X_j$ the discrete topology. The product space $X={\prod }_j{X}_j$ is compact by the Tychonoff theorem. For each $i\le j$, the compatibility condition $f_{ij}(x_j)=x_i$ defines a closed subset $C_{ij}\subseteq X$.

These sets have the finite intersection property. Indeed, for finitely many conditions choose an index $k$ above every index occurring in them, choose any $x_k\in X_k$, and use the transition maps from $k$ to define all required coordinates; choose the remaining coordinates arbitrarily. Compactness therefore givesThis is the nonemptiness theorem for inverse limits of finite sets.

Goldstine theorem states that the canonical image $J(B_X)$ of the closed unit ball of a normed vector space $X$ is weak-star dense in $B_{X^{\ast \ast }}$.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact in the weak-star topology. To prove it, map each $f\in B_{X^{\ast }}$ to its values inEvery factor is compact, so Tychonoff theorem makes $P$ compact. The image of $B_{X^{\ast }}$ is cut out by the closed linearity conditionsand is therefore closed in $P$. The product topology restricted to this image is exactly pointwise convergence on $X$, namely the weak-star topology. Hence the ball is compact.

The estimateshows $\Vert {\mu }_{x,y}\Vert \le \Vert x\Vert \Vert y\Vert$. For nonzero $x,y$, the rank-one operatorhas norm one and attains equality. The zero cases are immediate.

Embed the operator unit ball intoby $T\mapsto (⟨Tx,y⟩{)}_{x,y}$. The product is compact by Tychonoff theorem. A pointwise limit of these coordinates is a bilinear form $\theta$ satisfyingBy the stated representation theorem for bounded bilinear forms, $\theta (x,y)=⟨Tx,y⟩$ for a unique operator with $\Vert T\Vert \le 1$. The image is therefore closed and compact. Its product topology is precisely the weak operator topology $\sigma (\mathcal{B}(H),M)$.

The linear span $Z=\mathrm{span}M$ separates operators, and the preceding compactness lets part (a) identify $\mathcal{B}(H)$ isometrically with $Z^{\ast }$. Thus $\mathcal{B}(H)$ is a dual Banach space. This is the operator predual from matrix coefficients.