Proof of The Real Matrices with the Operator-Norm Distance Form a Complete Metric Space with Entrywise Convergence

Each result cited is universally quantified over the data in its own statement.

Let $r,s\in\mathbb{N}$; every clause is proved for these arbitrary sizes, using no others. Write $\mathcal{M}=\mathcal{M}_{r\times s}(\mathbb{R})$ and $d=d_{r\times s}$, so that $d(A,B)=\lVert A-B\rVert$ by The Distance between Real Matrices §distance. By Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §real-line, $d_{\mathbb{R}}$ is a metric on $\mathbb{R}$ with $d_{\mathbb{R}}(a,b)=|a-b|$ for $a,b\in\mathbb{R}$.

Entrywise identities. For $A,B,C\in\mathcal{M}$, comparing entries by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations, Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §zero and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality gives $B-A=(-1)(A-B)$, $A-C=(A-B)+(B-C)$ and $A=(A-B)+B$, and shows that $A-B=0$ if and only if $A=B$; the entries of $A-B$ are $A_{ij}-B_{ij}$.

Clause metric. $d$ is the map $\mathcal{M}\times\mathcal{M}\to\mathbb{R}$ of The Operator-Norm Distance between Real Matrices Is a Map §map, by The Distance between Real Matrices §distance. Let $A,B,C\in\mathcal{M}$. Nonnegative: $d(A,B)=\lVert A-B\rVert\ge0$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §nonnegative. Zero: by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §definite, $d(A,B)=0$ if and only if $A-B=0$, that is, if and only if $A=B$. Symmetric: $d(B,A)=\lVert(-1)(A-B)\rVert=|-1|\,\lVert A-B\rVert=d(A,B)$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §homogeneity. Triangle: $d(A,C)=\lVert(A-B)+(B-C)\rVert\le d(A,B)+d(B,C)$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §triangle. Hence $d$ is a metric on $\mathcal{M}$ by Metric Space §metric, and $(\mathcal{M},d)$ is a metric space by Metric Space §metric-space.

Clause entry-bound. Let $A,B\in\mathcal{M}$, $i\in[r]$ and $j\in[s]$. Since the $(i,j)$ entry of $A-B$ is $A_{ij}-B_{ij}$, The Operator Norm of Entries, Transposes, Multiples of the Identity, Orthogonal Factors and Diagonal Matrices §entry-bound gives $|A_{ij}-B_{ij}|\le\lVert A-B\rVert=d(A,B)$.

Clause norm-lipschitz. Let $A,B\in\mathcal{M}$. From $A=(A-B)+B$ and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §triangle, $\lVert A\rVert\le d(A,B)+\lVert B\rVert$; in the same way $\lVert B\rVert\le d(B,A)+\lVert A\rVert=d(A,B)+\lVert A\rVert$, by the clause metric (Metric Space §symmetric); so $\bigl|\lVert A\rVert-\lVert B\rVert\bigr|\le d(A,B)$.

Clause vector-bound. Let $A,B\in\mathcal{M}$ and $x\in\mathbb{R}^{s}$. From $A=(A-B)+B$ and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $Ax=(A-B)x+Bx$, so $Ax-Bx=(A-B)x$, comparing components by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §equality; hence $\lVert Ax-Bx\rVert\le\lVert A-B\rVert\,\lVert x\rVert=d(A,B)\,\lVert x\rVert$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §bound.

Entry sequences. A sequence $(A^{(k)})$ in $\mathcal{M}$ is a sequence in $\mathbb{R}^{[r]\times[s]}$, as $\mathcal{M}=\mathbb{R}^{[r]\times[s]}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §matrices; for $i\in[r]$ and $j\in[s]$, its entry sequence $(A^{(k)}_{ij})$, with $A^{(k)}_{ij}=A^{(k)}(i,j)$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §entries, is the sequence $k\mapsto A^{(k)}(i,j)$, a sequence in $\mathbb{R}$ (The Sequences k ↦ f_k(p) of a Sequence in ℝ^I, and the Map of Their Limits §sections). The set $[r]\times[s]$ is finite: $[r]=\{1,\dots,r\}$ and $[s]=\{1,\dots,s\}$ are finite intervals (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §naturals), and a product of finite sets is finite (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union).

Clause convergence. Let $(A^{(k)})$ be a sequence in $\mathcal{M}$ and $A\in\mathcal{M}$.

Suppose $(A^{(k)})$ converges to $A$ in $(\mathcal{M},d)$, and let $i\in[r]$, $j\in[s]$ and $\varepsilon>0$. The choices are made in this order: $i$, $j$ and $\varepsilon$, then $N$ depending only on $\varepsilon$, then $k\ge N$. By Convergent Sequence in a Metric Space §converges there is $N\in\mathbb{N}$ with $d(A^{(k)},A)<\varepsilon$ for every $k\ge N$; for these $k$, $d_{\mathbb{R}}(A^{(k)}_{ij},A_{ij})=|A^{(k)}_{ij}-A_{ij}|\le d(A^{(k)},A)<\varepsilon$ by the clause entry-bound. So $(A^{(k)}_{ij})$ converges to $A_{ij}$ in $(\mathbb{R},d_{\mathbb{R}})$, by the same definition.

Conversely, suppose $(A^{(k)}_{ij})$ converges to $A_{ij}$ in $(\mathbb{R},d_{\mathbb{R}})$ for all $i\in[r]$ and $j\in[s]$, and let $\varepsilon>0$; put $\eta=\varepsilon/(rs+1)>0$, where $rs$ and $rs+1$ are read in $\mathbb{R}$. The choices are made in this order: $\varepsilon$, then $\eta$, then one $N$ depending only on $\varepsilon$, then $k\ge N$.

The condition. Let $F:\mathbb{N}\to\mathcal{M}$, $k\mapsto A^{(k)}$, be the given sequence (Sequences §sequence). As $\mathcal{M}=\mathbb{R}^{[r]\times[s]}$ (entry sequences), the evaluation map of Projection, Evaluation and Currying Maps §evaluation, with $\mathbb{R}$ for its $Y$ and $[r]\times[s]$ for its $I$, is a map $e:\mathcal{M}\times([r]\times[s])\to\mathbb{R}$ with $e((B,q))=B(q)$ for $B\in\mathcal{M}$ and $q\in[r]\times[s]$. As $a-b\in\mathbb{R}$ for $a,b\in\mathbb{R}$, Maps and Relations Given by Formulas §binary gives a map $\delta:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ with $\delta((a,b))=a-b$, the difference map; let $\alpha:\mathbb{R}\to\mathbb{R}$ be the absolute-value map $a\mapsto|a|$ of Absolute Value and Squaring as Maps on the Reals, and Entrywise Absolute Values and Squares of a Real Matrix §absolute-value. The maps $F$, $A$, $e$, $\delta$ and $\alpha$, the number $\eta$, the sets $[r]\times[s]$ and $\mathbb{N}$, and the strict relation $<$ of $\mathbb{R}$, for which $a<b$ means $(a,b)\in{<}$, enter the formulas below as parameters: set variables, or for $<$ a class variable, to which they are assigned. Let $p$ and $k$ be distinct set variables, and let $\gamma$ be the guard $p\in[r]\times[s]\wedge k\in\mathbb{N}$ for them. Under $\gamma$: $p$ and $k$ are terms of type $[r]\times[s]$ and $\mathbb{N}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them §variable); $A^{(k)}=F(k)$ is of type $\mathcal{M}$, and $A(p)$ is of type $\mathbb{R}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them §value); $A^{(k)}(p)=e((F(k),p))$ is of type $\mathbb{R}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them §pair, then the clause value); the difference $A^{(k)}(p)-A(p)=\delta((A^{(k)}(p),A(p)))$ and its absolute value $|A^{(k)}(p)-A(p)|=\alpha(\delta((A^{(k)}(p),A(p))))$ are of type $\mathbb{R}$, by the clauses pair and value; and, $\eta$ being of type $\mathbb{R}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them §parameter), the pair $(|A^{(k)}(p)-A(p)|,\eta)$ is of type $\mathbb{R}\times\mathbb{R}$. Let $Q(p,k)$ be the condition "$p\in[r]\times[s]$, $k\in\mathbb{N}$ and $|A^{(k)}(p)-A(p)|<\eta$", that is, $\gamma\wedge(|A^{(k)}(p)-A(p)|,\eta)\in{<}$. By Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them §membership, with $<$ as its class $C$, $Q$ is expressed by a predicative formula.

One threshold. Let $p\in[r]\times[s]$, so $p=(i,j)$ with $i\in[r]$ and $j\in[s]$, and $A^{(k)}(p)=A^{(k)}_{ij}$ and $A(p)=A_{ij}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §entries. By the same definition of convergence, with $\eta$ in place of $\varepsilon$, there is $N_{p}$ with $N_{p}\in\mathbb{N}$ and $d_{\mathbb{R}}(A^{(k)}_{ij},A_{ij})=|A^{(k)}(p)-A(p)|<\eta$, that is, $Q(p,k)$, for every $k\in\mathbb{N}$ with $N_{p}\le k$. As $[r]\times[s]$ is finite, One Threshold for Finitely Many Monotone Conditions and for Tails §tails, with $[r]\times[s]$ for its $I$ and $k$ for its $n$, gives one common $N\in\mathbb{N}$ such that $Q(p,k)$ holds for every $p\in[r]\times[s]$ and every $k\in\mathbb{N}$ with $N\le k$; that is, $|A^{(k)}_{ij}-A_{ij}|<\eta$ for all $k\ge N$, $i\in[r]$ and $j\in[s]$. Let $k\ge N$. By The Operator Norm of Entries, Transposes, Multiples of the Identity, Orthogonal Factors and Diagonal Matrices §entry-sum applied to $A^{(k)}-A$, whose entries are $A^{(k)}_{ij}-A_{ij}$, and by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §constant in the ordered field $\mathbb{R}$, applied first to the sum over $j\in[s]$ for each $i\in[r]$ and then to the sum over $i\in[r]$,

$$d(A^{(k)},A)\le\sum_{i=1}^{r}\sum_{j=1}^{s}|A^{(k)}_{ij}-A_{ij}|\le\sum_{i=1}^{r}s\,\eta=rs\,\eta<\varepsilon,$$

the last step as $rs<rs+1$ and $\varepsilon>0$. So $(A^{(k)})$ converges to $A$ in $(\mathcal{M},d)$, by the same definition of convergence.

Claim (the matrix of limits). Let $(A^{(k)})$ be a sequence in $\mathcal{M}$ such that $(A^{(k)}_{ij})$ is a Cauchy sequence in $(\mathbb{R},d_{\mathbb{R}})$ for all $i\in[r]$ and $j\in[s]$. Then there is $L\in\mathcal{M}$ with $L_{ij}=\lim_{k\to\infty}A^{(k)}_{ij}$ for all $i\in[r]$ and $j\in[s]$, and $(A^{(k)})$ converges to $L$ in $(\mathcal{M},d)$.

Proof of the claim. As $(\mathbb{R},d_{\mathbb{R}})$ is complete by Convergence, Limits and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space, which is Complete §complete, each $(A^{(k)}_{ij})$ is convergent in $(\mathbb{R},d_{\mathbb{R}})$ by Complete Metric Space §complete. By The Sequences k ↦ f_k(p) of a Sequence in ℝ^I, and the Map of Their Limits §limit-map, applied with $I=[r]\times[s]$ to $(A^{(k)})$, a sequence in $\mathbb{R}^{[r]\times[s]}$ whose sequences $k\mapsto A^{(k)}(p)$ at the points $p$ are the entry sequences, there is $L\in\mathbb{R}^{[r]\times[s]}=\mathcal{M}$ such that, for all $i\in[r]$ and $j\in[s]$, $(A^{(k)}_{ij})$ converges to $L(i,j)=L_{ij}$ in $(\mathbb{R},d_{\mathbb{R}})$ and $L_{ij}=\lim_{k\to\infty}A^{(k)}_{ij}$. By the clause convergence, $(A^{(k)})$ converges to $L$ in $(\mathcal{M},d)$.

Clause cauchy. Let $(A^{(k)})$ be a sequence in $\mathcal{M}$.

Suppose $(A^{(k)})$ is Cauchy in $(\mathcal{M},d)$, and let $i\in[r]$, $j\in[s]$ and $\varepsilon>0$. The choices are made in this order: $i$, $j$ and $\varepsilon$, then $N$ depending only on $\varepsilon$, then $k,\ell\ge N$. By Cauchy Sequence in a Metric Space §cauchy there is $N\in\mathbb{N}$ with $d(A^{(k)},A^{(\ell)})<\varepsilon$ for all $k,\ell\ge N$, and then $d_{\mathbb{R}}(A^{(k)}_{ij},A^{(\ell)}_{ij})\le d(A^{(k)},A^{(\ell)})<\varepsilon$ by the clause entry-bound. So $(A^{(k)}_{ij})$ is Cauchy in $(\mathbb{R},d_{\mathbb{R}})$, by Cauchy Sequence in a Metric Space §cauchy.

Conversely, suppose $(A^{(k)}_{ij})$ is Cauchy in $(\mathbb{R},d_{\mathbb{R}})$ for all $i\in[r]$ and $j\in[s]$. By the claim, $(A^{(k)})$ converges in $(\mathcal{M},d)$ to some $L\in\mathcal{M}$. Let $\varepsilon>0$; the choices are made in this order: $\varepsilon$, then $N$ depending only on $\varepsilon$, then $k,\ell\ge N$. By the definition of convergence cited in the clause convergence, there is $N\in\mathbb{N}$ with $d(A^{(k)},L)<\varepsilon/2$ for every $k\ge N$, and for $k,\ell\ge N$ the clause metric (Metric Space §triangle and Metric Space §symmetric) gives $d(A^{(k)},A^{(\ell)})\le d(A^{(k)},L)+d(A^{(\ell)},L)<\varepsilon$. So $(A^{(k)})$ is Cauchy in $(\mathcal{M},d)$, by Cauchy Sequence in a Metric Space §cauchy.

Clause complete. Let $(A^{(k)})$ be a Cauchy sequence in $(\mathcal{M},d)$. By the clause cauchy, $(A^{(k)}_{ij})$ is Cauchy in $(\mathbb{R},d_{\mathbb{R}})$ for all $i\in[r]$ and $j\in[s]$, so by the claim $(A^{(k)})$ converges in $(\mathcal{M},d)$ to some $L\in\mathcal{M}$. So $(A^{(k)})$ is convergent in $(\mathcal{M},d)$ by Convergent Sequence in a Metric Space §convergent, and $(\mathcal{M},d)$ is complete by Complete Metric Space §complete.