Proof of Convergence in Euclidean Space is Coordinatewise Convergence

Throughout, $S$ is the successor map of Natural Numbers, so $S(j)=j+1$ by statement 1 of Arithmetic of Addition on the Natural Numbers. The order $\le$ on $\mathbb{R}$ is that of an ordered field and $s<t$ means $s\le t$ and $s\ne t$; $|\cdot|$ is the absolute value, so $d_{\mathbb{R}}(s,t)=|s-t|$, and $\lVert\,\cdot\,\rVert$ is the Euclidean norm. Let $\iota$ be the canonical map of $\mathbb{R}$.

By part 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, for every $m\in\mathbb{N}$ the difference $x^{(m)}-x$ is the point of $\mathbb{R}^n$ whose $i$-th coordinate is $x^{(m)}_i-x_i$, and by statement 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$,

Necessity. Suppose $(x^{(m)})_{m\in\mathbb{N}}$ converges to $x$ in $(\mathbb{R}^n,d_E)$, and let $i\in[n]$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By the definition of convergence there is $N\in\mathbb{N}$ with $d_E(x^{(m)},x)<\varepsilon$ for every $m\in\mathbb{N}$ with $N\le m$. Statement 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, applied to the point $x^{(m)}-x$, gives

so mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, gives $d_{\mathbb{R}}(x^{(m)}_i,x_i)<\varepsilon$ for every such $m$. Hence $(x^{(m)}_i)_{m\in\mathbb{N}}$ converges to $x_i$ in $(\mathbb{R},d_{\mathbb{R}})$.

Sufficiency. Suppose that for every $i\in[n]$ the sequence $(x^{(m)}_i)_{m\in\mathbb{N}}$ converges to $x_i$ in $(\mathbb{R},d_{\mathbb{R}})$, and let $\varepsilon$ be a real number with $0<\varepsilon$.

A tolerance for the coordinates. Statement 8 of Elementary Order Arithmetic in an Ordered Field gives $0<\varepsilon\cdot 2^{-1}$ and $\varepsilon\cdot 2^{-1}<\varepsilon$, and statement 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field gives $0<\iota(n)^{-1}$. Put

so $0<\delta$ by statement 5 of Elementary Order Arithmetic in an Ordered Field. Using the commutativity and associativity of multiplication in the field $\mathbb{R}$ together with $\iota(n)\,\iota(n)^{-1}=1$ and $a\cdot 1=a$,

A common threshold, by induction. Let $B$ be the set of $j\in\mathbb{N}$ for which there exists $N\in\mathbb{N}$ such that

Statement 4 of Properties of the Order on the Natural Numbers gives $1\le n$, so $1\in[n]$; and if $k\in[1]$ then $k\le 1$ and $1\le k$, so $k=1$ by antisymmetry, whence $[1]\cap[n]=\{1\}$. Convergence of the first coordinate supplies $N$ with $d_{\mathbb{R}}(x^{(m)}_1,x_1)<\delta$ for every $m$ with $N\le m$, so $1\in B$.

Suppose $j\in B$, with witness $N$. By statement 5 of Properties of the Order on the Natural Numbers, every $i\in[S(j)]$ satisfies $i\le j$ or $i=S(j)$. If $S(j)\notin[n]$, then no $i\in[S(j)]\cap[n]$ can equal $S(j)$, so $[S(j)]\cap[n]\subseteq[j]\cap[n]$ and the same $N$ witnesses $S(j)\in B$. If $S(j)\in[n]$, convergence of the coordinate $S(j)$ supplies $N'\in\mathbb{N}$ with $|x^{(m)}_{S(j)}-x_{S(j)}|<\delta$ for every $m$ with $N'\le m$. Put $N''=N+N'$; statement 6 of Properties of the Order on the Natural Numbers gives $N<N+N'$ and $N'<N'+N$, and addition on $\mathbb{N}$ is commutative by statement 4 of Arithmetic of Addition on the Natural Numbers, so $N\le N''$ and $N'\le N''$ by statement 1 of Properties of the Order on the Natural Numbers. Let $m\in\mathbb{N}$ with $N''\le m$ and let $i\in[S(j)]\cap[n]$. If $i\le j$ then $i\in[j]\cap[n]$ and $N\le m$ by transitivity, so the bound holds; if $i=S(j)$ then $N'\le m$ by transitivity and the bound holds. Hence $S(j)\in B$.

By Principle of Induction for the Natural Numbers, $B=\mathbb{N}$. Taking $j=n$ and noting $[n]\cap[n]=[n]$, there is $N\in\mathbb{N}$ such that

Conclusion. Let $m\in\mathbb{N}$ with $N\le m$. The $i$-th coordinate of $x^{(m)}-x$ is $x^{(m)}_i-x_i$, and $|x^{(m)}_i-x_i|\le\delta$ for every $i\in[n]$; since also $0\le\delta$, Coordinate Bounds Control the Euclidean Norm applied to the point $x^{(m)}-x$ gives

Combining with $\iota(n)\,\delta<\varepsilon$ by mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, gives $d_E(x^{(m)},x)<\varepsilon$. As $\varepsilon$ was arbitrary, $(x^{(m)})_{m\in\mathbb{N}}$ converges to $x$ in $(\mathbb{R}^n,d_E)$.