Proof of Bolzano-Weierstrass Theorem in Euclidean Space

Throughout, $[n]$ is the initial segment of $\mathbb{N}$ determined by $n$ and $S$ is the successor map of Natural Numbers, so $S(j)=j+1$ by statement 1 of Arithmetic of Addition on the Natural Numbers. Points of $\mathbb{R}^n$ are written with their coordinates, $x^{(m)}=(x^{(m)}_1,\dots,x^{(m)}_n)$. The order $\le$ on the real numbers is that of an ordered field, with the addition and additive inverses of the underlying field; $|\cdot|$ is the absolute value and $\lVert\,\cdot\,\rVert$ the Euclidean norm. Let $(\mathbb{R},d_{\mathbb{R}})$ be the real line.

Step 1: each coordinate sequence lies in a closed interval. Since $K$ is bounded there are a point $z\in\mathbb{R}^n$ and a real number $R>0$ with $d_E(z,y)\le R$ for every $y\in K$. Fix $m\in\mathbb{N}$ and $i\in[n]$. Condition 3 in the definition of a metric gives $d_E(x^{(m)},z)=d_E(z,x^{(m)})\le R$. By part 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ the $i$-th coordinate of $x^{(m)}-z$ is $x^{(m)}_i-z_i$, so statements 4 and 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give

Statement 6 of Properties of the Absolute Value in an Ordered Field therefore gives $-R\le x^{(m)}_i-z_i$ and $x^{(m)}_i-z_i\le R$. Applying statement 3 of Elementary Arithmetic in an Ordered Field to each of these, and rearranging with the commutativity and associativity of field addition together with Additive Cancellation and Elementary Additive Identities in a Field, we obtain

Taking $m=1$ and using transitivity gives $a_i\le b_i$, so the closed interval $[a_i,b_i]$ is defined, and $x^{(m)}_i\in[a_i,b_i]$ for every $m\in\mathbb{N}$ and every $i\in[n]$. By A Closed Interval is Sequentially Compact in the Real Line each $[a_i,b_i]$ is sequentially compact in $(\mathbb{R},d_{\mathbb{R}})$.

Step 2: extracting one coordinate at a time. Let $T$ be the set of $j\in\mathbb{N}$ for which there exist a strictly increasing sequence $(p_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ and a point $\ell\in\mathbb{R}^n$ such that, for every $i\in[j]\cap[n]$, the sequence $(x^{(p_k)}_i)_{k\in\mathbb{N}}$ converges to $\ell_i$ in $(\mathbb{R},d_{\mathbb{R}})$.

Base. 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. Hence $[1]\cap[n]=\{1\}$. The sequence $(x^{(m)}_1)_{m\in\mathbb{N}}$ has all its terms in the sequentially compact set $[a_1,b_1]$, so by the definition of sequential compactness there are a point $c\in[a_1,b_1]$ and a strictly increasing sequence $(p_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ with $(x^{(p_k)}_1)_{k\in\mathbb{N}}$ converging to $c$ in $(\mathbb{R},d_{\mathbb{R}})$. Let $\ell$ be the point of $\mathbb{R}^n$ whose first coordinate is $c$ and whose remaining coordinates are $0$. Then $1\in T$.

Step. Suppose $j\in T$, witnessed by $(p_k)_{k\in\mathbb{N}}$ and $\ell$. 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 element of $[S(j)]\cap[n]$ equals $S(j)$, so $[S(j)]\cap[n]\subseteq[j]\cap[n]$ and the same pair witnesses $S(j)\in T$.

Suppose instead $S(j)\in[n]$. The sequence $(x^{(p_k)}_{S(j)})_{k\in\mathbb{N}}$ has all its terms in the sequentially compact set $[a_{S(j)},b_{S(j)}]$, so there are a point $c\in[a_{S(j)},b_{S(j)}]$ and a strictly increasing sequence $(q_j')_{j'\in\mathbb{N}}$ in $\mathbb{N}$ such that $(x^{(p_{q_{j'}})}_{S(j)})_{j'\in\mathbb{N}}$ converges to $c$ in $(\mathbb{R},d_{\mathbb{R}})$. By statement 2 of A Subsequence of a Subsequence is a Subsequence the composite index sequence $(p_{q_{j'}})_{j'\in\mathbb{N}}$ is strictly increasing, and by statement 3 of the same result $(x^{(p_{q_{j'}})})_{j'\in\mathbb{N}}$ is a subsequence of $(x^{(m)})_{m\in\mathbb{N}}$.

Let $\ell'$ be the point of $\mathbb{R}^n$ whose $i$-th coordinate is $\ell_i$ for $i\ne S(j)$ and $c$ for $i=S(j)$. Let $i\in[S(j)]\cap[n]$. If $i=S(j)$ then $(x^{(p_{q_{j'}})}_i)_{j'\in\mathbb{N}}$ converges to $c=\ell'_i$ by construction. Otherwise $i\le j$, so $i\in[j]\cap[n]$ and $(x^{(p_k)}_i)_{k\in\mathbb{N}}$ converges to $\ell_i$; by statement 3 of A Subsequence of a Subsequence is a Subsequence the sequence $(x^{(p_{q_{j'}})}_i)_{j'\in\mathbb{N}}$ is the subsequence of $(x^{(p_k)}_i)_{k\in\mathbb{N}}$ determined by $(q_{j'})_{j'\in\mathbb{N}}$, so A Subsequence of a Convergent Sequence Has the Same Limit shows it converges to $\ell_i=\ell'_i$. Hence $(p_{q_{j'}})_{j'\in\mathbb{N}}$ and $\ell'$ witness $S(j)\in T$.

By Principle of Induction for the Natural Numbers, $T=\mathbb{N}$.

Step 3: conclusion. Since $n\in T$ and $[n]\cap[n]=[n]$, there are a strictly increasing sequence $(p_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ and a point $\ell\in\mathbb{R}^n$ such that $(x^{(p_k)}_i)_{k\in\mathbb{N}}$ converges to $\ell_i$ in $(\mathbb{R},d_{\mathbb{R}})$ for every $i\in[n]$. By Convergence in Euclidean Space is Coordinatewise Convergence, applied to the sequence $(x^{(p_k)})_{k\in\mathbb{N}}$ in $\mathbb{R}^n$, this subsequence converges to $\ell$ in $(\mathbb{R}^n,d_E)$.