For $n \ge 1$ every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence

Let $n \in \mathbb{N}$ with $n \ge 1$ and let $\bigl(x^{(j)}\bigr)_{j\in\mathbb{N}}$ be a sequence in $\mathbb{R}^{n}$ whose range $\{\, x^{(j)} : j \in \mathbb{N} \,\}$ is a bounded subset of $(\mathbb{R}^{n}, d_2)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it). Then there are a strictly increasing $i : \mathbb{N} \to \mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies $n_k \ge k$) and a point $p \in \mathbb{R}^{n}$ with

$x^{(i_j)} \longrightarrow p \quad \text{in } (\mathbb{R}^{n}, d_2)$

(Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$). By For $n \ge 1$ all norms on $\mathbb{R}^n$ are equivalent the same statement holds with $d_2$ replaced by the metric of any norm on $\mathbb{R}^{n}$, boundedness and convergence both being unchanged by that replacement (Equivalent norms, and the dictionary with equivalent metrics).

This is assembled from published theorems and is not proved again by bisection. The bisection is in Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, published at order 120; what is added here is the passage from compactness to sequential compactness and the reading of the conclusion in $\mathbb{R}^{n}$.

Choice cost: none. Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line is proved by bisection and uses no choice principle, and "compact implies sequentially compact" is a theorem of ZF (In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle). The five-way equivalence For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice is not used, precisely because it is stated under countable choice and dependent choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain) and would overcharge this corollary; the arrow-by-arrow account is What each implication between the compactness properties of a metric space costs: which are theorems of ZF, which use countable choice, and which use dependent choice.

• The case $n = 1$ and the published one-dimensional theorem. $\mathbb{R}^{1}$ is the set of functions $1 \to \mathbb{R}$ and is therefore not literally $\mathbb{R}$. The map $\theta : \mathbb{R} \to \mathbb{R}^{1}$ sending $t$ to the function with value $t$ at $0$ is a bijection, and $d_2(\theta(s),\theta(t)) = \sqrt{(s-t)^{2}} = |s-t|$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$, Basic properties of the absolute value), so $\theta$ is an isometric bijection onto $\mathbb{R}^{1}$ (Isometry, isometric embedding, and the subspace metric on a subset). Under that identification this corollary at $n = 1$ and the published Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence are the same statement, and neither is used to prove the other: the published theorem is proved on the real line, and the corollary above is proved from Heine-Borel in $\mathbb{R}^{n}$.

• Boundedness of the sequence is boundedness of its range, a set, and not a condition on each coordinate separately. The two do agree here: step 3.1 gives one direction, while the reverse follows from $\lVert y\rVert_2\le\iota(n)\lVert y\rVert_\infty$ in The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clause 3 after taking the maximum of the finitely many coordinate bounds. What does not follow from bounded coordinates is convergence, and the companion page carries that false statement.

• What sequential compactness gives and what it does not. It produces a convergent subsequence and says nothing about the original sequence. A bounded sequence need not converge, and a sequence with a convergent subsequence need not be bounded; both remarks are already recorded for the real line in Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence.