FALSE: a sequence in $\mathbb{R}^{n}$ whose coordinate sequences are each bounded converges

False claim: let $n \ge 1$ and let $\bigl(x^{(k)}\bigr)$ be a sequence in $\mathbb{R}^{n}$ such that every coordinate sequence $k \mapsto x^{(k)}_j$ $(j<n)$ is bounded (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then $\bigl(x^{(k)}\bigr)$ converges 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 $\mathbb{R}$, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it).

The claim conflates two theorems. What is true about boundedness is For $n \ge 1$ every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence: a bounded sequence has a convergent subsequence. What is true componentwise is For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 1, which is about convergence of the coordinate sequences and says nothing about boundedness. The false claim takes the hypothesis of the first and the conclusion of the second.

The witness is the smallest possible. Take $n = 1$ and let $x^{(k)} \in \mathbb{R}^{1}$ be the function $1 \to \mathbb{R}$ with value $\varepsilon_k$ at $0$, where $(\varepsilon_k)$ is the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$).

• The same witness separates the two notions on the real line. Boundedness of a real sequence gives a convergent subsequence and nothing more, and the alternating sequence has limit inferior $-1$ and limit superior $1$, so by A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$, and diverges to $\pm\infty$ iff both equal $\pm\infty$ it cannot converge. That is a second route to step 2.1; the one taken above uses only uniqueness of limits.

• Componentwise boundedness and boundedness agree, so nothing is gained by weakening the hypothesis. For $n \ge 1$ the comparison chain $\lVert y\rVert_\infty \le \lVert y\rVert_2 \le \iota(n)\lVert y\rVert_\infty$ (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) shows that a sequence in $\mathbb{R}^{n}$ has bounded range if and only if every coordinate sequence is bounded. So the refuted claim is exactly the claim that a bounded sequence converges, restated coordinatewise.

• The converse direction is fine. A convergent sequence in $\mathbb{R}^{n}$ does have bounded coordinate sequences, each coordinate sequence being convergent by For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 1 and a convergent real sequence being bounded (Every convergent sequence is bounded). Only the direction asserted above fails.