Let ${\{x^k\}}^{\infty}_{k=1}$ $\subset R^n$ be a convergent sequence, prove that there exists $B > 0$ such that $∥x^k∥ ≤ B$ for all $k$.

Question

Let ${\{x^k\}}^{\infty}_{k=1}$ $\subset R^n$ be a convergent sequence, prove that there exists $B > 0$ such that $∥x^k∥ ≤ B$ for all $k$.

I am struggling to prove this and would really appreciate help with proving it, thank you!

Answer 1

Start with the definition of convergence: since $\{x^k\}_{k=1}^\infty$ converges there is some $x$ that it converges to, and $\|x^k-x\|\rightarrow 0$ as $k\rightarrow \infty$. So there exists some $N\in {\mathbb N}$ such that $\| x^k - x\| < \epsilon$ whenever $k > N$ (clearly $\epsilon >0$ but otherwise we can choose $\epsilon$ to be anything we like).

Rearrange a little, and we get $\|x\|-\epsilon < \|x^k\| < \|x\| + \epsilon$

At this point, I think you can probably fill in the missing details for yourself :)