# Thread: Analysis proof - Convergence and Limit in R^n

1. ## Analysis proof - Convergence and Limit in R^n

Problem:
Suppose that ${ u_{k} }$ is a sequence of points in $\mathbb{R}^n$ that converges to the point u and that $\| u \| = r > 0$. Prove that there is an index K such that $\| u_{k} \| > \frac{r}{2}$ if $k \geq K$
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Attempt:
By given, $\lim_{k \to infty} u_{k} = u$ and $\| u \| = r >0$.

I believe I have to use following criterion, but do not know how to apply it to prove the problem.
Definition: A sequence of points ${ u_{k} }$ in $\mathbb{R}^n$ is said to converge componentwise to the point u for each index i with $1 \leq i \leq n$, then
$\lim_{k \to infty} p_{i}(u_{k}) = p_{i}(u)$.

Componentwise Convergence Criterion: Let ${ u_{k} }$ be a sequence in $\mathbb{R}^n$ and let $u \in \mathbb{R}^n$. Then ${ u_{k} }$ converges to u if and only if ${ u_{k} }$ converges componentwise to u.

Suppose that ${ u_{k} }$ is a sequence of points in $\mathbb{R}^n$ that converges to the point u and that $\| u \| = r > 0$. Prove that there is an index K such that $\| u_{k} \| > \frac{r}{2}$ if $k \geq K$
let $\epsilon = \frac{r}{2}.$ so there exists $K \in \mathbb{N}$ such that: $\forall k \geq K: \ ||u_k - u|| < \epsilon=\frac{r}{2}.$ thus: $||u|| - ||u_k|| \leq ||u_k-u|| < \frac{r}{2},$ which gives us: $||u_k|| > ||u|| - \frac{r}{2}=r-\frac{r}{2}=\frac{r}{2}. \ \ \ \square$