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

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

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

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 $\displaystyle { u_{k} }$ in $\displaystyle \mathbb{R}^n$ is said to converge componentwise to the point u for each index i with $\displaystyle 1 \leq i \leq n$, then
$\displaystyle \lim_{k \to infty} p_{i}(u_{k}) = p_{i}(u)$.

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

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