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Math Help - Analysis proof - Convergence and Limit in R^n

  1. #1
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    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
    ===========================
    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.

    Thank you for your help.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    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
    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
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