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I found this theorem in a book:

Theorem: Let $F: A \subset \mathbb{R}^n \to \mathbb{R}^m.$ Let $P \in \bar{A}$ and $L \in \mathbb{R}^m$. Then the following assertions are equivalent:

  1. $ \lim\limits_{X\to P} F(X) = L $
  2. For all ${P_k} \subset A$ such that $P_k \neq P\ \forall k $ and $P_k \to P$, $\lim\limits_{k \to \infty}F(P_k) = L$.

Proof:

$1 \implies 2)$ [...]

$2 \implies 1)$ Suppose that $\lim\limits_{X\to P} F(X) \neq L$. This would mean that there exists $\epsilon > 0$ such that for all $\delta > 0$ there exists $X \in A$ such that $0 < \|X-P\| < \delta$ and $\| F(X) - L \| \ge \epsilon $. Take $\delta = \frac1{k}$ with $k \in \mathbb{N}$ and let $P_k$ be the $k$-th term of the sequence. We see that $P_k \to P$ and $P_k \neq P$, but $\| F(P_k)-L\| \ge \epsilon$, which would mean that $\lim\limits_{k \to \infty} F(P_k) \neq L$, which contradicts the hypothesis.

It's the second part of the proof I'm interested in. I understand that asserting that $\lim\limits_{k \to \infty} F(P_k) \neq L$ means that if I make ball around $P$, no matter how small, there will be some $X$ inside that ball such that $F(X)$ stays far away from $L$. But I don't understand the proof. I understand each sentence separately, but I don't get how to connect them. Could anyone help me understand the whole argument?

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1 Answer 1

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If you understand each sentence separately. The only thing that you need undestand is that $ 2 \Rightarrow 1$ is equivalent to negartion of (1) implies the negation of (2). This was done.

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Oh, I get it now. I'm not sure if the argument I'm thinking of is the same one the author is describing, but whatever. Your answer sure helped, though! –  Javier Badia Jul 25 '12 at 20:02

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