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:
- $ \lim\limits_{X\to P} F(X) = L $
- 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?
