Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $P_k \to P$, $\lim\limits_{k \to \infty}F(P_k) = L$.


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

$2 \implies 1)$ 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?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.