I am reading from the book Multidimensional real analysis vol I by Duistermaat and Kolk and am trying to understand the following theorem from it:

Theorem 1.3.2. In the notation of Definition 1.3.1 we have $\lim_{x\to a}f(x)=b$ if and only if for every neighborhood $V$ of $b$ in $\mathbb R^p$ the inverse image $f^{-1}(V)$ is a neighborhood of $a$ in $A$.

Definition 1.3.1 is as follows:

Definition 1.3.1. Let $A \subset \mathbb{R}^n$ and let $a \in \overline{A}$; let $f \colon A \to \mathbb{R}^p$ and let $b \in \mathbb{R}^p$. Then the mapping $f$ is said to have a limit $b$ at $a$, whith notation $\lim_{x \to a} f(x) = b$, if for every $\epsilon > 0$ there exists $\delta > 0$ satisfying $$ x \in A \quad \text{and} \quad \| x - a \| < \delta \quad\implies\quad \| f(x) - b \| < \epsilon. $$

(Original image here.)

My question is: It is clear that the theorem implies that $\lim_{x\to a}f(x)=b\implies a\in A$, as $f^{-1}(V)$ is a neighborhood of $a$ in $A$ implies $a\in A$, which is not always true. I believe the theorem as stated is incorrect and this is a mistake in the book. How may it be rectified?

The definition of neighborhood in the book is as follows. A set $U$ is a neighborhood of $x$ in $A$ if $U$ contains an open set in $A$ which contains $x$.

  • $\begingroup$ I agree -- the book's definition is flawed, but the fix is simple. Just replace the condition $||x-a|| < \delta$ by $0 < ||x-a|| < \delta$. $\endgroup$ – quasi Aug 16 '18 at 16:35
  • $\begingroup$ Are you referring to the possibility that $a \in \bar{A} \setminus A$? $\endgroup$ – md2perpe Aug 16 '18 at 16:36
  • $\begingroup$ No, I'm considering the case $a\in A$. The limit of $f(x)$ as $x$ approaches $a$ is not dependent on the value of $f(a)$. $\endgroup$ – quasi Aug 16 '18 at 16:38
  • $\begingroup$ I am not sure how what you are saying is changing the problem in the theorem. In your definition as well, if $a\not\in A $ then how can we get a neighborhood of $a $ in $A $. (Also there are several books which use this definition as well. See math.stackexchange.com/questions/1979605/…) $\endgroup$ – Shahab Aug 16 '18 at 17:14

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