MHB Proper and Continuous Mappings in R^n .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Theorem 1.8.6 ... ...

Duistermaat and Kolk"s Theorem 1.8.6 and the preceding definition regarding proper mappings read as follows:View attachment 7731In the above proof we read the following:

" ... ... Indeed let $$( x_k )_{ k \in \mathbb{N} }$$ be a sequence of points in $$F$$ with the property that $$( f( x_k ) )_{ k \in \mathbb{N} }$$ is convergent with limit $$b \in \mathbb{R}^p$$. ... ... "My question is as follows:

How do we be sure that such a sequence exists in $$F$$?Help will be appreciated ... ...

Peter

 

[GJA]

Hi, Peter.

How do we be sure that such a sequence exists in $$F$$?

You are trying to prove that $f(F)$ is closed, which means you must show that $f(F)$ contains all of its limit points. This means you start with a sequence of points in $f(F)$ and assume it is convergent. The goal is then to show that the limit is in $f(F)$. Note the language of Lemma 1.2.12(iii) says "..sequence...that is convergent to a limit..."

Furthermore, the only subset of $\mathbb{R}^{n}$ that contains no convergent sequences is the empty set. If $S$ is a non-empty set with $p\in S$, then $x_{n}=p$ for all $n$ is a convergent sequence in $S$. So the only set that fails to admit the property of which you ask is the empty set.

 

[Math Amateur]

Hi, Peter.
You are trying to prove that $f(F)$ is closed, which means you must show that $f(F)$ contains all of its limit points. This means you start with a sequence of points in $f(F)$ and assume it is convergent. The goal is then to show that the limit is in $f(F)$. Note the language of Lemma 1.2.12(iii) says "..sequence...that is convergent to a limit..."

Furthermore, the only subset of $\mathbb{R}^{n}$ that contains no convergent sequences is the empty set. If $S$ is a non-empty set with $p\in S$, then $x_{n}=p$ for all $n$ is a convergent sequence in $S$. So the only set that fails to admit the property of which you ask is the empty set.

Thanks for the clear explanation ...

Appreciate your guidance and help ...

Thanks again ...

Peter