A Further Question on Proper and Continuous Mappings .... D&K Theorem 1.8.6 ....

[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 another 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 7732In the above proof we read the following:

" ... ... Thus for $$k$$ sufficiently large, we have $$f( x_k ) \in K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$, while $$K $$ is compact in $$\mathbb{R}^p$$. ... ... "I am confused by the above statement ... can someone please explain/clarify ... ...

Apologies in advance if I am missing something simple ... ...

Note that in particular I do not quite understand the statement $$f( x_k ) \in K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$ ... ... Hope someone can help ...

Peter***EDIT***

Oh ... !

$$f( x_k ) \in K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$

... probably means f( x_k ) \in K WHERE $$K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$

Is that right?

But then ... why is $$K$$ compact?

and

... why does $$x_k \in f^{-1} (K) \cap F$$ ... ... and further, why is $$f^{-1} (K) \cap F$$ compact?Hope someone can help with these further questions ...

Peter

 

[GJA]

Hi, Peter.

All good questions.

Oh ... !

$$f( x_k ) \in K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$

... probably means f( x_k ) \in K WHERE $$K = \{ y \in \mathbb{R}^p \mid \ \mid \mid y - b \mid \mid \le 1 \}$$

Is that right?

This is correct.

But then ... why is $$K$$ compact?

$K$ is a ball of unit radius centered at $b$. There are a number of ways to prove it is compact with one being to show that it is a closed, bounded subset of $\mathbb{R}^{p}$.

why does $$x_k \in f^{-1} (K) \cap F$$

$x_{k}\in F$ by assumption. Since $f(x_{k})\in K$, $x_{k}\in f^{-1}(K)$ by definition of the preimage of a set.

why is $$f^{-1} (K) \cap F$$ compact?

Since $K$ is compact and $f$ is proper, $f^{-1}(K)$ is a compact subset of $\mathbb{R}^{n}$. Compact subsets of $\mathbb{R}^{n}$ are necessarily closed. $F$ is closed (assumption), so the intersection $f^{-1}(K)\cap F$ is closed (intersection of closed sets is closed). Hence, $f^{-1}(K)\cap F$ is a closed subset of the compact set $f^{-1}(K)$, which makes it compact, too.

 

[Math Amateur]

Hi, Peter.

All good questions.
This is correct.
$K$ is a ball of unit radius centered at $b$. There are a number of ways to prove it is compact with one being to show that it is a closed, bounded subset of $\mathbb{R}^{p}$.
$x_{k}\in F$ by assumption. Since $f(x_{k})\in K$, $x_{k}\in f^{-1}(K)$ by definition of the preimage of a set.
Since $K$ is compact and $f$ is proper, $f^{-1}(K)$ is a compact subset of $\mathbb{R}^{n}$. Compact subsets of $\mathbb{R}^{n}$ are necessarily closed. $F$ is closed (assumption), so the intersection $f^{-1}(K)\cap F$ is closed (intersection of closed sets is closed). Hence, $f^{-1}(K)\cap F$ is a closed subset of the compact set $f^{-1}(K)$, which makes it compact, too.

Your replies to my questions were indeed helpful ... I now follow the theorem ...

Thanks!

Peter