Compactness and Continuity in R^n .... .... D&K Theorem 1.8.8 .... ....

[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 1.8.8 ... ...

Duistermaat and Kolk's Theorem 1.8.8 and its proof read as follows:View attachment 7740In the above proof we read the following:

" ... ... The definitions of supremum and of the compactness of $$f(K)$$ then give that $$\text{ sup } f(K) \in f(K)$$. ... ... " My question is as follows:

How, exactly, do the definitions of supremum and of the compactness of $$f(K)$$ imply that $$\text{ sup } f(K) \in f(K)$$. ... ... ?Hope someone can help ... ...

Peter=========================================================================================Members of MHB reading the above post may be helped by access to (i) D&K's definition of supremum, and (ii) D&K's definition of compactness plus their early results on compactness ... so I am providing the same ... as follows:View attachment 7741View attachment 7742Hope that helps ...

Peter

 

[castor28]

Hi Peter,

According to the definition used in Theorem 1.6.1, if $a = \sup A$, then for each $n>0$ we can find a number $x_n$ in the interval $(a-1/n,a]$. This gives a sequence that converges to $a$.

If $A$ is compact, the limit ($a$) must be an element of $A$ by definition 1.8.1.

 

[Math Amateur]

Hi Peter,

According to the definition used in Theorem 1.6.1, if $a = \sup A$, then for each $n>0$ we can find a number $x_n$ in the interval $(a-1/n,a]$. This gives a sequence that converges to $a$.

If $A$ is compact, the limit ($a$) must be an element of $A$ by definition 1.8.1.

Hi castor28 ...

Thanks for the help ... appreciate it ...

Peter