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Functions Continuous on Comapct Sets ... Apostol, Theorem 4.25 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...

I am focused on Chapter 4: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 4.25 ... ...


Theorem 4.25 (including its proof) reads as follows:



Apostol - 1- Theorem 4.25 ... PART 1 ...  .png
Apostol - 2 - Theorem 4.25 ... PART 2 ...  .png




In the above proof by Apostol we read the following:

" ... ... The sets \(\displaystyle f^{ -1 } (A)\) form an open covering of \(\displaystyle X\) ... ... "


Could someone please demonstrate an explicit formal and rigorous proof of this statement ....?

-----------------------------------------------------------------------------------------------------

My thoughts:

Since \(\displaystyle f\) is continuous we have that each set \(\displaystyle f^{ -1 } (A)\) is open

and ...

... for \(\displaystyle X \subseteq S\) we have

\(\displaystyle X \subseteq f^{ -1 } ( f(x) )\) ... ... (see Apostol Exercise 2.7 (a) Chapter 2, page 44 ...)

... and we also have \(\displaystyle f(X) \subseteq A_c\) where \(\displaystyle A_c = \bigcup_{ A \in F } A\) ...

Therefore \(\displaystyle X \subseteq f^{ -1 } ( f(x) ) \subseteq f^{ -1 } ( A_c )\) ....


Is that correct? ... Does that constitute a formal and rigorous proof?




Hope someone can help ...

Peter




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The above post refers to Apostol Exercise 2.7 so I am providing access to the same as follows:



Apostol -  Exercises 2.6 and 2.7 ... .png



Hope that helps ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,678
In the above proof by Apostol we read the following:

" ... ... The sets \(\displaystyle f^{ -1 } (A)\) form an open covering of \(\displaystyle X\) ... ... "


Could someone please demonstrate an explicit formal and rigorous proof of this statement ....?
I would put it like this:

If $x\in X$ then $f(x) \in f(X) \subseteq \bigcup_{A\in F}A$. Therefore $f(x)$ is in (at least) one of the sets $A \in F$, say $f(x)\in A_x$. Then $x \in f^{-1}(A_x)$. Since that holds for every $x$ in $X$, it follows that the sets \(\displaystyle f^{ -1 } (A)\) form a covering of \(\displaystyle X\).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I would put it like this:

If $x\in X$ then $f(x) \in f(X) \subseteq \bigcup_{A\in F}A$. Therefore $f(x)$ is in (at least) one of the sets $A \in F$, say $f(x)\in A_x$. Then $x \in f^{-1}(A_x)$. Since that holds for every $x$ in $X$, it follows that the sets \(\displaystyle f^{ -1 } (A)\) form a covering of \(\displaystyle X\).


Thanks for a very convincing proof Opalg ...

Appreciate your help ...

Peter