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

#### Peter

##### Well-known member
MHB Site Helper
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:  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 ....?

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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: Hope that helps ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
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
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 ...