Composition of Two Continuous Functions ... Browder, Proposition 3.12

  • #1
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Summary:

I need help with an aspect of Andrew Browder's proof that the composition of two continuous functions is continuous ...

Main Question or Discussion Point

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help in understanding the proof of Proposition 3.12 ...


Proposition 3.12 and its proof read as follows:


Browder - Proposition 3.12 ... .png



In the above proof by Browder we read the following:

" ... ... Since ##f(I) \subset J##, ##f^{ -1 } ( g^{ -1 }(V) ) = f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)## ... ... "


My question is as follows:

Can someone please explain exactly why/how ##f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)## ... ...



Help will be much appreciated ...

Peter
 

Answers and Replies

  • #2
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You have an equation of the type ##A \cap B = A##. If you draw a Venn diagram, you will see that this is equivalent to ##A \subseteq B##. Now ##I \subseteq f^{-1}(J)## but ##I## is all we have, so ##I=f^{-1}(J)##.

More interesting is why ##f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)##. Can you tell why this is true?
 
  • #3
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You have an equation of the type ##A \cap B = A##. If you draw a Venn diagram, you will see that this is equivalent to ##A \subseteq B##. Now ##I \subseteq f^{-1}(J)## but ##I## is all we have, so ##I=f^{-1}(J)##.

More interesting is why ##f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)##. Can you tell why this is true?

Thanks fresh_42 .... can now see why ##f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)## ... ...

As for ##f^{-1}(U\cap J) =f^{-1}(U) \cap f^{-1}(J)## ... I know this is true for any sets U and J (See Topology by Munkres, Exercise 2(c) page 20 ...) ... but I think you are asking me to give a sense of why it is true ... but I have not yet got any ideas regarding this ...

Thanks again for your help ...

Peter
 
  • #4
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Such equations are often shown elementwise: choose an arbitrary element $x\in f^{-1}(A\cap B)$ and show that it is in ##f^{-1}(A)## and in ##f^{-1}(B)##, too. For the other direction, choose an element ##x\in f^{-1}(A)\cap f^{-1}(B)## and show that it is an element of ## f^{-1}(A\cap B)##. As long as it is true for an arbitrary element, as long is it true for all elements.
 
  • #5
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Such equations are often shown elementwise: choose an arbitrary element $x\in f^{-1}(A\cap B)$ and show that it is in ##f^{-1}(A)## and in ##f^{-1}(B)##, too. For the other direction, choose an element ##x\in f^{-1}(A)\cap f^{-1}(B)## and show that it is an element of ## f^{-1}(A\cap B)##. As long as it is true for an arbitrary element, as long is it true for all elements.

Oh ... so you were asking for a formal proof ...

That, I think would proceed as follows:

To prove ... ##f^{-1} (A \cap B) =f^{-1} (A) \cap f^{-1} (B)##

Now ... ##x \in f^{-1}(A \cap B)##

##\Longrightarrow f(x) \in A \cap B##

##\Longrightarrow f(x) \in A## and ##f(x) \in B##

##\Longrightarrow x \in f^{-1} (A)## and ##x \in f^{-1} (B)##

##\Longrightarrow x \in f^{-1} (A) \cap f^{-1} (B)##

So ... we have ##f^{-1} (A \cap B) \subset f^{-1} (A) \cap f^{-1} (B)## ... ... ... ... ... (1)


Now ... ##x \in f^{-1} (A) \cap f^{-1} (B)##

##\Longrightarrow x \in f^{-1} (A)## and ##x \in f^{-1} (B)##

##\Longrightarrow f(x) \in A## and ##f(x) \in B##

##\Longrightarrow f(x) \in A \cap B##

##\Longrightarrow x \in f^{-1}(A \cap B)##

So ... we have ##f^{-1} (A) \cap f^{-1} (B) \subset f^{-1} (A \cap B)## ... ... ... ... ... (2)

Now ... (1) and (2) imply ##f^{-1} (A \cap B) =f^{-1} (A) \cap f^{-1} (B)##

Is that correct?


Thanks again for your help ...

Peter
 
  • #6
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Is that correct?
Yep, correct. It does not generally work for ##f##, where we have ##f(A\cap B) \subseteq f(A) \cap f(B)## but not the other direction.
 

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