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Thread: A question about image of inverse functions

  1. #1
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    A question about image of inverse functions

    I was reading a problem and I saw this:

    Let f and g be functions with range of $\displaystyle \mathbb {R} \cup \{ \infty \} $

    Then $\displaystyle (fg)^{-1} ( \{ \infty \} ) = [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{ - \infty \} ) \cap g^{-1}(0, \infty ] ] $$\displaystyle \cup [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} ) ] \cup [ f ^{-1}(0, \infty ] \cap g^{-1} ( \{ - \infty \} ) ]$

    Why is this true? I'm really confuse about that.

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    I was reading a problem and I saw this:

    Let f and g be functions with range of $\displaystyle \mathbb {R} \cup \{ \infty \} $

    Then $\displaystyle (fg)^{-1} ( \{ \infty \} ) = [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{ - \infty \} ) \cap g^{-1}(0, \infty ] ] $$\displaystyle \cup [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} ) ] \cup [ f ^{-1}(0, \infty ] \cap g^{-1} ( \{ - \infty \} ) ]$

    Why is this true? I'm really confuse about that.

    Thank you.
    I have a problem with this. The reason $\displaystyle \infty$ has to be added to $\displaystyle \mathbb{R}$ is that, of course, $\displaystyle \infty$ is not normally a member of $\displaystyle \mathbb{R}$. And that is true because arithmetic operations cannot be extended to $\displaystyle \infty$. So my question is, since, normally, fg(x)= (f(x))(g(x)), what does "fg" mean if either f(x) or g(x) is $\displaystyle \infty$?

    I guess we could assume that $\displaystyle (a)(\infty)= \infty$ for any $\displaystyle a\ne 0$ but what is $\displaystyle (0)(\infty)$? In order for the equation you give to be true, it must be some number other than $\displaystyle \infty$ or simply left undefined. But that should have been said.

    Even then, I have a problem with $\displaystyle f^{-1}(-\infty)$. "$\displaystyle -\infty$" is not in the range of f! In order for that to make sense, you must have meant that f and g have ranges $\displaystyle \mathbb{R}\cup \{\infty\}\cup\{-\infty\}$ but then the equation you give is not correct. The first set on the right, $\displaystyle f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0)) $ consistes of numbers, x, such that $\displaystyle fg(x)= -\infty$, not $\displaystyle \infty$.

    I can only conclude that the "$\displaystyle -\infty$" terms are incorrect and your statement should have been $\displaystyle (fg)^{-1}(\infty)= [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{ \infty \} ) \cap g^{-1}(0, \infty ] ] $$\displaystyle \cup [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} ) ] \cup [ f ^{-1}(0, \infty ] \cap g^{-1} ( \{\infty \} ) ]$.

    That, with the definition of "product" I provided, is a true statement.

    For any function f, $\displaystyle f^{-1}(A)= {x | f(x)\in A}$. That is, it is the set of all values of x that give a value for f(x) contained in A.

    So $\displaystyle (fg)^{-1}(\{\infty\})= {x| f(x)g(x)= \infty$. And, of course, that is true if either $\displaystyle f(x)= \infty$ and g(x) is non-zero or $\displaystyle g(x)= \infty$ and f(x) is non-zero.

    That's exactly what "$\displaystyle [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{\infty \} ) \cap g^{-1}(0, \infty ] ] $$\displaystyle \cup [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} ) ] \cup [ f ^{-1}(0, \infty ] \cap g^{-1} ( \{\infty \} ) ]$" says.

    The first term, $\displaystyle f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))], is all x such that $\displaystyle f(x)= \infty$ and g(x)< 0. Such a product, by the definitions I gave, is $\displaystyle \infty$ The second term, $\displaystyle f^{-1} ( \{\infty \} ) \cap g^{-1}(0, \infty ]$, is all x such that f(x)= \infty$ and g(x)> 0[/tex]. The third term $\displaystyle [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} )$ is the set of all x such that f(x)< 0 and $\displaystyle g(x)= \infty$. The final set, f ^{-1}(0, \infty ] \cap g^{-1} ( \{ - \infty \} ) [/tex] is the set of all x such that f(x)> 0 and $\displaystyle g(x)= \infty$.


    The reason for that union of four sets was to separate $\displaystyle f^{-1}((-\infty,0))$ from $\displaystyle f^{-1}((0, \infty])$ and to separate $\displaystyle g^{-1}((-\infty, 0))$ from [tex]g^{-1}((0, \infty]). That is to say "all x except x such that f(x)= 0 or g(x)= 0".
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