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Math Help - 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  \mathbb {R} \cup \{ \infty \}

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

    Then (fg)^{-1} ( \{ \infty \} ) = [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{ - \infty \} ) \cap g^{-1}(0, \infty ] ] \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 \infty has to be added to \mathbb{R} is that, of course, \infty is not normally a member of \mathbb{R}. And that is true because arithmetic operations cannot be extended to \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 \infty?

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

    Even then, I have a problem with f^{-1}(-\infty). " -\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 \mathbb{R}\cup \{\infty\}\cup\{-\infty\} but then the equation you give is not correct. The first set on the right, f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0)) consistes of numbers, x, such that fg(x)= -\infty, not \infty.

    I can only conclude that the " -\infty" terms are incorrect and your statement should have been (fg)^{-1}(\infty)= [ f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))] \cup [ f^{-1} ( \{  \infty \} ) \cap g^{-1}(0, \infty ] ] \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, 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 (fg)^{-1}(\{\infty\})= {x| f(x)g(x)= \infty. And, of course, that is true if either f(x)= \infty and g(x) is non-zero or g(x)= \infty and f(x) is non-zero.

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

    The first term, img.top {vertical-align:15%;} f(x)= \infty and g(x)< 0. Such a product, by the definitions I gave, is \infty The second term, f^{-1} ( \{\infty \} ) \cap g^{-1}(0, \infty ], is all x such that f(x)= \infty" alt="f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))], is all x such that f(x)= \infty and g(x)< 0. Such a product, by the definitions I gave, is \infty The second term, f^{-1} ( \{\infty \} ) \cap g^{-1}(0, \infty ], is all x such that f(x)= \infty" /> and g(x)> 0[/tex]. The third term [ f^{-1} [ - \infty ,0) \cap g^{-1} ( \{ \infty \} ) is the set of all x such that f(x)< 0 and 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 g(x)= \infty.


    The reason for that union of four sets was to separate f^{-1}((-\infty,0)) from f^{-1}((0, \infty]) and to separate 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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