1. ## 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.

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, $f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))], is all x such that 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".