discrete question

• Nov 28th 2005, 01:12 PM
ummmm
help, i'd be impressed
Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
[ f(f-1)) is f and inverse of f]
• Nov 28th 2005, 01:42 PM
CaptainBlack
Quote:

Originally Posted by ummmm
Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
[ f(f-1)) is f and inverse of f]

If $\displaystyle f(A)\ \neq \ B$, $\displaystyle \ f^{-1}(S)$ may not even be defined! (As we may choose $\displaystyle S$ to be $\displaystyle B-f(A)$)

Or have I missed something.

Of course I have missed something:

choose $\displaystyle S$ to be $\displaystyle B-f(A)$,

then:

$\displaystyle f^{-1}(S)\ =\ f^{-1}(B-f(A))\ =\ \emptyset$ -the empty set.

Back to the problem:

(we need only consider the case where $\displaystyle f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)

$\displaystyle f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,

so if $\displaystyle a\ \epsilon \ f^{-1}(S)$ then $\displaystyle f(a)\ \epsilon \ S$ and $\displaystyle f(a)\ \epsilon \ f(A)$

$\displaystyle \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $\displaystyle f(f^{-1}(S))\ \subseteq\ f(A)$,

i.e. $\displaystyle \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$

Now suppose $\displaystyle s \epsilon\ (S\cap f(A))$ then clearly
$\displaystyle s\ =\ f(f^{-1}(s))$ and so $\displaystyle \ s \epsilon f(f^{-1}(S))\$

$\displaystyle \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$

hence:

$\displaystyle S\cap f(A)\ =\ f(f^{-1}(S))$

RonL
• Nov 28th 2005, 02:48 PM
ummmm
i think f(a)=b
i think f(a) has to equal b because there if a function from a to b.
• Nov 28th 2005, 05:00 PM
ummmm
anyone?
anyone?
• Nov 29th 2005, 12:40 AM
lewisje
B is merely the codomain for f
f(A) may be a proper subset of B

All that is necessary is that if x and y are in A and f(x) and f(y) are distinct, then x and y are distinct.
• Dec 1st 2005, 09:07 PM
CaptainBlack
The proof on its own:

(we need only consider the case where $\displaystyle f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)

$\displaystyle f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,

so if $\displaystyle a\ \epsilon \ f^{-1}(S)$ then $\displaystyle f(a)\ \epsilon \ S$ and $\displaystyle f(a)\ \epsilon \ f(A)$

$\displaystyle \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $\displaystyle f(f^{-1}(S))\ \subseteq\ f(A)$,

i.e. $\displaystyle \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$

Now suppose $\displaystyle s \epsilon\ (S\cap f(A))$ then clearly
$\displaystyle s\ =\ f(f^{-1}(s))$ and so $\displaystyle \ s \epsilon f(f^{-1}(S))\$

$\displaystyle \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$

hence:

$\displaystyle S\cap f(A)\ =\ f(f^{-1}(S))$

RonL