# Equality of sets

• Dec 15th 2009, 06:58 AM
oldguynewstudent
Equality of sets
I have a final tonight and I forgot how to do this:

Let f: X $\displaystyle \longrightarrow$ Y be a function. Show that

(a) f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T)

(b) If f is 1-1, then show that the above inclusion is actually an equality of sets.

(a) f(S $\displaystyle \cap$ T) $\displaystyle \equiv$ { y | y = f(x), x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T} $\displaystyle \equiv$ {y | y = f(x), x $\displaystyle \epsilon$ S} and {y | y = f(x), x $\displaystyle \epsilon$ T} $\displaystyle \subseteq$ f(S) $\displaystyle \cap$ f(T)

(b) Is the following correct?

If f 1-1 then f($\displaystyle x_1$) = f($\displaystyle x_2$) $\displaystyle \longrightarrow$ $\displaystyle x_1$ = $\displaystyle x_2$

We have already shown that f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T). It is left to show that f(S) $\displaystyle \cap$ f(T) $\displaystyle \subset$ f(S $\displaystyle \cap$ T)).

If f(x) $\displaystyle \epsilon$ f(S) $\displaystyle \cap$ f(T), then f(x) $\displaystyle \epsilon$ f(S) and f(x) $\displaystyle \epsilon$ f(T). Because f is 1-1, x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T. This means x $\displaystyle \epsilon$ S $\displaystyle \cap$ T. Again because f is 1-1, f(x) $\displaystyle \epsilon$ f(S $\displaystyle \cap$ T). This proves each is a subset of the other so they are equivalent.
• Dec 15th 2009, 07:21 AM
Plato
Quote:

Originally Posted by oldguynewstudent
We have already shown that f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T). It is left to show that f(S) $\displaystyle \cap$ f(T) $\displaystyle \subset$ f(S $\displaystyle \cap$ T)).

If f(x) $\displaystyle \epsilon$ f(S) $\displaystyle \cap$ f(T), then f(x) $\displaystyle \epsilon$ f(S) and f(x) $\displaystyle \epsilon$ f(T). Because f is 1-1, x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T. This means x $\displaystyle \epsilon$ S $\displaystyle \cap$ T. Again because f is 1-1, f(x) $\displaystyle \epsilon$ f(S $\displaystyle \cap$ T). This proves each is a subset of the other so they are equivalent.

The part in red has a logical error but the idea is correct.
$\displaystyle \begin{gathered} z \in f(S) \cap f(T) \hfill \\ z \in f(S)\;\& \;z \in f(T) \hfill \\ \left( {\exists s \in S} \right)\left[ {f(s) = z} \right]\;\& \;\left( {\exists t \in T} \right)\left[ {f(t) = z} \right] \hfill \\ z = f(s) = f(t)\; \Rightarrow \;s = t \hfill \\ \; \Rightarrow \;z \in f(S \cap T) \hfill \\ \end{gathered}$