Equality of sets

**oldguynewstudent** · December 15th 2009, 06:58 AM

I have a final tonight and I forgot how to do this:

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

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

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

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

(b) Is the following correct?

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

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

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

**Plato** · December 15th 2009, 07:21 AM

Originally Posted by oldguynewstudent

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

If f(x) $\epsilon$ f(S) $\cap$ f(T), then f(x) $\epsilon$ f(S) and f(x) $\epsilon$ f(T). Because f is 1-1, x $\epsilon$ S and x $\epsilon$ T. This means x $\epsilon$ S $\cap$ T. Again because f is 1-1, f(x) $\epsilon$ f(S $\cap$ 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.
$\begin{gathered} z \in f(S) \cap f(T) \hfill \\ z \in f(S)\;\& \;z \in f(T) \hfill \\ \left( {\exists s \in S} \right)\left[ {f(s) = z} \right]\;\& \;\left( {\exists t \in T} \right)\left[ {f(t) = z} \right] \hfill \\ z = f(s) = f(t)\; \Rightarrow \;s = t \hfill \\ \; \Rightarrow \;z \in f(S \cap T) \hfill \\ \end{gathered}$

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