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Let $f$ be a function from the set $A$ to the set $B$. Let $S$ adn$T$ be subsets of $A$ .Show that

- $f(S \cup T) = f(S) \cup f(T)$
- $f(S \cap T) = f(S) \cap f(T)$

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Let y be an arbitrary element.

(a) let y $\in$ f(SUT)

f$^{-1}$y $\in$ SUT

f$^{-1}$y $\in$ S or f$^{-1}$y$\in$ T

y $\in$ f(S) or y $\in$ f(T)

y $\in$ f(S)Uf(T)......means every y which belongs to f(SUT) also belongs to f(S)Uf(T), hence f(SUT) is subset of f(S)Uf(T) -(1)

let y $\in$ f(S)Uf(T)

y $\in$ f(S) or y $\in$ f(T)

f$^{-1}$y $\in$ S or f$^{-1}$y $\in$ T

f$^{-1}$y $\in$ SUT

y $\in$ f(SUT).....hence f(S)Uf(T) is subset of f(SUT) -(2)

from (1) and (2), f(SUT)=f(S)Uf(T)

(b)let y$\in$ f(S$\cap$T)

f$^{-1}$y $\in$S$\cap$T

f$^{-1}$y $\in$S and f$^{-1}$y $\in$T

y$\in$ f(S) and y $\in$f(T)

y $\in$f(S)$\cap$f(T) .....hence f(S$\cap$T) is subset of f(S)$\cap$f(T) -(1)

let y$\in$ f(S)$\cap$f(T)

y $\in$f(S) and y $\in$f(T)

f$^{-1}$y $\in$S and f$^{-1}$y$\in$ T .....this step is only true if f is one-one.It says that if y is in both f(S) and f(T) then its preimage will be common in both S and T. If f is not one one then this is false because there can be an element x1 $\in$S and x2 $\in$T such that x1$\neq$x2 and f(x1)=f(x2)=y

So f(S)$\cap$f(T) is subset of f(SUT) only if f is one-one -(2)

if f is one-one then from (1) and (2) ,f(S)$\cap$f(T) =f(SUT)

If f is not one-one then f(S$\cap$T) is subset of f(S)$\cap$f(T) and not vice versa.

(a) let y $\in$ f(SUT)

f$^{-1}$y $\in$ SUT

f$^{-1}$y $\in$ S or f$^{-1}$y$\in$ T

y $\in$ f(S) or y $\in$ f(T)

y $\in$ f(S)Uf(T)......means every y which belongs to f(SUT) also belongs to f(S)Uf(T), hence f(SUT) is subset of f(S)Uf(T) -(1)

let y $\in$ f(S)Uf(T)

y $\in$ f(S) or y $\in$ f(T)

f$^{-1}$y $\in$ S or f$^{-1}$y $\in$ T

f$^{-1}$y $\in$ SUT

y $\in$ f(SUT).....hence f(S)Uf(T) is subset of f(SUT) -(2)

from (1) and (2), f(SUT)=f(S)Uf(T)

(b)let y$\in$ f(S$\cap$T)

f$^{-1}$y $\in$S$\cap$T

f$^{-1}$y $\in$S and f$^{-1}$y $\in$T

y$\in$ f(S) and y $\in$f(T)

y $\in$f(S)$\cap$f(T) .....hence f(S$\cap$T) is subset of f(S)$\cap$f(T) -(1)

let y$\in$ f(S)$\cap$f(T)

y $\in$f(S) and y $\in$f(T)

f$^{-1}$y $\in$S and f$^{-1}$y$\in$ T .....this step is only true if f is one-one.It says that if y is in both f(S) and f(T) then its preimage will be common in both S and T. If f is not one one then this is false because there can be an element x1 $\in$S and x2 $\in$T such that x1$\neq$x2 and f(x1)=f(x2)=y

So f(S)$\cap$f(T) is subset of f(SUT) only if f is one-one -(2)

if f is one-one then from (1) and (2) ,f(S)$\cap$f(T) =f(SUT)

If f is not one-one then f(S$\cap$T) is subset of f(S)$\cap$f(T) and not vice versa.