Function
Properties of Function
Proof for f( S ∪ T ) ⊆ f(S) ∪ f(T) :
Let y be an arbitrary element of f( S ∪ T ).
Then there is an element x in S ∪ T such that y = f(x).
If x is in S, then y is in f(S). Hence y is in f(S) ∪ f(T) .
Similarly y is in f(S) ∪ f(T) if x is in T.
Hence if y ∈ f( S ∪ T ), then y ∈ f(S) ∪ f(T).
QED
Proof for f(S) ∪ f(T) ⊆ f( S ∪ T ) :Hence Property 1 has been proven.
Let y be an arbitrary element of f(S) ∪ f(T) .
Then y is in f(S) or in f(T).
If y is in f(S), then there is an element x in S such that y = f(x).
Since x ∈ S implies x ∈ S ∪ T, f(x) ∈ f( S ∪ T ).
Hence y ∈ f( S ∪ T ).
Similarly y ∈ f( S ∪ T ) if y ∈ f(T) .
QED
Let y be an arbitrary element of f( S ∩ T ).Note here that the converse of Property 2 does not necessarily hold.
Then there is an element x in S ∩ T such that y = f(x), that is
there is an element x which is in S and in T, and for which y = f(x) holds.
Hence y ∈ f(S) and y ∈ f(T), that is y ∈ f(S) ∩ f(T) . QED