up:: Preimage of Function
Let be a function and subsets .
Then let a point in ‘s preimage. Then
By the definition of the preimage of a function, we have that
Thus, and .
Therefore, .
up:: Preimage of Function
Let f:X→Y be a function and subsets C,D⊂Y.
Then let x∈f−1(C∪D) a point in f‘s preimage. Then
x∈f−1(C∪D)⟺f(x)∈C∪D⟺(f(x)∈C)∨(f(x)∈f(D))By the definition of the preimage of a function, we have that
y∈f−1(C∪D)⟺(x∈f−1(C))∨(x∈f−1(D))⟺y∈f−1(C)∪f−1(D)Thus, f−1(C∪D)⊆f−1(C)∪f−1(D) and f−1(C∪D)⊇f−1(C)∪f−1(D).
Therefore, f−1(C∪D)=f−1(C)∪f−1(D).