please help on this bijection question?

Question

Let f: X --> Y be a function.

Prove that f[Bc] = (f[B])c for all B is a improper subset of X iff f is a bijection, where c is complement.

please show all the step. thank you.

Answer 1

Well... this is sorta obvious...
For any x in X, x belongs to B if and only if f[b] belongs to f[B]
Thus, if y in Y does not belong to f[B] (i.e., belongs to f[B]c), then it is necessary that g[y] (where g is the inversion of f - which exists, because f is a bijection) does not belong to B - i.e., belongs to Bc.
Q.E.D.

Answer 2

Suppose for all B < X, f[B*] = (f[B])*. We must prove that f is injective and surjective.

Surjective: choose B = { }, the empty set, then we have f[X] = Y.

Injective: suppose f(x) = y. Let B = {x}, then (f[B])* = Y - {y} = f[B*] = f[X - {x}], showing that if x' <> x then f(x')<> y.

===

Suppose that f is bijective. We must prove that f[B*] < (f[B])* and f[B*] > (f[B])* for all B < X.

"<": Let y be a member of f[B*], then there is x not in B such that f(x) = y. Because f is injective, y is not the image of any other x, in particular not an element of f[B], and therefore y is an element of (f[B])*.

">": Let y be a member of (f[B])*, then there is no x in B such that f(x) = y. Because f is surjective, there must be an x in the complement of B such that f(x) = y, therefore y is contained in the image f[B*].

Answer 3

(1) Suppose f[Bc] = (f[B])c for all subsets B of X. For any element x1 of X,

f[{x1}c] = (f(x1))c

Hence for any element x2 of X, where x2 is in {x1}c, f(x2) is in (f(x1))c, so f(x2) is different from f(x1). That is, f is one-to-one. Also,

f[X] = f[0c] = 0c = Y (where 0 is the empty set)

so f is onto. Hence, f is a bijection.

(2) Suppose f is a bijection. For any subset B of X,

f[B] and f[Bc] are disjoint

since f is one-to-one. But,

f[B] U f[Bc] = Y

since f is onto. Hence f[Bc] = (f[B])c.