Multiple choice

Let f : A $\rightarrow$ B be an injective (one-to-one) function. Define g : 2A $\rightarrow$ 2B as: g(C) = {f(x) | x $\epsilon$ C}, for all subsets C of A. Define h : 2B$\rightarrow$2A as: h(D) = {x | x $\epsilon$ A, f(x) $\epsilon$ D}, for all subsets D of B.

Which of the following statements is always true?

  1. g(h(D)) $\subseteq $ D
  2. g(h(D)) $\supseteq $ D
  3. g(h(D)) $\cap$ D = $\phi$
  4. g(h(D)) $\cap$ (B—D) $\ne$$\phi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

F: A $\rightarrow$B be an injective (one- to - one) function A $\rightarrow$B means A $\supseteq $B Define g: 2A $\rightarrow$2B as: g (C) = {f(x) | x $\epsilon$ C}, for all subset C of A. g(C) $\subseteq $B Define : n : 2B $\rightarrow$ 2A as: h(D) = { x | x $\epsilon$A, f(x) $\epsilon$D}, for all subsets D of B 2B $\subseteq $2A h(D) $\subseteq $ 2A 2(h(D)) $\subseteq $C