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?
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g(h(D)) $\subseteq $ D
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g(h(D)) $\supseteq $ D
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g(h(D)) $\cap$ D = $\phi$
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g(h(D)) $\cap$ (B—D) $\ne$$\phi$
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