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I m stacked in one prove which dealt with sets and functions. I m concerned to prove that: $$f \circ g ( X \cap Y) \subseteq (f \circ g)( X) \cap (f \circ g) (Y)$$ Assume that $g$ is function from $A$ to $B$ and $f$ is from $B$ to $C$, Furthermore $X \subseteq A$ and $Y \subseteq A$.

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Why not prove that $h(X \cap Y) \subset h(X) \cap h(Y)$ first? Suppose $x \in h(X \cap Y)$. Where does that lead you? – copper.hat Nov 6 '13 at 16:49
Notice that the composition of functions is not important. If you define $h = f \circ g: A \to C$, then you are trying to show that $h(X \cap Y) \subseteq h(X) \cap h(Y)$. – Sammy Black Nov 6 '13 at 16:53
Whoa, @copper.hat. Synchronicity. – Sammy Black Nov 6 '13 at 16:54

1 Answer 1

Let $g : A \to B$ and $f : B \to C$ where $X, Y \subseteq A$.

Define $h : A \to C$ by $g = f \circ g$.

Let $b \in h(X \cap Y)$, then there exists $a \in X \cap Y$ such that $h(a) = b$.

Since $a \in X$, then $b = h(a) \in h(X)$. Since $a \in Y$, then $b = h(a) \in h(Y)$.

Hence $b \in h(X) \cap h(Y)$.

We've shown that an arbitrary element in $h(X \cap Y)$ is in $h(X) \cap h(Y)$, which means $$(f \circ g)(X \cap Y) = h(X \cap Y) \subseteq h(X) \cap h(Y) = (f \circ g)(X) \cap (f \circ g)(Y).$$

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