# Function Proof that deals with Set Theory

How to prove that this does not hold if $H$ is not injective or how to show that this equation is true just for injective functions? $$H(X\cap Y)=H(X)\cap H(Y)$$

• If this is homework, please show your working. – Shaun Nov 6 '13 at 18:54

If $H$ is not injective, that means that there are two points $x,y$ such that $H(x) = H(y)$.
Then let $X=\{x\}, Y=\{y\}$. $X \cap Y = \emptyset$, so $H(X \cap Y) = \emptyset$, but $H(X) =\{H(x)\}$, $H(Y) = \{H(y\})= \{H(x)\}$, and so $H(X) \cap H(Y) = \{H(x)\} \ne \emptyset$.
Hint: Consider $H$ to be a constant function from natural numbers into the natural numbers (or any set with at least two elements).