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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? $$H(X\cap Y)=H(X)\cap H(Y)$$

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  • $\begingroup$ If this is homework, please show your working. $\endgroup$ – Shaun Nov 6 '13 at 18:54
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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 $H(X) \cap H(Y) = \{H(x)\} \ne \emptyset$.

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Hint: Consider $H$ to be a constant function from natural numbers into the natural numbers (or any set with at least two elements).

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