This question already has an answer here:

Here's my attempt:

  • f(A∩B) = f({x|x∈A∧x∈B}) = {f(x)|x∈{x|x∈A∧x∈B}}
  • f(A)∩f(B) = f({x|x∈A}) ∩ f({x|x∈B}) = {f(x)|x∈{x|x∈A}} ∩ {f(x)|x∈{x|x∈B}} = {x|x∈{f(x)|x∈{x|x∈A}}∧x∈{f(x)|x∈{x|x∈B}}}

And now I'm stuck. Please help.


marked as duplicate by Martin Sleziak, Claude Leibovici, PhoemueX, Travis, Mark Fantini Sep 21 '14 at 9:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


You have $$ A\cap B\subset A\implies f(A\cap B)\subset f(A),\\ A\cap B\subset B\implies f(A\cap B)\subset f(B) $$ so it follows that $f(A\cap B)\subset f(A)\cap f(B)$.

  • $\begingroup$ This is great. But can you explain or prove why if $X \subset Y$, then $f(X) \subset f(Y)$? $\endgroup$ – FutureTrillionaire Sep 21 '14 at 3:41
  • $\begingroup$ If $y\in f(X)$ then by definition $y=f(s)$ for $s\in X$, but $X\subset Y$, so $s\in Y$ as well, which, together with $y=f(s)$, implies that $y\in f(Y)$. $\endgroup$ – Kim Jong Un Sep 21 '14 at 3:43

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