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.
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Khashayar Baghizadeh
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In this post you can find links to several posts about this fact: http://math.stackexchange.com/questions/228613/is-this-a-valid-proof or http://math.stackexchange.com/questions/225333/is-this-proof-correct-to-prove-fa-cap-fb-subseteq-fa-cap-b-for-all-fun or http://math.stackexchange.com/questions/231145/prove-fs-cap-t-subseteq-fs-cap-ft or http://math.stackexchange.com/questions/170725/ or http://math.stackexchange.com/questions/239783/ or http://math.stackexchange.com/questions/144870/ – Martin Sleziak Sep 21 '14 at 06:12
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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 $f(A\cap B)\subset f(A)\cap f(B)$.
Kim Jong Un
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This is great. But can you explain or prove why if $X \subset Y$, then $f(X) \subset f(Y)$? – FutureTrillionaire Sep 21 '14 at 03:41
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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 $y\in f(Y)$. – Kim Jong Un Sep 21 '14 at 03:43