How to disprove, if $f$ is a function, $f(A \cap B) != f(A) \cap f(B)?$


marked as duplicate by user63181, Dan Rust, Davide Giraudo, Norbert, TZakrevskiy Dec 20 '13 at 12:45

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Counterexample: Let $f\colon\{1,2\}\rightarrow\{1\}$ be given by $f(1)=1,f(2)=1$ and let $A=\{1\},B=\{2\}$.

To see why this is a counter example, note that $A\cap B=\emptyset$ and so $f(A\cap B)=\emptyset$, but $f(A)\cap f(B)=\{1\}\cap\{1\}=\{1\}$ and so the LHS is not equal to the RHS.

  • $\begingroup$ Thanks Daniel!!! But I was looking for an answer along the lines of math.stackexchange.com/questions/105956/… . I mean I do get it from examples, but I am unable to understand it in mathematical terms. At which step, does this equation fall? $\endgroup$ – user1063185 Sep 3 '13 at 3:23
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    $\begingroup$ @user1063185 It's not an equation because equality does not hold, so it's not a well-posed question to ask when it fails. Note that this is more than 'an example'. It is precisely a proof that not for all $A$ and $B$ does $f(A\cap B)= f(A)\cap f(B)$. $\endgroup$ – Dan Rust Sep 3 '13 at 10:22
  • $\begingroup$ Are there any "pretty" functions that anyone knows that can be used as a suitable counter example as well? $\endgroup$ – Valentino Feb 1 '15 at 0:40

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