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

Andrés E. Caicedo
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user1063185
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1 Answers1

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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.

Dan Rust
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  • Thanks Daniel!!! But I was looking for an answer along the lines of http://math.stackexchange.com/questions/105956/proof-of-intersection-of-inverse-images-of-functions . 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? – user1063185 Sep 03 '13 at 03:23
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    @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 $f(A\cap B)= f(A)\cap f(B)$. – Dan Rust Sep 03 '13 at 10:22
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    Are there any "pretty" functions that anyone knows that can be used as a suitable counter example as well? – Valentino Feb 01 '15 at 00:40
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    @Valentino How about $f \colon \mathbb{R} \to \mathbb{R} \colon x \mapsto x^2$ with $A = (-\infty,0)$ and $B = (0,\infty)$. Then $f(A \cap B) = f(\emptyset) = \emptyset$ but $f(A) \cap f(B) = (0,\infty) \cap (0,\infty) = (0,\infty)$. – Dan Rust Mar 23 '21 at 11:06