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Possible Duplicate:
Conditions Equivalent to Injectivity

Let $A$ := "$f$ is injective" and $B$ := "$f(X \cap Y) = f(X) \cap f(Y)$".

My first idea is to show $B \implies A$ through contraposition, so $\lnot A \implies \lnot B$. Would it then be enough if I say: $f$ is not injective and then show an example where the equation in $B$ is wrong? Would it be a proof then?

Martin Sleziak
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Jay
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  • It is not an exact duplicate question, my question is more general, my question is that if want to show not A => not B is proved if i say f is not injective and have an example where b is not true, if that is enough to count as a proof – Jay Nov 05 '11 at 22:04
  • I bumped up the post @Jay. Let's see if anyone wants the post reopened. – Srivatsan Nov 05 '11 at 22:46
  • Jay: If you mean that for every non-injective $f$ you find an example of sets $X$, $Y$ for which the equality fails, then yes; this is a correct proof. – Martin Sleziak Dec 09 '11 at 12:44

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