Let $f$ be a function. Prove that $f$ is injective iff $f(A\cap B)=f(A)\cap f(B)$ for all subsets $A$ and $B$ of the domain of $f$.
I realize that this is a duplicate question, however the past post on this question doesn't make sense to me so I would like more clarification. Here is the post that I have gone over:
$f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective.
What I am not understanding is the general process of how to approach this proof. I would think that the steps to proving this would go something like this:
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Assume $f$ is injective. Show that this implies that $f(A\cap B)\subset f(A)\cap f(B)$ and then show that it also implies that $f(A\cap B)\supset f(A)\cap f(B)$. Then double containment has been shown, so $f$ being injective implies that $f(A\cap B)=f(A)\cap f(B)$.
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Now assume that $f(A\cap B)=f(A)\cap f(B)$ is true. Show that this implies that $f$ is injective.
I wanted to approach this in two steps because of the "if and only if" statement. However, in the other example, only part number 1 is addressed. Whynot number 2?
Best Answer
I do not know the other examples, but I assumed they use $\Leftrightarrow$ between each assertion – making sure that each assertions are equivalent. So they are doing 1 and 2 simultaneously, reading 1 in one way and 2 the other way. In my opinion it is not a good habit think with equivalence through a problem as it can be difficult to spot if you have mistaken an implication for a equivelence etc.
You way of proof is perfectly right.
Have a nice day.