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Let $f:X \to Y$ be a function.

Then $f$ is one-to-one iff for all subsets $A$ and $B$ of $X$, $f(A\cap B) = f(A) \cap f(B)$.

Any proofs or guidance would be greatly appreciated

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    $\begingroup$ What have you tried so far? $\endgroup$ – cmk May 31 '19 at 15:43
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    $\begingroup$ Begin by examining simple examples. Let $X = \{x_1, x_2, x_3\}$, $Y = \{y_1, y_2, y_3\}$, $A = \{x_1, x_2\}$, $B = \{x_2, x_3\}$. First, consider $$ f(x_{k}) = y_{k}, \quad k = 1, 2, 3. $$ Is it 1-to-1? Does it preserve intersections? Now, what about $$ f(x_1) = f(x_2) = y_1, \quad f(x_{3}) = y_3 ? $$ $\endgroup$ – avs May 31 '19 at 15:58
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An incomplete proof of the --> part of the biconditional

enter image description here

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Hint for one side:

If $f$ is not one-to-one then $X$ has two distinct elements $u,v$ with $f(u)=f(v)$.

Now let $A$ and $B$ both be specific singletons and see what happens.

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In case your question in on the logical method : Different ways to prove an iff statement.

With : Proposition A = f is a one to one function

Proposition B = for all subsets ... the image of the intersection under f is the intersection of the images of the subsets.

(1) Prove A -> B

(2) Prove B -->A

(3) Conclude : A <--> B


(1) Prove A -> B

(2) Prove ~A --> ~B

(3) Use contraposition : on (2) : B --> A

(4) Conclude : A <--> B


(1) Suppose A is true and that B is false ( or the inverse: A false, B true) that is suppose : ~ (A <--> B)

(3) Derive a contradiction.

(3) Conclude that : (A <--> B)


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