# Proof Regarding an iff Statement [duplicate]

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

• What have you tried so far? – cmk May 31 '19 at 15:43
• 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 $$f(x_1) = f(x_2) = y_1, \quad f(x_{3}) = y_3 ?$$ – avs May 31 '19 at 15:58

An incomplete proof of the --> part of the biconditional

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.

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)