1. ## HELP with this Proof PLEASE

Hello,

Can anyone help me with this question please ?

Let X and Y be sets, and let f : X → Y be a function. If A c ("Contained in") X, write f[A] for the set {f(x) | x ∈ A}. Prove that f is a surjection if and only if Y − f[A] c ("Contained in") f[X − A] for all A c ("Contained in") X.

Note: c means "contained in", I am sorry because I don't know how to type this symbol in and "c" is the closest representation to it.

2. You need this lemma. $\left( {E \in P(X)} \right) \Rightarrow f\left[ X \right]\backslash f\left[ E \right] \subseteq f\left[ {X\backslash E} \right]$

If f is surjective then by definition f[X]=Y.
So by the lemma we have at once $X\backslash f\left[ A \right] = Y\backslash f\left[ A \right] \subseteq f\left[ {X\backslash A} \right]$.

Now if $\left( {\forall A \in P(X)} \right)\left[ {Y\backslash f\left[ A \right] \subseteq f\left[ {X\backslash A} \right]} \right]$ we need to show onto.
Suppose that f is not onto then $\left( {\exists q \in Y} \right)\left[ {q \notin f[X]} \right]$.
But that means $\left( {\forall A \in P(X)} \right)\left[ {Y\backslash f\left[ A \right] \not \subset f\left[ {X\backslash A} \right]} \right]$ because $\left[ {q \notin f[X\backslash A]} \right]$.