Inclusion of sets

Question

let $F$ be a multi application :$F: X\leadsto Y$

I have this definitions

"If $B\subset Y$ is an non empty set we have : $F^{-1}(B)=\lbrace x\in X, F(x)\subset B\rbrace $ $F^{-1}_+(B)=\lbrace x\in X, F(x)\cap B \neq \emptyset \rbrace$ "

How to proove that $X\setminus F^{-1}(B)=F^{-1}_+(Y\setminus B)$ et $X\setminus F^{-1}_+(B)=F^{-1}(Y\setminus B)$

please ,help me

thank you

Answer 1

The two equalities you wish to prove are (essentially) the same, and once one has been proven the other should be quite easy (following the same steps).

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To prove the first equality, you basically just follow the definitions, and recall that to show that two sets $U$ and $V$ are equal, it is enough to show $U \subseteq V$ and $V \subseteq U$. I'll give you one of the set inclusions.

Suppose that $x \in X \setminus F^{-1} ( B )$. This means that $x \notin F^{-1} ( B )$, and so by definition it follows that $F(x) \nsubseteq B$. Since $F(x) \nsubseteq B$ there must be an element $y$ of $F(x)$ which is not an element of $B$; that is, $y \in F(x) \setminus B$. Note that since $F(x) \subseteq Y$ this means that $y \in Y \setminus B$, and so we actually have that $y \in F(x) \cap ( Y \setminus B )$, and therefore $F(x) \cap ( Y \setminus B ) \neq \varnothing$. By definition of $F^{-1}_+ ( \cdot )$ it follows that $x \in F^{-1}_+ ( Y \setminus B )$. Therefore $X \setminus F^{-1} ( B ) \subseteq F^{-1}_+ ( Y \setminus B )$.

If you understand how these steps follow one another, you shouldn't have too much difficulty reversing them, and proving the opposite set inclusion (quite likely in a less wordy manner than I did above).