What happens if the image of a set is empty? Can someone give me hint how to prove that for a function $f:X\rightarrow Y$ and $B\subseteq X$ we have $$f(B)=\emptyset \Rightarrow B=\emptyset \ ?$$ 
Is my idea that, supposing $x\in B\neq \emptyset$, we would have to have, since $f$ is a function, a $y$ such that $f(x)=y$ and therefore $y\in f(B)=\emptyset$, which is a contradiction, correct ?
 A: Yes.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
A: Here is a slightly longer answer. :-)
A straightforward calculation shows that both parts are actually equivalent: we start with the left hand side, expand the definition of $\;\cdot[\cdot]\;$, and then simplify.
\begin{align}
& f[B] = \varnothing \\
\equiv & \;\;\;\;\;\text{"basic property of empty set"} \\
& \langle \forall y :: y \not\in f[B] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cdot[\cdot]\;$"} \\
& \langle \forall y :: \lnot \langle \exists x : x \in B : f(x) = y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"simplify: apply DeMorgan to $\;\exists x\;$"} \\
& \langle \forall y :: \langle \forall x : x \in B : f(x) \not= y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"rearrange quantifications -- to prepare for one-point rule"} \\
& \langle \forall x,y : y = f(x) : x \not\in B \rangle \\
\equiv & \;\;\;\;\;\text{"one-point rule"} \\
& \langle \forall x :: x \not\in B \rangle \\
\equiv & \;\;\;\;\;\text{"basic property of empty set"} \\
& B = \varnothing \\
\end{align}
