# 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 ?

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You don't need a hint; that is a correct proof. – MJD Mar 27 '12 at 17:15
That is correct. – Isaac Solomon Mar 27 '12 at 17:16
You can cast it as a proof by contrapositive (which it is) instead of a proof by contradiction: $B\neq\varnothing\implies f(B)\neq\varnothing$. – Arturo Magidin Mar 27 '12 at 17:22
Correct proof. Try formalizing it more though. – ThisIsNotAnId Mar 27 '12 at 17:24
$Y$ can be the empty set if $X$ is also. – MJD Mar 27 '12 at 18:06

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And answers have to be at least $40$. But those excess characters don't have to actually end up displaying anything. – Chris Eagle Mar 28 '12 at 10:11
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}