Show that: $R(Z) = R(X \cap Z) \subseteq Y$ and $R^{-1}(Z) = R^{-1}(Z \cap Y) \subseteq X$.

Question

Let R be a nonempty relation from X to Y, and let Z be any set.
(X and Y are non-empty sets)

Show that:

1. $R(Z) = R(X \cap Z) \subseteq Y$
   
2. $R^{-1}(Z) = R^{-1}(Z \cap Y) \subseteq X$

where,

$R(Z) := ${$y : (z, y) \in R$ for some $z \in Z$}

I've got an intuitive understanding. Since R is a relation from X to Y, the element z we're taking should be present in X. So, $Z \subseteq X$ and therefore, $R(Z) = R(X \cap Z)$.

And since $X \cap Z \subseteq X$,
$R(Z) \subseteq Y$.

A similar argument would work for the second one. But how do I write a formal proof?

Answer 1

Careful! It may not be the case that $Z\subseteq X$, since $Z$ can be any set.

Suppose $R$ to be a relation on numbers, and let $Z = X \cup \{\text{sheep}\}$.

Clearly $Z \not\subseteq X$. But $R(Z) = R(X \cap Z)$ still holds, since intuitively, the sheep does not contribute to $R(Z)$, or formally, $R(\{\text{sheep}\})=\emptyset$.

To show the equation, we pick any $y \in R(Z)$. By definition of $R(\,\cdot \,)$, there is some $x\in Z$ such that $(x,y)\in R \subseteq X\times Y$.

This forces $x\in X$, and hence $x \in X \cap Z$.

Since $(x,y)\in R$, by definition of $R(\,\cdot \,)$ again, $y\in R(X\cap Z)$.

This shows that $R(Z) \subseteq R(X\cap Z)$. The reverse inclusion is similar, albeit more straightforward.