Exercise 3.7

Let \(R\) and \(A\) be sets. Show that if \(R\) is a relation, then \(\operatorname{Im}_R(A)\) is a set.

Proof. Suppose \(R\) is a relation, as \(R\) is a set by exercise 3.5 \(\operatorname{ran}(R)\) is a set. Because \(\operatorname{Im}_R(A) \subseteq \operatorname{ran}(R)\), by rule 4 \(\operatorname{ran}(R) \cap \operatorname{Im}_R(A) = \operatorname{Im}_R(A)\) is a set too. \(\square\)

Exercise 3.8

Show that the class \(\mathbf{U} = \left\{ x : \exists a \exists b \left[ x = \left\langle a,b \right\rangle \right] \right\}\) is not a set.

Proof. Suppose for a contradiction \(\mathbf{U}\) is a set. Let \(R\) be the Russell class defined in Theorem 3.1, that is let \(R = \left\{ x : x\text{ is a set }\land x \notin x \right\}\). Consider the class \(R\times R = \left\{ x : \exists a\in R \exists b\in R \left[ x = \left\langle a,b \right\rangle \right] \right\}\) which is a collection of sets. Since \(\mathbf{U}\) is a set, \((R\times R) \cap \mathbf{U} = R\times R\) is a set by rule 4. By exercise 3.5, \(\operatorname{dom}(R\times R) = R\) is a set too, contradicting theorem 2.1. \(\square\)

Exercise 3.9

Let \(F\) be a class. Show that if \(F\) is a function and if \(\operatorname{dom}(F)\) is a set, then \(F\) is a set.

Proof. \(F\) is a function and \(\operatorname{dom}(F)\) is a set, so by axiom of replacement (rule 5), \(\operatorname{Im}_F(\operatorname{dom}(F)) = \operatorname{ran}(F)\) is a set. Applying Example 3.4, \(\operatorname{dom}(F) \times \operatorname{ran}(F)\) is also a set. Since \(F\) is a function, \(F \subseteq \operatorname{dom}(F) \times \operatorname{ran}(F)\), then by rule 4, \(F = F \cap (\operatorname{dom}(F) \times \operatorname{ran}(F))\) is a set. \(\square\)