The ZFC Axioms and the Basic Set Constructions: Examples and Counterexamples

$\mathcal{P}(\mathrm{\varnothing })=\{\mathrm{\varnothing }\}$ is the general identity applied at $\mathrm{\varnothing }$.

A set $z$ satisfies $z\subseteq \{\mathrm{\varnothing }\}$ exactly when every element of $z$ equals $\mathrm{\varnothing }$, that is, exactly when $z$ has no element or $\mathrm{\varnothing }$ is its only element; those two sets are $\mathrm{\varnothing }$ and $\{\mathrm{\varnothing }\}$, and both are indeed included in $\{\mathrm{\varnothing }\}$. Hence $\mathcal{P}(\{\mathrm{\varnothing }\})=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$.

A set $z$ satisfies $z\subseteq A$ exactly when every element of $z$ is $\mathrm{\varnothing }$ or $\{\mathrm{\varnothing }\}$, so $z$ is determined by whether it has $\mathrm{\varnothing }$ as an element and whether it has $\{\mathrm{\varnothing }\}$ as an element; the four resulting sets are $\mathrm{\varnothing }$, $\{\mathrm{\varnothing }\}$, $\{\{\mathrm{\varnothing }\}\}$ and $A$, and each is included in $A$. Hence $\mathcal{P}(A)=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\},\{\{\mathrm{\varnothing }\}\},A\}$.

The four are pairwise distinct: $\mathrm{\varnothing }$ has no element, $\{\mathrm{\varnothing }\}$ has $\mathrm{\varnothing }$ as its only element, $\{\{\mathrm{\varnothing }\}\}$ has $\{\mathrm{\varnothing }\}$ as its only element, and $A$ has both; moreover $\mathrm{\varnothing }\mathrm{\ne }\{\mathrm{\varnothing }\}$, since the second has an element and the first does not.

The three power sets are computed and the four members of $\mathcal{P}(A)$ are distinct, which is the example.

$\mathrm{\varnothing }\mathrm{\ne }\{\mathrm{\varnothing }\}$, because $\{\mathrm{\varnothing }\}$ has $\mathrm{\varnothing }$ as an element and the empty set has none.

$(\mathrm{\varnothing },\mathrm{\varnothing })=\{\{\mathrm{\varnothing }\},\{\mathrm{\varnothing },\mathrm{\varnothing }\}\}$ by the definition, and $\{\mathrm{\varnothing },\mathrm{\varnothing }\}=\{\mathrm{\varnothing }\}$, so both members of the outer pair are $\{\mathrm{\varnothing }\}$ and the outer pair is $\{\{\mathrm{\varnothing }\}\}$.

$(\mathrm{\varnothing },\{\mathrm{\varnothing }\})=\{\{\mathrm{\varnothing }\},\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}\}$ directly by the definition, with no collapse, since $\mathrm{\varnothing }\mathrm{\ne }\{\mathrm{\varnothing }\}$ keeps the inner pair a set with two elements.

The two pairs are distinct: their first coordinates agree but their second coordinates are $\mathrm{\varnothing }$ and $\{\mathrm{\varnothing }\}$, which differ, so the characterising property forbids equality. This is visible in the computed sets as well, since $\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$ is an element of the second and not of the first.

$(a,b)$ is the unordered pair whose members are $\{a\}$ and $\{a,b\}$, so $\bigcap (a,b)=\{a\}\cap \{a,b\}$; and $z$ lies in that intersection exactly when $z=a$, and $z=a$ or $z=b$, which is exactly $z=a$. Hence $\bigcap (a,b)=\{a\}$.

Likewise $\bigcup (a,b)=\{a\}\cup \{a,b\}$, and $z$ lies in that union exactly when $z=a$, or $z=a$ or $z=b$, which is exactly $z=a$ or $z=b$. Hence $\bigcup (a,b)=\{a,b\}$.

Applying $\bigcup$ to step 1.1 gives $\bigcup \bigcap (a,b)=\bigcup \{a\}=a$, so the first coordinate is recovered.

By steps 1.1 and 1.2, $\bigcup (a,b)\setminus \bigcap (a,b)=\{a,b\}\setminus \{a\}$, whose elements are the $z$ with $z=a$ or $z=b$, and $z\mathrm{\ne }a$; that is, it is $\mathrm{\varnothing }$ when $b=a$ and has $b$ as its only element when $b\mathrm{\ne }a$.

Both coordinates are therefore determined by the set $(a,b)$, which is the content of the characterising property made explicit.

$A\cup B=B$: an element of $A\cup B$ is $\mathrm{\varnothing }$ or is an element of $B$, and $\mathrm{\varnothing }\in B$, so the two sets have the same elements.

The only element of $A$ is $\mathrm{\varnothing }$, and the elements of $B$ are $\mathrm{\varnothing }$ and $\{\mathrm{\varnothing }\}$, so the pairs with first coordinate in $A$ and second in $B$ are exactly $(\mathrm{\varnothing },\mathrm{\varnothing })$ and $(\mathrm{\varnothing },\{\mathrm{\varnothing }\})$; unfolding the definition of the ordered pair, these are $\{\{\mathrm{\varnothing }\}\}$ and $\{\{\mathrm{\varnothing }\},\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}\}$.

Hence $A\times B$ has exactly those two elements, and they are distinct because $\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$ belongs to the second and not to the first.

Both elements lie in $\mathcal{P}(\mathcal{P}(B))$: this is the general fact applied with $A\cup B=B$, and it is also visible directly, since each is a set of subsets of $B$.

The product is listed in full and its two elements are exhibited inside the ambient double power set that makes the separation legitimate.

Neither of the two sets is included in the other. The only element of $a$ is $\mathrm{\varnothing }$ and the only element of $b$ is $\{\mathrm{\varnothing }\}$, and these differ because $\{\mathrm{\varnothing }\}$ has an element while $\mathrm{\varnothing }$ has none; so $\mathrm{\varnothing }\in a$ with $\mathrm{\varnothing }\mathrm{\notin }b$, and $\{\mathrm{\varnothing }\}\in b$ with $\{\mathrm{\varnothing }\}\mathrm{\notin }a$.

The general inclusion holds, and it is an equality exactly when one of the two sets is included in the other; by step 1.1 that fails here, so the inclusion is proper.

The witnessing element is $a\cup b$ itself: it is a subset of $a\cup b$, hence an element of $\mathcal{P}(a\cup b)$, whereas $a\cup b\subseteq a$ would force $b\subseteq a$ and $a\cup b\subseteq b$ would force $a\subseteq b$, both excluded by step 1.1; so $a\cup b$ is in neither $\mathcal{P}(a)$ nor $\mathcal{P}(b)$.

$(\mathrm{\varnothing },\mathrm{\varnothing })=\{\{\mathrm{\varnothing }\},\{\mathrm{\varnothing },\mathrm{\varnothing }\}\}=\{\{\mathrm{\varnothing }\}\}$, which has $\{\mathrm{\varnothing }\}$ as an element; $\mathrm{\varnothing }$ has no element, so $(\mathrm{\varnothing },\mathrm{\varnothing })\mathrm{\ne }\mathrm{\varnothing }$.

Since $\mathrm{\varnothing }$ is the only element of each of $A$, $B$, $C$, the product $A\times B$ has $(\mathrm{\varnothing },\mathrm{\varnothing })$ as its only element, so $((\mathrm{\varnothing },\mathrm{\varnothing }),\mathrm{\varnothing })$ is an element of $(A\times B)\times C$.

Every element of $A\times (B\times C)$ has the form $(x,y)$ with $x\in A$, hence with $x=\mathrm{\varnothing }$; if $((\mathrm{\varnothing },\mathrm{\varnothing }),\mathrm{\varnothing })$ were such an element then the characterising property would give $(\mathrm{\varnothing },\mathrm{\varnothing })=\mathrm{\varnothing }$, which step 1.1 refutes.

The set $(A\times B)\times C$ therefore has an element that $A\times (B\times C)$ does not, so the two products are different.

$\mathrm{\varnothing }\mathrm{\ne }\{\mathrm{\varnothing }\}$, since the second has an element and the first has none; so $\mathrm{\varnothing }\mathrm{\notin }B$ and $\{\mathrm{\varnothing }\}\mathrm{\notin }C$, while $\mathrm{\varnothing }\in A$ and $\{\mathrm{\varnothing }\}\in D$.

The inclusion from left to right always holds: an element of $A\times C$ is a pair $(x,y)$ with $x\in A$ and $y\in C$, hence with $x\in A\cup B$ and $y\in C\cup D$, so it lies in $(A\cup B)\times (C\cup D)$; the same argument applies to $B\times D$.

The pair $(\mathrm{\varnothing },\{\mathrm{\varnothing }\})$ lies in $(A\cup B)\times (C\cup D)$, because $\mathrm{\varnothing }\in A$ gives $\mathrm{\varnothing }\in A\cup B$ and $\{\mathrm{\varnothing }\}\in D$ gives $\{\mathrm{\varnothing }\}\in C\cup D$.

It lies in neither product on the left. Membership in $A\times C$ would give $\{\mathrm{\varnothing }\}\in C$, and membership in $B\times D$ would give $\mathrm{\varnothing }\in B$; the characterising property makes the coordinates unambiguous, and step 1.1 rules out both.

The inclusion of step 1.2 therefore omits the element exhibited at step 2.1, so it is proper.

Suppose the schema holds for every formula of the language.

Instantiate it at the formula $\phi (x):=x\mathrm{\notin }x$: there is a set $R$ such that, for every set $x$, $x\in R$ holds if and only if $x\mathrm{\notin }x$.

No such set exists, so the supposition fails and the schema is false. Instantiating instead at $\phi (x):=x=x$ produces a set with every set as an element, which is impossible for the same underlying reason.

Suppose the condition defining the intersection determines a set at $\mathrm{\varnothing }$, and call it $c$.

$\mathrm{\varnothing }$ has no elements, so every set $x$ satisfies "$x$ belongs to every element of $\mathrm{\varnothing }$" vacuously; hence every set is an element of $c$.

No set has every set as an element, so no such $c$ exists; in particular the equation of the claim asserts something of an object that is not there.

The claim also fails on its own terms: were $c$ equal to $\mathrm{\varnothing }$ it would have no elements, while step 2.1 puts $\mathrm{\varnothing }$ itself among its elements. Both readings collapse, so $\bigcap \mathrm{\varnothing }$ is left undefined rather than assigned the value $\mathrm{\varnothing }$.

Take $a=b=c:=\{\mathrm{\varnothing }\}$.

$a\setminus b$ has no elements, since $z\in a\setminus b$ requires $z\in a$ and $z\mathrm{\notin }b$ while $a$ and $b$ are the same set; so $a\setminus b=\mathrm{\varnothing }$, and $(a\setminus b)\setminus c$ likewise has no elements and equals $\mathrm{\varnothing }$.

By the same computation $b\setminus c=\mathrm{\varnothing }$, and $z\in a\setminus \mathrm{\varnothing }$ requires only $z\in a$, so $a\setminus (b\setminus c)=a=\{\mathrm{\varnothing }\}$.

The two sides are $\mathrm{\varnothing }$ and $\{\mathrm{\varnothing }\}$, which differ because the second has an element and the first has none; the claim is therefore false.