Tao Analysis I - 3.1.2

Exercise 3.1.2

Using only Axiom 3.2, Axiom 3.1, Axiom 3.3, and Axiom 3.4, prove that the sets $\emptyset$, $\{ \emptyset \}$, $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$ are all distinct (i.e., no two of them are equal to each other).

Let's remind ourselves of Axioms 3.1-3.4,

• Axiom 3.1 (Sets are objects). If A is a set, then A is also an object.
• Axiom 3.2 (Equality of sets). Two sets $A$ and $B$ are equal, $A=B$, iff every element of $A$ is an element of $B$ and vice versa.
• Axiom 3.3 (Empty set). There exists a set $\emptyset$, known as the empty set, which contains no elements, i.e., for every object $x$ we have $x \not \in \emptyset$.
• Axiom 3.4 (Singleton sets and pair sets). If $a$ is an object, then there exists a set $\{a\}$ whose only element is $a$, i.e., for every object $y$, we have $y \in \{a\}$ iff $y = a$; we refer to $\{a\}$ as the singleton set whose element is $a$. Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a, b\}$ whose only elements are $a$ and $b$; i.e., for every object $y$, we have $y \in \{a, b\}$ iff $y = a$ or $y = b$; we refer to this set as the pair set formed by $a$ and $b$.

Show $\emptyset$ is different from $\{ \emptyset \}$, $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$.

Let's first show that $\emptyset$ is not equal to any of $\{ \emptyset \}$, $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$. 

Axiom 3.3 defines the empty set as having no members. That means no element of any other set can be a member of $\emptyset$.

The other sets do have members by Axiom 3.4. To be specific, $\emptyset \in \{ \emptyset \}$, $\{ \emptyset \} \in \{\{ \emptyset \}\}$ and $\emptyset \in \{ \emptyset, \{ \emptyset \} \}$. 

Axiom 3.2 on set equality requires equal sets to have all their members in common. So $\emptyset$ is different from all the others - because all the other sets have members, and $\emptyset$ has no members.

Having dealt with $\emptyset$, we can now remove it from the list and consider the remaining sets $\{ \emptyset \}$, $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$.

For the rest of this exercise we won't explicitly state that we're using Axiom 3.4 to say that a singleton set $\{a\}$ has a member $a$, and that a pair set $\{a,b\}$ has members $a$ and $b$.

Show $\{\emptyset\}$ is different from $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$.

Let's show that $\{\emptyset\}$ is not equal to any of $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$. 

The element $\{ \emptyset \}$ of $\{\{ \emptyset \}\}$ is not a member of $\{\emptyset\}$. Similarly, the element $\emptyset$ of $\{ \emptyset, \{ \emptyset \} \}$ is not a member of $\{ \emptyset \}$. So by Axiom 3.2, $\{\emptyset\}$ is not equal to any of $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$.

Having dealt with $\{\emptyset\}$, we can now remove it from the list and consider the remaining sets $\{\{ \emptyset \}\}$ and $\{ \emptyset, \{ \emptyset \} \}$.

Show $\{\{ \emptyset \}\}$ is different from $\{ \emptyset, \{ \emptyset \} \}$.

Let's show that $\{\{ \emptyset \}\}$ is not equal to $\{ \emptyset, \{ \emptyset \} \}$.

We could say the sets are not equal because their cardinalities are different. That is, $|\{\{ \emptyset \}\}|=1$ and $|\{ \emptyset, \{ \emptyset \} \}|=2$. However we haven't covered cardinality in the book yet so we must rely only on the Axioms 3.1-3.4, 

We can see the element $\emptyset$ of $\{ \emptyset, \{ \emptyset \} \}$ is not a member of $\{\{ \emptyset \}\}$. So by Axiom 3.2, $\{\{ \emptyset \}\}$ is not equal to $\{ \emptyset, \{ \emptyset \} \}$.

All the results above show that no two of the given sets are equal. $\square$