Exercise 3.1.2 (Empty set)

Question

Using only Definition 3.1.4, Axiom 3.1, Axiom 3.2, and Axiom 3.3, prove that the sets $\emptyset$, $\{\emptyset\}$, $\{\{\emptyset\}\}$, and $\{\emptyset, \{\emptyset\}\}$ are all distinct (i.e., no two of them are equal to each other).

Il-Young Son · Accepted Answer

By Axiom 3.2, $\emptyset$ is a set containing no objects.  By Axiom 3.1, $\emptyset$ are
    $\{\emptyset\}$ are also objects that can be members of sets and by Axiom 3.3, $\{\emptyset\}$
    and $\{\{\emptyset\}\}$ are singleton sets whose only member is $\emptyset$ and
    $\{\emptyset\}$, respectively and that $\{\emptyset, \{\emptyset\}\}$ is a pair set whose only
    members are $\emptyset$ and $\{\emptyset\}$.  Since, $\emptyset
otin\emptyset$ (it has no
    members) and $\emptyset\in\{\emptyset\}$, $\emptyset
eq\{\emptyset\}$ by Definition
    3.1.4. Similarly, $\emptyset
eq\{\{\emptyset\}\}$ and
    $\emptyset
eq\{\emptyset,\{\emptyset\}\}$.  Since $\emptyset
eq\{\emptyset\}$, we also know,
    by Definition 3.1.4, that $\{\emptyset\}
eq\{\{\emptyset\}\}$,
    $\{\emptyset\}
eq\{\emptyset, \{\emptyset\}\}$, and that
    $\{\{\emptyset\}\}
eq\{\emptyset,\{\emptyset\}\}$.

Khagan Gulusoy · Answer

$
ewcommand{
otimplies}{\centernot\implies}$
\begin{remark}
What does it mean for two sets not to be equal? We use the laws of logic (See Appendix A.2). Applying de Morgan's laws to Definition 3.1.4, we get
$$
eg\left(\forall x: x \in A \implies  x \in Bight) \quad 	ext{ OR } \quad 
eg\left(\forall x: x\in B \implies x\in Aight).$$
Equivalently,
$$\exists x: x \in A 
otimplies  x \in B \quad 	ext{ OR } \quad \exists x: x\in B 
otimplies x\in A.$$
$x \in A 
otimplies x \in B$ is the same as $x \in A$ and $x 
ot \in B$; similarly, we get $x 
ot\in A$ and $x \in B$ (cf. Section A.2).
ewline 
In other words, to prove that $A 
eq B$, we must either find an element of $A$ which is not contained in $B$ or vice versa.
\end{remark}
\begin{proof}
\begin{enumerate}[label=(\alph*)]
    \item ($\emptyset
eq \{\emptyset\}$) By the remark above, to prove that $\emptyset
eq \{\emptyset\}$, we have to prove that there is an element inside $\{\emptyset\}$ which is not contained in $\emptyset$ or vice versa, i.e.,
    $$\exists x: x\in \emptyset \implies x
otin \{\emptyset\} \quad 	ext{OR} \quad \exists x: x\in \{\emptyset\}\implies x
otin \emptyset.$$
    The only element $x$ that exists in $\{\emptyset\}$ is $\emptyset$ (Axiom 3.3). Since $\emptyset$ is an object (Axiom 3.1) and since for all objects $x$, we have $x 
ot\in \emptyset$ (Axiom 3.2), we see that in particular $\emptyset 
otin \emptyset$. In other words, $\emptyset$ is an element of $\{\emptyset\}$ but not an element of $\emptyset$.
    
    \item ($\{\emptyset\}
eq \{\{\emptyset\}\}$) Similarly, to prove that $\{\emptyset\}
eq \{\{\emptyset\}\}$, we have to prove that there is an element inside $\{\emptyset\}$ which is not contained in $\{\{\emptyset\}\}$ or vice versa, i.e.,
    $$\exists x: x\in \{\{\emptyset\}\} \implies x
otin \{\emptyset\} \quad 	ext{OR} \quad \exists x: x\in \{\emptyset\}\implies x
otin \{\{\emptyset\}\}.$$
    By Axiom 3.3, $\{\emptyset\}\in \{\{\emptyset\}\}.$ But $\{\emptyset\}
otin \{\emptyset\},$ (i.e, the element $\{\emptyset\}$ of the first set $\{\{\emptyset\}\}$ is not contained in the second set $\{\emptyset\}$) because otherwise, by Axiom 3.3, $\emptyset$ would be equal to $\{\emptyset\},$ which contradicts the previous part (a). Thus, $\{\emptyset\}$ is an element of $\{\{\emptyset\}\}$ but not an element of $\{\emptyset\}.$
    
    \item ($\{\emptyset, \{\emptyset\}\}
eq \{\{\emptyset\}\}$) Similarly, to prove that $\{\emptyset, \{\emptyset\}\}
eq \{\{\emptyset\}\}$, we have to prove that there is an element inside $\{\emptyset, \{\emptyset\}\}$ which is not contained in $\{\{\emptyset\}\}$ or vice versa, i.e.,
    $$\exists x: x\in \{\emptyset,\{\emptyset\}\} \implies x
otin \{\{\emptyset\}\} \quad 	ext{OR} \quad \exists x: x\in \{\{\emptyset\}\}\implies x
otin \{\emptyset,\{\emptyset\}\}.$$
    $\emptyset$ is not contained in $\{\{\emptyset\}\}$  because otherwise, by Axiom 3.3, $\emptyset$ would be equal to $\{\emptyset\},$ which contradicts the part (a). But $\{\emptyset,\{\emptyset\}\}$ contains $\emptyset,$ (by Axiom 3.3) i.e, the element $\emptyset$ of the first set $\{\emptyset,\{\emptyset\}\}$ is not contained in the second set $\{\{\emptyset\}\}$.
    
    \item The rest ($\emptyset 
eq \{\{\emptyset\}\}, \emptyset
eq \{\emptyset,\{\emptyset\}\}, \{\emptyset\}
eq \{\emptyset, \{\emptyset\}\}$) follows using the same line of argumentation.
\end{enumerate}
\end{proof}

Exercise 3.1.2 (Empty set)

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Exercise 3.1.2 (Empty set)

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