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\begin{document}
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\section{Set Paradoxes}

\begin{enumerate}

\item \textbf{Cantor's Paradox.}  {\color{blue} Let $\mathcal{V}$ be the set of all sets.}  Since every subset of $\mathcal{V}$ is a set, $\wp(\mathcal{V}) \subseteq \mathcal{V}$.  This implies  $|\wp(\mathcal{V}) |\leq |\mathcal{V}|$.  However, by Cantor's Theorem, $|\mathcal{V}| < |\wp(\mathcal{V})|$.  This contradicts the Law of Trichotomy for Cardinal Numbers.

\item \textbf{Russell's Paradox.}  {\color{blue} Let $\mathcal{Z}$ be the set of all sets that are not elements of themselves}:  that is, $\mathcal{Z} = \{X \in \mathcal{V} \, | \, X \not \in X \}$.  Then both $\mathcal{Z} \in \mathcal{Z}$ and $\mathcal{Z} \not \in \mathcal{Z}$ lead to contradiction.


\item \textbf{Burali-Forti Paradox.}  {\color{blue} Let $\mathcal{O}$ be the set of all ordinal numbers.}  By Theorem 5.2, $\mathcal{O}$ is a well-ordered set, so $\text{ord}(\mathcal{O}) = \lambda$ for some ordinal number $\lambda \in \mathcal{O}$.  By Theorem 5.5, $\lambda = \text{ord}(s(\lambda))$, meaning that $\text{ord}(\mathcal{O})=\text{ord}(s(\lambda))$, which in turn means that $\mathcal{O} \simeq s(\lambda)$.  However, by Corollary 4.12, a well-ordered set cannot be order-isomorphic to one of its initial segments.


\item   {\color{blue} Let $\mathcal{K}$ be the set of all cardinal numbers.}  By the Axiom of Choice, for each $\alpha \in \mathcal{K}$, we can choose a representative set $A_\alpha$ with $|A_\alpha|=\alpha$.  Let $A= \bigcup \{A_\alpha \, | \, \alpha \in \mathcal{K}\}$.  Consider its power set, $\wp(A)$.  Since $|\wp(A)| \in \mathcal{K}$, there exists a corresponding representative set $A_{|\wp(A)|}$, which by definition is  a subset of $A$.  Hence, $|\wp(A)| = |A_{|\wp(\mathcal{A})|}| \leq |A|$.  However, by Cantor's Theorem, $|A| < |\wp(A)|$.

\item {\color{blue} Let $A$ be any nonempty set and let $\mathfrak{A}$ be the set of all sets equipotent to $A$}; that is, $\mathfrak{A}$ is the equivalence class of all sets having cardinal number $|A|$.  Let $I$ be any other set, and for each $i \in I$, let $A_i = A \times \{i\} = \{ (a, i) \mid a \in A \}$.  Then $A \sim A_i$ for all $i \in I$, and moreover $|\{A_i \mid i \in I \}| = |I|$.  Since $\mathfrak{A}$ is a set, it has a power set, $\wp(\mathfrak{A})$.  Replace $I$ with $\wp(\mathfrak{A})$ in the construction above to obtain a set $\{A_i \mid i \in \wp(\mathfrak{A}) \}$ of cardinality $|\wp(\mathfrak{A})|$.  Since each $A_i$ is equipotent to $A$, by definition each $A_i \in \mathfrak{A}$.  Thus $\{ A_i \mid i \in \wp(\mathfrak{A}) \} \subseteq \mathfrak{A}$.  This implies $|\{ A_i \mid i \in \wp(\mathfrak{A}) \}| = |\wp(\mathfrak{A})| \leq |\mathfrak{A}|$.  However, by Cantor's Theorem, 
$ |\mathfrak{A}|<|\wp(\mathfrak{A})|$.

\item {\color{blue} Let $A$ be any nonempty well-ordered set and let $\mathfrak{W}$ be the set of all well-ordered sets order-isomorphic to $A$}; that is, $\mathfrak{W}$ is the equivalence class of all well-ordered sets having ordinal number $\text{ord}(A)$.  Let $I$ be any other set, and for each $i \in I$, let $A_i = A \times \{i\} = \{ (a, i) \mid a \in A \}$, ordered by $(a, i) \prec (b, i) \iff a \prec b$.  Then $A \simeq A_i$ for all $i \in I$, and moreover $|\{A_i \mid i \in I \}| = |I|$. (Note that these are cardinalities, not ordinalities!)  Since $\mathfrak{W}$ is a set, it has a power set, $\wp(\mathfrak{W})$.  Replace $I$ with $\wp(\mathfrak{W})$ in the construction above to obtain a set $\{A_i \mid i \in \wp(\mathfrak{M}) \}$ of cardinality $|\wp(\mathfrak{W})|$.  Since each $A_i$ is order-isomorphic to $A$, by definition each $A_i \in \mathfrak{W}$.  
Thus $\{ A_i \mid i \in \wp(\mathfrak{W}) \} \subseteq \mathfrak{W}$.  This implies $|\{ A_i \mid i \in \wp(\mathfrak{W}) \}| = |\wp(\mathfrak{W})| \leq |\mathfrak{W}|$. However, by Cantor's Theorem, $|\mathfrak{W}| < |\wp(\mathfrak{W})|$.
\end{enumerate}

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\textbf{IMPORTANT NOTES.}  {\color{blue}Every one of these paradoxes arises as a result of assuming the existence of a  set consisting of an infinite collection of sets.  Four of these paradoxes involve a contradiction of the Law of Trichotomy for Cardinal Numbers by way of Cantor's Theorem.}
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