Set Theory HT25, Basic axioms
Flashcards
Axioms
@State the extensionality axiom.
@State the pairing axiom.
@State the union axiom.
@State the powerset axiom.
@State the comprehension axiom.
For any formula $\phi(z)$ with free variable $z$,
\[\forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land \phi(z)))\]Notation
@Define the notation $\bigcup x$ and $\bigcup \{x, y\}$.
$\bigcup x$ is the set guaranteed to exist by the union axiom.
$\bigcup \{x, y\}$ is the union of $x$ and $y$.
@Define the notation $x \subseteq y$.
@Define the notation $\bigcap x$.
Consequences
@Justify that there is a unique set with no elements.
Existence follows from the empty set axiom.
For uniqueness, suppose $x$ and $y$ both have no elements so that $\forall z \text{ } z \notin x$ and $\forall z \text{ } z \notin y$. But then by the extensionality axiom
\[\forall x \forall y (x = y \leftrightarrow \forall z(z \in x \leftrightarrow z \in y))\]it follows that $x = y$.
@Justify that for any sets $x$ and $y$, there is a unique set whose elements are precisely $x$ and $y$.
Existence follows from the pairing axiom.
For uniqueness, suppose $a$ and $b$ both have this property. But then extensionality implies $a = b$.
What is
\[\mathcal P(\mathcal P(\mathcal P(\emptyset)))\]
?
@Prove that there is no set of all sets, i.e. no set $\Omega$ such that $\forall x \text{ } x \in \Omega$.
Suppose $\Omega$ exists. Then by comprehension, consider $R := \{x \in \Omega \mid x \notin x\}$. Then $R \in \Omega$, so $R \in R \iff R \notin R$, a contradiction.