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10.13 Basic Axioms

Axiom of extension

Two sets are equal if and only if they have the same elements.

\(x\in A\iff x\in B\), this defines equality of sets

  • Reflexivity: \(A=A\)
  • Symmetry: \(A=B\iff B=A\)
  • Transitivity: \(A=B\wedge B=C\implies A=C\)

Remark

\(x \in B \Rightarrow x \in A\), this defines \(B \subset A\) (\(B\) is a subset of \(A\))

Also \((B \subset A)\) and \((A \subset B) \Leftrightarrow A=B\).

  • \(B \subset B\) (reflexive)
  • \(B \subset A\) and \(A \subset B \Rightarrow A=B\) (antisymmetry)
  • \(B \subset A\), \(C \subset B \Rightarrow C \subset A\) (transitive)

Axiom of specification

If \(A\) is a set and \(S(x)\) is a sentence, there exists a set s.t. \(\{x\in A:S(x)\text{ is true}\}\) whose elements are those elements of \(A\) which satisfy the property \(S\)

Positive real numbers: \(\{x\in \R:x>0\}\)

Let \(A\) be a set and \(S(x)\) be property or sentence.

A sentence is built out of:

  • \(A\in B\)
  • \(A=B\)

Anything that can be obtained by the logical operations from these:

  • not (\(A\notin/\in B,\,\,A\neq /=B\))
  • and \(\Big((x\in A)\) and \((x\in B)\Big)\)
  • or \(\Big((A=B)\) and \((A= C)\Big)\)
  • if ...., then ..... \((A\in B)\implies(B\in C)\)
  • equivalent to \((x\in A\iff x\in B)\)
  • there exists \((\exists x\in A)\)
  • for any \((\forall x\in A)\)

\(x \in A\) depends on \(x\) and on \(A\). \(\exists x \in A\) depends on \(A\) (not on \(x\)). \(\forall x \in A\) depend on \(A\) (not on \(x\)).

If a sentence involves \(x\), it depends on it unless \(x\) appears in \(\forall x\) or \(\exists x\)

Proposition

Under no circumstance, there exists a set containing all sets

Proof

Let \(A\) be a set and \(B=\{x\in A:x\notin x\}\) where \(B\subset A\), then we will never have \(B\in A\)

Because if \(B\in A\), either \(B\in B\) or \(B\notin B\), then either \(B\notin B\) or \(B\in B\), then contradiction

Thus if \(A\) is the set of all sets, \(B\notin A\implies\)there is a set not in \(A\)\(\implies\)\(A\) is not the set of all sets

Axiom of existence: Empty set exists

there is a set with no elements (the empty set exists)

Proof: Assume \(A\) is a set. Let \(\empty = \{x \in A: x \neq x \}\), exists by Axiom of Specification.

Note: If \(B \neq A\) and \(\empty' = \{x \in B : x \neq x \}\), then \(\empty=\empty'\) by the Axiom of Extension.

Thus \(\emptyset\) is unique

Axiom of pairing

If \(A,B\) are sets, there exist a set \(X\) with \(A,B\) as elements (\(A \in X, B \in X\)).

Corollary: It follows that there is a set whose elements are precisely \(A\) and \(B\).

Proof

Take \(X\) as in the axiom (\(A \in X, B \in X\)) and let \(\{A, B\} = \{x \in X: (x=A) \text{ or } (x=B)\}\). \(\exists\) by Axiom of Specification

\(\{A, B\}=\{B,A\}\) is the unordered pair of A and B.

If \(A = B\) we have \(\{A, A\}=\{x \in X : x = A\}\) we also write it as \(A\).

\(\{ \empty, \empty \} = \{ \empty \} = \{ x \in X: x = \empty \}\) this set is not empty!

Then we can get \(\empty\), \(\{ \empty \}\), \(\{ \empty, \{ \empty \} \}\), \(\{ \empty, \{ \empty \}, \{ \empty, \{ \empty \} \} \}\) by axiom of pairing \(\Rightarrow\) there are many sets

For any set \(A\), \(\emptyset \subset A\), every element of \(\emptyset\) is an element of \(A\).

Because there is no element of \(\emptyset\) which would not be an element of \(A\).