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10.16 Basic Axioms

Intersection

Let \(A,B\) be two sets, then \(A\cap B=\{x\in A:x\in B\}\checkmark\quad \{x:x\in A\text{ and }x\in B\} \times\)

Notation: Suppose that \(S(x)\) is a property and all sets \(x\) that satisfy this property are elements of a set \(A\), then \(\{x:S(x)\}=\{x\in A:S(x)\}\)

Property

  • \(A\cap B=B\cap A\)
  • \((A\cap B)\cap C=A\cap (B\cap C)\)
  • \(A\cap \empty=\empty\)
  • \(A\subset B\iff A\cap B=A\)

Proof

Use the axiom of extension since all these four items involves equality of set

  1. \(A \cap B = \{x : x \in A \text{ and } x \in B\} = \{x : x \in B \text{ and } x \in A\} = B \cap A\)

  2. \((A \cap B) \cap C = \{x : x \in A \cap B \text{ and } x \in C\} = \{x : (x \in A \text{ and } x \in B) \text{ and } x \in C\}\)
    ​\(= \{x : x \in A \text{ and } (x \in B \text{ and } x \in C)\} = A \cap (B \cap C)\).

  3. \(A \cap \emptyset = \{x \in A : x \in \emptyset\} = \{x \in A : \text{FALSE}\} = \emptyset\).

  4. We know \(A \subset B \iff (x \in A \implies x \in B)\) and \(A \cap B = \{x \in A : x \in B\} =A\)

    Then \(A \cap B = A\iff \{x \in A : x \in B\} = A = \{x \in A : \text{TRUE}\} \implies \forall x \in A, x \in B\)

Union

Let \(\mathcal{C}\) be a collection of sets. Then there exists a set \(\bigcup\mathcal{C}\) (also denoted by \(\bigcup_{A \in \mathcal{C}} A\)) such that \((A \in \mathcal{C}, x \in A) \iff x \in \bigcup\mathcal{C}\) where "collection of sets" means "set"

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"Weak" Version: If \(\mathcal{C}\) is a collection of sets \(\exists\) a set \(B\) s.t. \((A \in \mathcal{C}, x \in A) \implies x \in B\).

This set \(B\) may contain some 'rubbish'

\(\bigcup\mathcal{C} = \{x \in B: \exists A \in \mathcal{C} \text{ with } x \in A \}\) (Axiom of Specification)

Then we can create a large set.

\(A, B, C\) sets. Define a set \(\{A,B,C\} = \{x: x =A \text{ or }x = B \text{ or }x = C\}\)
Need the axiom of union.
​\(\{A,B,C\} = \cup \{\{A,B\} ,\{B,C\}\}\)
​\(\{A,B,C\} = \cup \mathcal{C}: \mathcal{C} = \{\{A\},\{B\},\{C\}\}\)
​\(\{A,B\} = \cup \mathcal{C}: \mathcal{C} = \{\{A\},\{B\}\}\) why we need axiom of pairing?

Because we need the axiom of pairs to create \(\{A\}\)
\(\cup \{A,B\} = A \cup B\), \(\cup \{A,B,C\} = A \cup B \cup C\)

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