📂Set Theory

Union axiom

Axiom

$$\forall X \left( \exists U \left( \forall a \left( a \in x \land x \in X \implies a \in U \right) \right) \right)$$ For any set $X$, there exists a set $U$ that contains all the elements of the elements of $X$.

Definition of Union ¹

The axiom of union guarantees the existence of the union defined as follows: $$x \in A \lor x \in B \iff x \in A \cup B$$ For any two sets $A$ and $B$, the set containing elements that belong to at least one of them is called the union of $A$ and $B$, and is represented as $A \cup B$.

Explanation

The axiom of union and the definition of union are distinctly different. Although a definition is merely a statement of a concept, and the axiom guarantees its existence, the phrase “containing the elements of the elements” in the explanation of the axiom of union differs. If one speaks of the elements of element1 as element2, then element1 is necessarily a set, and the shape of element1 has been guaranteed existence through the pairing axiom and appears similar to $\left\{ A, B \right\}$’s $A$ and $B$. In other words, the union can be seen as being created through an operation $\cup$ between $A$ and $B$, and the concept of union as intended by the axiom of union refers to something like $U(X) := \left\{ a \in x : x \in X \right\}$ when a set of sets like $X = \left\{ A, B \right\}$ is given.

While this distinction might not mean much when dealing with mathematics below the undergraduate level, one should precisely understand and move on if they wish to understand the axiom out of curiosity or on the rare occasion it is needed.

Basic Properties

For the subsets $A$, $B$, and $C$ of the set $X$, the following hold:

• [1] Identity Law: $$A \cup \emptyset = A \\ A \cap X = A$$
• [2] Idempotent Law: $$A \cup A = A \\ A \cap A = A$$
• [3] Commutative Law: $$A \cup B = B \cup A \\ A \cap B = B \cap A$$
• [4] Associative Law: $$A \cup ( B \cup C) = (A \cup B) \cup C \\ A \cap (B \cap C) = ( A \cap B ) \cap C$$
• [5] Distributive Law: $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \\ A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
• [6] De Morgan’s Theorem: $$(A \cup B)^{c} = A^{c} \cap B^{c} \\ (A \cap B)^{c} = A^{c} \cup B^{c}$$
• [7] $$(A \setminus B)^{c} = A^{c} \cup B$$

Proof

[7]

$$\begin{align*} x \in (A \setminus B)^{c} &\iff x \notin A \setminus B \\ &\iff x \notin A \text{ or } x \in B \\ &\iff x \in A^{c} \text{ or } x \in B \\ &\iff x \in A^{c} \cup B \end{align*}$$

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