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The \emph{union} of two sets $A$ and $B$ is the set which contains all $x \in A$ and all $x \in B$, denoted $A \cup B$. We can extend this to any (finite or infinite) family $(A_i)_{i\in I}$, writing $\bigcup_{i\in I}A_i$ for the union of this family. Formally, for a family $(A_i)_{i\in I}$ of sets: The \emph{union} of two sets $A$ and $B$ is the set which contains all $x \in A$ and all $x \in B$, denoted $A \cup B$. We can extend this to any (finite or infinite) family $(A_i)_{i\in I}$, writing $\bigcup_{i\in I}A_i$ for the union of this family. Formally, for a family $(A_i)_{i\in I}$ of sets:
\[ x \in \bigcup_{i\in I}A_i\; \Leftrightarrow \;\bigvee_{i\in I}\, (x\in A_i) \] \[ x \in \bigcup_{i\in I}A_i\; \Leftrightarrow \;\bigvee_{i\in I}\, (x\in A_i) \]
Alternatively, and equivalently, Alternatively, and equivalently,
\[x \in \bigcup_{i\in I}A_i\; \Leftrightarrow \;\exists i\in I\text{ such that } x\in A_i\] \[x \in \bigcup_{i\in I}A_i\; \Leftrightarrow \;\exists i\in I\text{ such that } x\in A_i\]
This characterization makes it much clearer that if $I$ is itself the empty set (that is, if we are taking the union of an empty family), then the union is empty; that is, This characterization makes it much clearer that if $I$ is itself the empty set (that is, if we are taking the union of an empty family), then the union is empty; that is,
\[\bigcup_{i\in\emptyset}A_i=\emptyset\] \[\bigcup_{i\in\emptyset}A_i=\emptyset\]
Often elements of sets are taken from some universe $U$ of elements under consideration (for example, the real numbers $\Reals$, or living things on the planet, or words in a particular book). When this is the case, it is meaningful to discuss the \emph{complement} of a set: if $A$ is a set of elements from some universe $U$, then the complement of $A$ is the set Often elements of sets are taken from some universe $U$ of elements under consideration (for example, the real numbers $\Reals$, or living things on the planet, or words in a particular book). When this is the case, it is meaningful to discuss the \emph{complement} of a set: if $A$ is a set of elements from some universe $U$, then the complement of $A$ is the set
\[A^C = U\backslash A= \{x\in U\suchthat x\notin A\}\] \[A^C = U\backslash A= \{x\in U\suchthat x\notin A\}\]
From an axiomatic point of view, the existence of the union is guaranteed by the axiom of union. From an axiomatic point of view, the existence of the union is guaranteed by the axiom of union.
Note that the sets $A_i$ may be, but need not be, disjoint. Unions satisfy some basic properties that are obvious from the definitions: Note that the sets $A_i$ may be, but need not be, disjoint. Unions satisfy some basic properties that are obvious from the definitions:
\begin{itemize} \begin{itemize}
\item Idempotency: $A \cup A = A$ \item Idempotency: $A \cup A = A$
\item $A \cup A^C = U$ where $U$ is the \emph{universe} of $A$ \item $A \cup A^C = U$ where $U$ is the \emph{universe} of $A$
\item Commutativity: $A \cup B = B \cup A$ \item Symmetry: $A \cup B = B \cup A$
\item Associativity: $(A \cup B) \cup C = A \cup (B \cup C)$ \item Associativity: $(A \cup B) \cup C = A \cup (B \cup C)$
\end{itemize} \end{itemize}
Here are some examples of set unions: Here are some examples of set unions:
\begin{gather*} \begin{gather*}
\{1,2\}\cup\{3,4\} = \{1,2,3,4\}\\ \{1,2\}\cup\{3,4\} = \{1,2,3,4\}\\
\{1,2\}\cup\{1,4\} = \{1,2,4\}\\ \{1,2\}\cup\{1,4\} = \{1,2,4\}\\
\{blue, green\}\cup\emptyset = \{blue, green\}\\ \{blue, green\}\cup\emptyset = \{blue, green\}\\
\{x\in\Ints\ \mid\ x\geq 1\}\cup\{x\in\Ints\ \mid\ x\leq\-1\} = \{x\in\Ints\ \mid\ x\neq 0\}\\ \{x\in\Ints\ \mid\ x\geq 1\}\cup\{x\in\Ints\ \mid\ x\leq\-1\} = \{x\in\Ints\ \mid\ x\neq 0\}\\
\{x\in\Reals\ \mid\ x\geq 1\}\cup\{x\in\Reals\ \mid\ x\leq\-1\} = \{x\in\Reals\ \mid\ -1<x<1\} = (-1,1)\\ \{x\in\Reals\ \mid\ x\geq 1\}\cup\{x\in\Reals\ \mid\ x\leq\-1\} = \{x\in\Reals\ \mid\ -1<x<1\} = (-1,1)\\
\{x\in\Reals\ \mid\ x\geq 2\}\cup\{x\in\Reals\ \mid\ x\leq 2\} = \Reals\\ \{x\in\Reals\ \mid\ x\geq 2\}\cup\{x\in\Reals\ \mid\ x\leq 2\} = \Reals\\
\bigcup_{\substack{n\in\Ints\\n>0}} \{x=p/q\in\Rats\ \mid\ q<n\text{ when }p/q\text{ is in lowest terms }\} = \Rats \bigcup_{\substack{n\in\Ints\\n>0}} \{x=p/q\in\Rats\ \mid\ q<n\text{ when }p/q\text{ is in lowest terms }\} = \Rats
\end{gather*} \end{gather*}
The first, third, fourth and fifth of these are the union of disjoint sets, while the second, sixth and seventh are not - in those cases, the sets overlap each other. The first, third, fourth and fifth of these are the union of disjoint sets, while the second, sixth and seventh are not - in those cases, the sets overlap each other.