complement dualizes union and intersection

1. Proposition

Let \(X\) be a set, \(U_i \subseteq X\) be subsets. Then

\begin{align*} \bigcap (X \setminus U_i) =& X \setminus \bigcap U_i \bigcup (X \setminus U_i) =& X \setminus \bigcap U_i \\ \end{align*}

2. Proof

2.1. 1)

We note that the following are equivalent by definition:

  1. \(x \in \bigcap (X \setminus U_i)\).
  2. \(x \in X \setminus U_i\) for each \(i \in I\).
  3. \(x \in X\) and \(x \not\in U_i\) for each \(i \in I\).
  4. \(x \in X\) and \(x \not\in \bigcup U_i\)
  5. \(x \in X \setminus \bigcup U_i\)

2.2. 2)

We note that the following are equivalent by definition:

  1. \(x \in \bigcup (X \setminus U_i)\).
  2. \(x \in X \setminus U_i\) for some \(i \in I\).
  3. \(x \in X\) and \(x \not\in U_i\) for some \(i \in I\).
  4. \(x \in X\) and \(x \not\in \bigcap U_i\)
  5. \(x \in X \setminus \bigcap U_i\)

Date: nil

Author: Anton Zakrewski

Created: 2024-10-25 Fr 20:18