Set stuff

there is some intresting question about discrete math

why \( \mathcal{P}(\emptyset) = \lbrace \emptyset \rbrace \)

consider \( \mathcal(S) \) where \( S \) is a set, for example

• if \( S = \lbrace a \rbrace \), then \( \mathcal{P}(S) = \lbrace \emptyset, \lbrace a \rbrace \rbrace \), same as \( \mathcal{P}(a) \)
• if \( S = \lbrace a, b \rbrace \), then \( \mathcal{P}(S) = \lbrace \emptyset, \lbrace a \rbrace, \lbrace b \rbrace, \lbrace a, b \rbrace \rbrace \), same as \( \mathcal{P}(a, b) \)
• if \( S = \lbrace \emptyset \rbrace \), then \( \mathcal{P}(S) = \lbrace \emptyset, \lbrace \emptyset \rbrace \rbrace \), same as \( \mathcal{P}(\emptyset) \)

so, because \( \lbrace \rbrace \) usually hidden, \( \mathcal{P}(\emptyset) = \lbrace \emptyset \rbrace \)

let \( A = \emptyset, B = \emptyset \), how to \( A \times B \)

Consider \( A \times B = \lbrace (a,b) | a \in A, b \in B \rbrace \)

because

• A is empty, no elements can be paired
• B is also empty

conlusion: no paired element = empty

proof by cardinality \( |A| = 0, |B| = 0, \text{so, when } |A \times B| = |A| \times |B| = 0 \times 0 = 0 \)

does \( P(\emptyset) = \lbrace \lbrace \rbrace \rbrace = \lbrace \emptyset \rbrace \) ?

true, this is due \( \lbrace \emptyset \rbrace \) same as \( \lbrace \lbrace \rbrace \rbrace \), then consider example \( P(a) = \lbrace \lbrace \rbrace, \lbrace a \rbrace\rbrace \)

Suppose X and Z are independent of each other

• \( | \mathcal{P}(X \cap Z)| \)
• \( | X - Z | \)
• \( | X \oplus Z | \)
• \( | X \cap Z | \)
• \( | \mathcal{P}(X) \cup \mathcal{P}(Z)| \)
• \( | \overline{\mathcal{P}(X) \cup \mathcal{P}(Z)}| \)

nb: bagian ini belum jadi