- #1
evinda
Gold Member
MHB
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Hello! (Wave)
Sentence:
If $A,B$ are sets, there is the (unique) set, of which the elements are exactly the following: $\langle a,b\rangle: a \in A \wedge b \in B$.
Proof:
Remark: $\langle a,b\rangle:=\{ \{a\},\{a,b\}\}$
If $a \in A$, then $\{ a \} \subset A \rightarrow \{ a \} \in \mathcal{P}A \rightarrow \{a\} \in \mathcal{P}(A \cup B)$
If $b \in B$, then $\{a,b\} \subset A \cup B \rightarrow \{a,b\} \in \mathcal{P}(A \cup B)$
Therefore, $\{ \{a\},\{a,b\}\} \in \mathcal{P}\mathcal{P}(A \cup B)$
Therefore, from the theorem:
"Let $\phi$ type. If there is a set $Y$, such that $\forall x(\phi(x) \rightarrow x \in Y)$, there there is the set $\{x:\phi(x)\}$."
we conclude that there is the set $\{\langle a,b \rangle : a \in A \wedge b \in B \}$
Could you explain me the above proof?
Sentence:
If $A,B$ are sets, there is the (unique) set, of which the elements are exactly the following: $\langle a,b\rangle: a \in A \wedge b \in B$.
Proof:
Remark: $\langle a,b\rangle:=\{ \{a\},\{a,b\}\}$
If $a \in A$, then $\{ a \} \subset A \rightarrow \{ a \} \in \mathcal{P}A \rightarrow \{a\} \in \mathcal{P}(A \cup B)$
If $b \in B$, then $\{a,b\} \subset A \cup B \rightarrow \{a,b\} \in \mathcal{P}(A \cup B)$
Therefore, $\{ \{a\},\{a,b\}\} \in \mathcal{P}\mathcal{P}(A \cup B)$
Therefore, from the theorem:
"Let $\phi$ type. If there is a set $Y$, such that $\forall x(\phi(x) \rightarrow x \in Y)$, there there is the set $\{x:\phi(x)\}$."
we conclude that there is the set $\{\langle a,b \rangle : a \in A \wedge b \in B \}$
Could you explain me the above proof?