Is $ \bigcup_{ a\in A}\{a\} = \{a\mid a\in A\} = A$ true

Question

Of course the question is trivial if the answer is yes to the question above. But if not - than its possible to prove that with the other defention of ordered pair - $(a,b) = \{\{x\}, \{x,y\}\}$ And $\bigcup_{a\in A} \{a\}$ is of course = to the right side of the ordered pair as needed.!

Answer 1

Yes!

recall that $a \in A \cup B$ if and only if $a \in A$ or $a \in B$. More generally, $a \in \bigcup_{i \in I} A_i$ if and only if there exists some $i$ so that $a \in A_i$.

the equality $\{a \mid a \in A\}=A$ is trivial. On the other hand, if $a \in A$, then $a \in \{a\}$ by definition, so $a \in \bigcup_{a \in A}\{a\}$. On the other hand, if $b \in \bigcup_{a \in A} \{a\}$, then there exists some $\{x\}$ so that $b \in \{x\}$, so $b=x$, and hence $b$ is in $A$.