For the following:
Question: Use the universal property of coproducts to show that, for three sets $A, B$ and $C$, the coproducts $(A\sqcup B)\sqcup C$ and $A\sqcup (B \sqcup C)$ are equipollent. Therefore we can write $A\sqcup B\sqcup C$.
What I would like to know is what would be a good notation for denoting $(A\sqcup B)\sqcup C=A\sqcup (B \sqcup C),$ in terms of set builder notation, where $A=\{(a ,i )| a\in A , i\in I\}$, $B =\{(b ,j )| b\in B, j\in J\}$ and $C =\{(c ,k )| c\in C, k\in K \}$. Would it be something like the following cases, where the first is only for the case of three sets, and the second is for index family of sets:
$(A\sqcup B)\sqcup C=(\{(a ,i )\}\cup \{(b ,j )\})\cup \{(c ,k )\}$ and $A\sqcup (B \sqcup C)=\{(a ,i )\}\cup (\{(b ,j )\}\cup \{(c ,k )\})$. But what happens if we have in the case of index family of sets where $M\sqcup \{N_i\}_{i\in I}=\{N_i\}_{i\in I}\sqcup M$, then $M=\{m|m\in M\}$, $\{N_i\}_{i\in I}=\{(n,i)|n\in N_i, i\in I\}$ $ M\sqcup \{N_i\}_{i\in I}=\{m\}\cup \{(n,i)\}$ similarly $\{N_i\}_{i\in I}\sqcup M=\{(n,i)\}\cup \{m\}$
Lastly here is a concrete example for three sets:
Let $A=\{a,b\}$ $B=\{b,c,d\}$ $C=\{c\}$
then
$A\sqcup B=\{(a,1),(b,1)\}\cup \{(b,2),(c,2), (d,2)\} =\{(a,1),(b,1),(b,2),(c,2), (d,2)\} $
and $(A\sqcup B)\sqcup C=\{(a,1),(b,1),(b,2),(c,2), (d,2)\}\cup \{{c}\}=\{(a,1),(b,1),(b,2),(c,2), (d,2),c\}$
Similarly $B\sqcup C=\{(b,2),(c,2), (d,2)\}\cup \{{c}\}$
and $A \sqcup (B\sqcup C)=\{(a,1),(b,1)\}\cup \{(b,2),(c,2), (d,2)\}\cup \{{c}\}=\{(a,1),(b,1),(b,2),(c,2), (d,2),c\}$
Thank you in advance