Proving that in $S \sqcup T,$ it must be that $S \cup T = S \sqcup T$.

Question

I was reading the hints of @GregMartin on solving $(c)$ of this problem The sum of two sets and the disjoint union.

I am convinced about it. but I am unable to write a proof, could anyone help me in doing so, please?

Answer 1

That question uses sloppy notation. It is not true that $S\cup T=S\sqcup T$ but rather that after identifying $S$ and $T$ with corresponding subsets $\textrm{in}_S(S)$ and $\textrm{in}_T(T)$ we have that $$ S\sqcup T=\textrm{in}_S(S)\cup \textrm{in}_T(T). $$

To see why, you can consider the following concrete realization of $S\sqcup T$. It equals the subset of $(S\sqcup T)\times \{1,2\}$ where $S$ is identified with $S\times \{1\}$ and $T$ is identified with $T\times \{2\}$. This construction is known as the "tagged union", where you treat $1$ and $2$ as labels to tell apart the sets $S$ and $T$, even if they intersect originally.