Define an operation $\square$ on two sets $S$ and $T$, such that if $|S| = n$ and $|T| = m$, then $|S\mathbin\square T| = n + m$

Question

So what I came up with, similar to Cartesian product, is that for each element in S, you make an ordered pair with the whole set T. For each element in T, you make a pair with the whole set in S. Defined as follows:

$$S\mathbin{\square} T := \{(s,T) \mid \forall s \in S\} \cup \{(t,S) \mid \forall t \in T\}$$

where the ordered pair is defined as follows: $(x,t) = \{\{x\}, \{x,y\}\}$

To prove it is correct, I need to prove that the sets that are being unioned are disjoint, but not sure how to do that. Any help would be appreciated

Answer 1

To disjointify them use a similar idea: define $0 = \emptyset$ as usual and $1= \{\emptyset\}$. These are clearly different.

Then $S\boxplus T := \cup \{(s,0): s \in S\} \cup \{(t,1): t \in T\}$ will do I think.