Suppose that $S\sim T$, $P\sim Q$, $S⋂P=\varnothing $ and $T⋂Q=\varnothing$. Prove that $(S∪P) \sim (T∪Q)$.

Question

By definition of equivalent sets, a set $S$ is equivalent to set $T$ if and only if the function $f:S\to T$ is one-to-one and onto. A set $P$ is equivalent to set $Q$ if and only if the function $f:P\to Q$ is one-to-one and onto.

How do we prove $(S∪P)\sim (T∪Q)$?

Answer 1

Let $f : S \to T$ and $g : P \to Q$ be a one-to-one and onto functions. To show that $(S∪P)\sim (T∪Q)$ it suffices to give a one-to-one and onto function $h : S∪P \to T∪Q$. Define $h :S∪P \to T∪Q$ to be given by: $$h(x) =\begin{cases} f(x) &\text{if} &x \in S\\ g(x) &\text{if} &x \in P.\end{cases}$$

Now it is routine to check that $h$ is one-to-one and onto, using the facts that $f$ and $g$ are and also that $S⋂P=\varnothing = T \cap Q$.