If two sets say $S$ and $T$ are equal is it true that $|2^{S}| =|2^{T}|$.
Here is the motivation. Suppose that $S$ has infinite (or countable) order but that is is written as the union of a finite number of mutually exclusive sets say $S = \bigcup_{i}^{n} R_{i}$. Then at least one of the $R_{i}$ has infinite order else $S$ is finite, a contradiction - I am using the "pigeonhole principle" here. Now look at $2^{S}$ and $2^{\bigcup_{i}^{n} R_{i}}$. Is it true that \begin{equation}|2^{S}| = |2^{\bigcup_{i}^{n} R_{i}}| \end{equation}
I am thinking that $2^{\bigcup_{i}^{n} R_{i}}$ has order $2^n$ which is finite. But on the other hand $2^{S}$ order is uncountable.
Here is an example. Suppose that we write $\mathbb{N} = \bigcup_{i=1}^{2} R_{i}$. We can for instance write $R_{1} = \{1,2,3\}$ and $R_{2}=\{4,5,6,7,8,...\}$ so that $R_{2}$ has infinite order. Now $2^{\bigcup_{i=1}^{2}R_{i}}$ is the set $\{\emptyset, \{R_{1}\}, \{R_{2}\}, \{R_{1}, R_{2}\}$ and has order $4$. But $2^{\mathbb{N}}$ is uncountable.