Originally Posted by

**Plato** Well you yourself said it does in the OP.

The function $\displaystyle g$ maps the **subsets** of $\displaystyle [2]$

to the integers. The power set of $\displaystyle [2]$ is the set of all subsets of $\displaystyle [2]$.

From the OP the function $\displaystyle g=\{(\emptyset,0),(\{1\},1),(\{2\},1),(\{1,2\},2)$.

So it has everything to do with power sets.

But once again you notation is confused.

$\displaystyle 2^{[2]}$ is the set of all functions from the set $\displaystyle [2]$ to the set $\displaystyle \{0,1\}$.

However, that is not the way $\displaystyle g$ was defined.

If I were you, I would get a textbook that uses standard notation.