How many colors do we need?

Question

How many different colors do we need so that the set of all possible colorings of $\mathbb{R}^3$ is greater than the powerset of $\mathbb{R}$. Countably many doesn't seem to be enough and even $|\mathbb{R}|$ seems insufficient. I asked this on stackexchange but there is no answer. Is the powerset of $\mathbb{R}$ the least cardinality of the set of colors for this to hold?

Answer 1

You are asking for the smallest cardinal $\lambda$ for which $2^{|\newcommand\R{\mathbb{R}}\R|}<\lambda^{|\R^3|}$.

The answer is $\lambda=(2^{|\R|})^+$.

First of all, this cardinal works, since $\lambda$ by itself is large enough to be larger than $2^{|\R|}$, and so $\lambda^{|\R^3|}$ is as well.

Conversely, $\lambda=2^{|\R|}$ is clearly not large enough, since $\lambda^{|\R|}=(2^{|\R|})^{|\R|}=2^{|\R|}$, which is not large enough.

So $\lambda=(2^{|\R|})^+$ is optimal.