Let $X$ be a finite set with at least 2 elements. Then the set of all finite-length "strings", $$X^* = \bigcup_{L \in \mathbb{Z}^+} \prod_{i=1}^L X_i = \{ (x_1, \ldots, x_L) : L \in \mathbb{Z}^+ \text{and } x_i \in X\}$$ is countably infinite. If we make the strings infinitely long through a series of Cartesian products: $$X^{\mathbb{N}} = \prod_{i \in \mathbb{N}} X_i = \{ (x_1, \ldots): x_i \in X \}$$ then the resulting Carteisan product is uncountable and equal in cardinality to the real numbers.
When $X$ is countably infinite, I know from this question that $X^\mathbb{N}$ is still equal in cardinality to the real numbers, but I'm not sure about $X^*$.
- What is the cardinality of $\mathbb{N}^*$?
The next obvious question is what happens when $X$ is uncountably infinite and equal in cardinality to the real numbers. It will certainly be uncountable, but I want to know how it compares to the cardinality of the reals.
- What are the cardinalities of $\mathbb{R}^*$ and $\mathbb{R}^\mathbb{N}$?