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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 $X^*$.

  1. 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.

  1. What are the cardinalities of $\mathbb{R}^*$ and $\mathbb{R}^\mathbb{N}$?
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$\mathbb{N}^*$ is countably infinite, so it's cardinality is equal to the cardinality of $\mathbb{N}$. The reason for this is that $\mathbb{N}^*$ is a countable union of countable sets. More generally, we may consider any countably or uncountably infinite set $X$. If $X^*$ consists of finite sequences of elements from $X$, then the cardinality of $X^*$ is equal to the cardinality of $X$. See: Countable_Union_of_Countable_Sets_is_Countable.

For $\mathbb{R}^\mathbb{N}$, its cardinality is equal to the reals, which says that there are as many infinite sequences of real numbers as there are real numbers. Details: cardinality-of-all-real-sequences.

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