Which of the following set is countable? Let $X$ be the set roots of unity in $\mathbb C$. Let $S(X)$ be the set of all sequences of elements in $X$. Which of the following subset of $S(X)$ is countable ?


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*The set $ \ \ A \ \  $ of all $(x_n) \in  S(X)$ such that $(x_n)$ is an eventually constant sequence .

*The set $ \ \ B \ \ $ of all $(x_n) \in  S(X)$ such that $x_n = 1$, whenever n is prime number.

*The set $ \ \ C \ \ $ of all $(x_n) \in  S(X)$ such that each $x_n$ is 26th root of unity .

*The set $ \ \ D \ \ $ of all $(x_n) \in  S(X)$ such that $x_{2n} = 1$ for all n$\geq$ 1 .
I think answer is (3), because  number of 26th root of unity is finite. so $C$ is countable

Even if $x_n$ was limited to second roots of unity (or primitive third roots) there would be an uncountably infinite number of such sequences (think binary representations of numbers between $0$ and$1$).
The answer is 1. Group all the sequences by when they get constant, and note that in each category there are countably many sequences, since for the purpose of counting them they might as well be finite.
Of course, you still need to show that 2 and 4 are uncountable.
