# Accumulation points question (Real Analysis)

Given a sequence of real numbers $\{x_n\}_{n\geq 0}$, assume that there is a finite collection of subsequences of $\mathbb N$, $\{n_k^i \}_{k\geq 0 }$, $i\in I$, where $I$ is a finite index set, and $A_i$, $i\in I$ is the set of accumulation points of $\{x_{n_k^i}\}_{k\geq 0 }$. Assume further that $\bigcup \limits_{i\in I}\{n_k^i\}_{k\geq 0}=\mathbb N \setminus B$ , where $B$ is a finite set, possibly empty.

- Prove that the set of all accumulation points of the initial sequence is $A=\bigcup\limits _{i\in I}A_i$.
- Is this true for an infinite family $I$? Why or why not?

Emma

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