To prove the given statements, we must carefully analyze two main points:

1. If the sequence $(a_n)_{n \in \mathbb{N}}$ converges to $c$, then $c$ is a limit point of the sequence.
2. $c$ is the only limit point of $(a_n)_{n \in \mathbb{N}}$, meaning there are no other limit points of the sequence.

Part 1: $c$ is a limit point of $(a_n)_{n \in \mathbb{N}}$

Definition of Convergence: A sequence $(a_n)$ converges to a limit $c$ if for every $\epsilon > 0$, there exists an index $N \in \mathbb{N}$ such that for all $n \geq N$, we have: $|a_n - c| < \epsilon.$

Now, we recall the definition of a limit point of a sequence. A point $c$ is a limit point of the sequence $(a_n)$ if for any $\epsilon > 0$, there exists an infinite number of terms of the sequence $(a_n)$ within the $\epsilon$-neighborhood of $c$, i.e., $\forall \epsilon > 0, \ \exists \text{ infinitely many } n \text{ such that } |a_n - c| < \epsilon.$

Since $(a_n)$ converges to $c$, by the definition of convergence, we know that for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $a_n$ lies within $\epsilon$-distance from $c$, i.e., $|a_n - c| < \epsilon \quad \forall n \geq N.$ Thus, for any $\epsilon > 0$, there are infinitely many $a_n$'s within $\epsilon$-distance from $c$ (since all terms beyond the index $N$ are within $\epsilon$ of $c$). This proves that $c$ is a limit point of $(a_n)$.

Part 2: $c$ is the only limit point of $(a_n)_{n \in \mathbb{N}}$

To show that $c$ is the only limit point of $(a_n)$, we proceed by contradiction.

Assume that there exists another limit point $L \neq c$. By the definition of a limit point, for any $\epsilon > 0$, there are infinitely many terms of the sequence within $\epsilon$-distance of $L$.

Since $(a_n)$ converges to $c$, we know that for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $a_n$ satisfies $|a_n - c| < \frac{|L - c|}{2}.$ This implies that beyond a certain index $N$, all the terms of the sequence are within a very small distance from $c$, and thus, no terms of the sequence can lie within a neighborhood of $L$, contradicting the assumption that $L$ is a limit point.

Thus, $L \neq c$ cannot be a limit point, meaning $c$ is the only limit point of the sequence.

Conclusion:

• We have shown that if a sequence $(a_n)$ converges to $c$, then $c$ is a limit point of $(a_n)$.
• We also proved that $c$ is the only limit point of $(a_n)$, as no other point can satisfy the conditions of being a limit point.

Would you like further clarification on any of these steps?

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Here are 5 related questions:

1. What is the definition of a subsequential limit?
2. How does a limit point differ from an accumulation point?
3. Can a sequence have multiple limit points if it does not converge?
4. What happens if a sequence converges to a point, but some subsequences converge elsewhere?
5. How does this proof change if we remove the assumption that the sequence converges?

Tip: Understanding the relationship between subsequences and limit points is key to mastering convergence and accumulation point concepts.