If $(p_n)$ is a convergent sequence in $\mathbb{R}$, then $A=\{p\} \cup \{p_n\mid n \in \mathbb N\}$ is compact. Question:
Suppose that $(p_n)$ is a convergent sequence in $\mathbb{R}$ with $\lim_{n\to\infty} p_n =p$. Prove that $A=\{p\} \cup \{p_n:n\in \mathbb{N}\}$ is a compact set of $\mathbb{R}$.
Not really an attempt (because I don't have an idea at all..):
-I'm going to attempt to use the Heine-Borel theorem (every closed and bounded interval is compact).
Since $(p_n)$ is convergent sequence, the sequence must be bounded. I just need to find the closed interval which will lead to compactness. I don't know how to choose this interval but I'd assume that since we have $\{p\} \cup \{p_n:n\in \mathbb{N}\}$ , then I can choose the smallest point and largest point in $(p_n)$ or maybe even that $p$ itself a boundary. Yep, I'm lost.
 A: Compact is equivalent to closed and bounded. Any convergent sequence is bounded, and adding the limit $p$ of the sequence will still be bounded.  A convergent sequence has only one limit point, for if there were others, say $l \not \in \{p_j\}$, then we could find disjoint neighborhoods around $l$ and $p$ and notice there are infinitely many members of the sequence around $l$.  Therefore for any point not in $\{p\} \cup \{p_j\}$ there is an open neighborhood of that point which has finitely many members of $\{p\} \cup \{p_j\}$, therefore we can look at the minimum distance and find an open neighborhood which doesn't contain $\{p\} \cup \{p_j\}$.  This means that the complement of the set is open, and so the set is both closed and bounded.
A: You can use the fact that in $\mathbb R^n $ , the compact sets are precisely those that are closed and bounded. You can show a convergent sequence is bounded; after the first few points, all points will be (by convergence) within $e>0$ of the limit. Now, you can show that the sequence is a closed subset, since it contains all of its limit points--a sequence can converge to at most one point in the Reals (or, e.g., in any Hausdorff space), and, if the sequence is convergent, it will have just one limit point.
