Practice of exam 2

2. Closure is closed

Let $(x_n)\subset \mathbb{R}$. Show that

$$S=\{x_n|n\in \mathbb{N}\}\cup Lim(x_n)$$

is closed. Show that $S$ is compact $⟺$ $(x_n)$ is bounded.
Firstly, we can prove $S$ is closed by contradiction: suppose there exists $(x_{n_j})\subseteq S$ such that $\lim (x_{n_j})\notin S$, we have two cases:

1. If $(x_{n_j})\subseteq (x_n)$, it leads to a contradiction with $lim(x_{n_j})\notin S$ due to $Lim(x_n)\subset S$ and $\lim (x_{n_j})\in Lim(x_n)$.
2. If $A=(x_{n_j})\backslash (x_n)\ne \varnothing$ and $\lim (x_{n_j})=x^{\prime }$, we can say $A\subset Lim(x_n)$ and $\lim (x_{n_j})=\lim (A)\in Lim(x_n)$ which leads to a contradiction with $\lim (x_{n_j})\notin S$.
   Hence in both cases we can say $S$ is closed proved by contradiction. $\square$