Show that $\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}$ is closed Let $a_n$ be a sequence in set of real numbers. Assume $a_n$ converges to $a$ in $\mathbb{R}$. Show that $\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}$ is closed
Recall. $X\in\mathbb{R}$ is closed if and only if the complement of $X$ is open.
Let $X=\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}$ be a set. 
I need to show that $X^c$ is open. Let $t\in X^c$. Then, $t\not\in X$. 
So what should I do?
 A: Let $A=\{ a_n : n\in \mathbb N \} \cup \{ a\}$. 
For any $x\in \mathbb R \backslash A$, let $\epsilon = |x-a|>0$. 
Since $a_n \rightarrow a$, there exists an $N$ such that for all $n> N$, 
$|a_n -a|<\frac{\epsilon}{2}$.
Let $\delta = \min \{ \frac{\epsilon}{2}, |x-a_1|, \cdots, |x-a_N|\}$. Then open ball $\{s\in \mathbb R: |x-s|<\delta\}\subset \mathbb R\backslash A$.
A: Since $(a_n)$ is convergent, therefore it is bounded. Thus, there exists $M>0$ such that $|a_n| \leq M$ for all $n \in \mathbb N.$ As $a_n \to a,$ so we also have that $|a| \leq M$. We thus have, $$X\subseteq [-M,M]\\(-\infty,-M)\cup (M,\infty) \subseteq X^c.$$
Case $1: t<-M.$ Then $t \in (-\infty,-M) \subseteq X^c.$ 
Case $2: t>M.$ Then $t \in (M,\infty)\subseteq X^c.$ 
Case $3: t \in [-M,M].$ Let $\epsilon_1=|a-t|>0.$ Since, $a_n \to a,$ therefore, there exists $n_0\in \mathbb N$ such that $|a_n-a|<\frac{\epsilon_1}{2}$ for all $n \geq n_0.$ Let $\epsilon_2= \min_{n<n_0}|a_n-t|>0$ and $\epsilon=\frac{\min \{\epsilon_1,\epsilon_2\}}{2}.$ Then $t \in (t-\epsilon,t+\epsilon)\subseteq X^c.$ 
It follows that $X^c$ is open.
