Show that a set is closed. The sequence $x_n\;n\in\mathbb{N}$ converges to $x$ over $\mathbb{R}$.
Define $A=\left \{ x_n\mid n\in\mathbb{N} \right \}$
How do I show that $A\cup \left \{ x \right \}$  is closed?
 A: Hint: In a metric space closed is equivalent to sequentially closed. Hence, it suffices to show that for any convergent sequence in $A\cup\{x\}$, its limit is also in $A\cup\{x\}$. 
Full solution: Let $(a_n)_n$ be a convergent sequence in $A\cup\{x\}$ with limit $a$. We need to show $a\in A\cup\{x\}$.  If there exists an $n_0$ s.t. $a_n=x$ for all $n\geq n_0$, it follows $a=x\in A\cup\{x\}$ and we are done. Now suppose the opposite. Then, $a_n\neq x$ for infinitely many $n$'s. Hence, there exists a subsequence $(a_{n_k})_k$ such that $a_{n_k}\neq x$ for all $k$. We know that for each $k$ there exists an $m_k$ s.t. $a_{n_k}=x_{m_k}$. This gives us a sequence of indices $(m_k)_k$. Now, pick an increasing subsequence of these indices $(m_{k_j})_j$. Then $(a_{n_{k_j}})_j=(x_{m_{k_j}})_j$ is a subsequence of $(x_n)_n$ and hence converges to the same limit. Since it is also a subsequence of the convergent sequence $(a_n)_n$, we get $a=x\in A\cup\{x\}$. 
A: It is not just closed it is compact.  And, every compact set is closed.
A set is compact if every open cover has a finite sub cover.
In any open cover of A, there is an open set that includes $x.$  It will also cover points in A that are close to $x.$
This leaves finitely many points in $A$ that are outside this set, these can be covered with finitely many sets.
A: A point-set proof.
Let $B = A \cup \{x\}$ and $ C = \mathbb{R} \setminus B$: we have to prove that $C$ is open, i.e. every element in $C$ has a neighbourhood that is disjoint from $B$.
Let $y \in C$ and $d = \| y - x \|$, the distance from $x$: it is positive $d \gt 0$ because $y \neq x$. Let $U = \{a \in \mathbb{R} \mid \| a - x \| \lt \dfrac{1}{2}d \}$, a neighbourhood of $x$.
We have that:


*

*by definition of $x$ as limit of $x_n$, there exists $N \in \mathbb{N}$ such that for all $n \gt N$, $x_n \in U$;

*therefore $B \setminus U$ has finitely many points: $\{x_n\}_{n=0,\ldots, N}$.


For $n=0,\ldots, N$, let $d_n = \| y - x_n \|$ the distances between $x_n$ and $y$: they are positive $d_n \gt 0$ because $y \neq x_n$.
We are considering finitely many numbers, so there exists a minimum:
$k = min\{d_0,\ldots,d_N,\frac{1}{2}d\}$
$k$ is positive $k \gt 0$ because all $d_n,d$ are positive. Let $r = \dfrac{1}{2}k$: we have that all $d_0,\ldots,d_N,\frac{1}{2}d > r$.
Let $V = \{a \in \mathbb{R} \mid \| a - y \| \lt r \}$: a neighbourhood of $y$ and we claim it is disjoint from B:


*

*$x \notin V$ because $\| y - x \| = d > r$;

*for $n=0,\ldots, N$, the number $x_n \notin V$ because $\| y - x_n \| = d_n > r$;

*for $n \gt N$, the number $x_n \notin V$ because


*

*$\| x_n - x \| \lt \dfrac{1}{2}d$ because $x_n \in U$;

*$d = \| y - x \| \lt \| y - x_n \| + \| x_n - x \|$ by triangular inequality;

*therefore $\| y - x_n \| \gt d - \| x_n - x \| \gt d - \dfrac{1}{2}d = \dfrac{1}{2}d > r$.


