A closed discrete set 
Let $V$ be a normed vector space.
Let $(b_n)\subseteq V, b_n \to b\in V.$  Show that $B := \{b,b_1,b_2\dots\}$ is closed.

I know that if $b_n\to b,$ then $b_n$ is Cauchy. That is, $\forall \epsilon > 0, \exists N\in\mathbb{N}, n,m\geq N\Rightarrow ||b_n-b_m|| < \epsilon.$ Also, if $(x_n)\subseteq B, x_n \to x$, then if $x\not\in B, ||x_n-x|| > 0\,\forall n.$ But how can I use the fact that $b_n$ is convergent to show that $B$ is closed? I think I can use the convergence of $(b_n)$ to show that $\exists \epsilon_0 > 0$ such that $||x_{n_k}-x|| \geq \epsilon_0\,\forall k$, where $(x_{n_k})$ is a subsequence of $(x_n).$ Also, it may be easier to show that $V\backslash B$ is open.
 A: Suppose that $x\notin B$. Then in particular $x\ne b$, so let $\epsilon=\frac12\|b-x\|$. There is an $n_0\in\Bbb N$ such that $b_n\in B_\epsilon(b)$ for all $n\ge n_0$, and $B_\epsilon(b)\cap B_\epsilon(x)=\varnothing$, so $b_n\notin B_\epsilon(x)$ when $n\ge n_0$. You now have a nbhd of $x$ that excludes all but finitely many points of $b$. Let $\delta=\min\{\|x-b_k\|:k<n_0\}$; can you see how to use $\delta$ to get a nbhd of $x$ that is disjoint from $B$?
A: Let $x \in V\setminus B$ and $r=\frac{\|x-b\|}{2}>0.$ Then there exists $n_0 \in \mathbb{N}$ such  that $\|b_n-b\|<r$ for all $n \geq n_0.$ So
$$r=\|b-x\|-r<\|b-x\|-\|b_n-b\|\leq \|b_n-x\|$$ for all $n\geq n_0.$ Let $$\epsilon=\min\{r,\|b_i-x\|:1\leq i \leq n_0\}.$$ It follows that $$\|b_n-x\|\geq \epsilon$$ for all $n \in \mathbb{N}.$ Additionally $$\|b-x\|=2r>r\geq \epsilon.$$ So $$B_{\epsilon}(x)\subseteq V\setminus B.$$ We may hence conclude that $V\setminus B$ is open. In other words, $B$ is closed.
A: Let's prove something more general. If $(X,d)$ is a metric space and $x_n \to x$, then $\{x_n:n \geq 1\}\cup \{x\}$ is compact (hence closed).
Proof: Let $\{G_k: k \in K\}$ be an open cover of your set. Then $x \in G_k$ for some $k \in K$. Since $G_k$ is open, there is some $\epsilon > 0$ such that the ball $B_M(x, \epsilon) \subseteq G_k$. This is good news: because $x_n \to x$, there is $n_0 \geq 1$ such that $d(x_n, x) < \epsilon$ if $n \geq n_0$ and thus $x_n \in G_k$ if $n \geq n_0$. Now, choose indices $k_1, \dots, k_{n_0-1}$ such that $x_i \in G_{k_i}$ for $i=1, \dots, n_0-1$. Then
$$\{G_k, G_{k_1}, \dots, G_{k_{n_0}-1}\}$$
is a finite subcover, proving that your set is compact. $\quad \square$
A: Let $c\in V$ \ $B.$ Take $U,$ a nbhd of $b$ and take $V,$ a nbhd of $c,$ with $U, V$ disjoint. 
The set $S=\{n\in \Bbb N:b_n\not \in U\}$ is finite because $b_n\to b.$ So the set $D=\{b_n: n\in S\}=B\setminus U$ is a finite subset of $B.$ 
Now $c\not \in D$ and $D$ is finite so $c$ has a nbhd $V'$ which is disjoint from $D.$ So $V\cap V'$ is a nbhd of $c$ which is disjoint from $B.$ So $c \not\in \overline B.$
We could also say that  $c\not \in \overline {B\cap U}$ because $c\not \in \overline U,$ and that $c\not \in \overline D=\overline {B\setminus U}$, so $c\not\in \overline {B\cap U}\cup \overline {B\setminus U}=\overline {(B\cap U)\cup (B\setminus U)}=\overline B.$ 
