Countable Local Basis in Compact Complement Topology

Theorem
Let $T = \left({\R, \tau}\right)$ be the compact complement topology on $\R$.

Let $p \in \R$.

Then sets of the form:
 * $\left({-\infty \,.\,.\, -n}\right) \cup \left({p - \dfrac 1 n \,.\,.\, p + \dfrac 1 n}\right) \cup \left({n \,.\,.\, \infty}\right)$

form a countable local basis for $p$.

Proof
Let $p \in \R$.

Let:
 * $\mathcal B_p = \left\{{\left({-\infty \,.\,.\, -n}\right) \cup \left({p - \dfrac 1 n \,.\,.\, p + \dfrac 1 n}\right) \cup \left({n \,.\,.\, \infty}\right): n \in \N}\right\}$

Let $n \in \N$ and $P_n \in \mathcal B$, so that:


 * $P_n = \left({-\infty \,.\,.\, -n}\right) \cup \left({p - \dfrac 1 n \,.\,.\, p + \dfrac 1 n}\right) \cup \left({n \,.\,.\, \infty}\right)$

Then:
 * $\R \setminus P_n = \left[{-n \,.\,.\, p - \dfrac 1 n}\right] \cup \left[{p + \dfrac 1 n \,.\,.\, n}\right]$

which is the union of two compact sets in $\R$ and therefore itself compact.

Clearly $p \in P_n$.

So $\mathcal B$ is a set of open neighborhoods of $p$ in $T$

Now let $U \in \tau$ such that $p \in U$.

Then $V := \R \setminus U$ is compact and is therefore bounded.

Suppose $\sup V = M, \inf V = m$.

Then $\exists n_B \in \N: n_T > M, -n_B > m$.

But as $p \notin V$ we have:
 * $\exists \epsilon \in \R: p + \epsilon \notin V, p - \epsilon \notin V$

So $\exists n \in \N: \epsilon > \dfrac 1 n$

and so by making $n$ large enough you can fix it so that:
 * $\left({-\infty \,.\,.\, -n}\right) \cup \left({p - \dfrac 1 n \,.\,.\, p + \dfrac 1 n}\right) \cup \left({n \,.\,.\, \infty}\right) \subseteq U$

so fulfilling the conditions for $\mathcal B$ to be a local basis for $p$ which is countable.