Countable Local Basis in Compact Complement Topology

Theorem

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.

Let $p \in \R$.

Then sets of the form:

$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$

form a countable local basis for $p$.

Proof

Let $p \in \R$.

Let:

$\BB_p = \set {\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to: n \in \N}$

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

$P_n = \openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$

Then:

$\R \setminus P_n = \closedint {-n} {p - \dfrac 1 n} \cup \closedint {p + \dfrac 1 n} n$

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

Clearly $p \in P_n$.

So $\BB$ 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:

$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to \subseteq U$

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

$\blacksquare$