A question about one point compactifications. Let S be a topological space that is locally compact, Hausdorff and second countable-but not compact. Let S* be the one point compactification  of S. Then S* is compact Hausdorff. But is S* always necessarily second countable?
 A: You need a countable neighborhood base at the point at infinity $\infty$. But $S$ is locally compact second countable so we may choose a countable collection $\mathcal C$ of opens with compact closure covering $S$. Thus, if $K\subset S$ is compact, then it is covered by finitely many elements of $\mathcal C$. So if we take $\mathcal D$ to be the collection of the closures of all finite unions of elements from $\mathcal C$ then $\mathcal D$ is countable and by construction there is a $D_n\in \mathcal D$ s.t. $S\cup \left \{  \infty  \right \} \setminus D_n\subseteq S\cup \left \{ \infty  \right \} \setminus K$.
A: The answer is yes. To show the compactification $S \cup \{\infty\}$ is first-countable at $\infty$ let $\mathcal B$ be a countable base for $S$. Then let $ \mathcal B' \subset \mathcal B$ be the collection of all $B \in \mathcal B$ such that $S-B$ is compact. Then you need to show $\mathcal D = \{(S-B) \cup \{ \infty \} \colon B \in \mathcal B'\}$ is a neighborhood basis for $\infty$. This should be easy to do from the definition of the one-point compactification. Finally show that $ \mathcal D \cup \mathcal B$ is a basis for $S \cup \{\infty\}$. 
Edit: You might consider instead to use $\mathcal D' = \{D_1 \cup \ldots \cup D_n \colon D_i \in \mathcal D\}$. However we have $\mathcal D' = \mathcal D$ because $D_1 \cup \ldots \cup D_n \cup \{\infty\} = (S-B_1) \cup \ldots \cup (S-B_n) \cup \{\infty\} = (S - B_1 \cap \ldots \cap B_n) \cup \{\infty\}$ and $\mathcal B'$ is closed under intersections, so $D_1 \cup \ldots \cup D_n $ is already an element of $\mathcal D$.
