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I am currently preparing for an exam, and I stumbled across this question that deals with severalt topics I am trying to focus on. In particular, properties shared by locally compact spaces and their one-point compactifications. The question is as follows:

Let $X$ be a locally compact space and $X^{+}$ be its one-point compactification.

(a) Suppose $X$ has a countable base. Is is true that $X^{+}$ has a countable base?

(b) Suppose $X$ is first countable. Is it true that $X^{+}$ is first countable?

Thanks in advance for your help!

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What have you tried? What examples of locally compact (Hausdorff) spaces and their one-point compactifications are you familiar with, and which ones have you tested to see if they are counterexamples? –  Qiaochu Yuan Jan 11 '13 at 3:17
    
I forgot about the one-point compactification of the naturals. –  josh Jan 12 '13 at 20:16
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1 Answer 1

up vote 2 down vote accepted

Assume that $X$ is locally compact and Hausdorff (for non-Hausdorff spaces there are non-equivalent definitions of local compactness and stuff is not as nice..).

If $B_n$, $n \in \mathbb{N}$ is a countable base for $X$, then let $\mathcal{C}$ be the collection all closures of $B_n$ provided these are compact, and also make sure $\mathcal{C}$ is closed under finite unions. Local compactness ensures lots of these exist and a simple counting argument shows that $\mathcal{C}$ is also a countable family of compact sets.

Then the $B_n$, together with the complements (in $X^{+}$) of the members of $\mathcal{C}$, form a countable base for $X^{+}$. To see that we have a local base for the point $\infty$, consider a set of the form $\{\infty\} \cup X\setminus K$, where $K \subset X$ is compact, which is a basic standard type of open set in $X^{+}$. We can cover $K$ by finitely many sets $B_k$ with compact closure (using local compactness of $X$ and compactness of $K$) and their union $C$ is in $\mathcal{C}$. Then $X^{+} \setminus C \subset (\{\infty\} \cup X \setminus K)$.

Note that a countable base for $X$ is also a necessary condition for $X^{+}$ to have a countable base, as $X$ embeds into $X^{+}$ and having a countable base is hereditary.

The second assertion fails, though: consider an uncountable discrete space $X$. The basic open subsets around $\infty$ will be the co-finite subsets of $X^{+}$ that contain it. A standard argument, similar to why an uncountable co-finite topology space is not first countable, will show that $X^{+}$ is not first countabel at $\infty$ as well.

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