I have managed to show 1. and 2. but I am not sure how to go about the last part about showing that the space is not paracompact. How should this be done?
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It is sufficient to show that $\beta \mathbb{N} \times (\beta \mathbb{N}\setminus{p})$ is not normal. – Henno Brandsma May 26 '19 at 15:30
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Regarding (2): In any space $X$ we have the theorem that if $V$ is a locally finite family of closed subsets of $X$ then $\cup V$ is closed. Now let $U$ be any locally finite family. It is easy to see that $V={\bar u: u\in U}$ is also locally finite. So $\overline {\cup U}\subset$ $ \overline {\cup V}=\cup V=$ ${\bar u:u\in U}\subset \overline {\cup U}.$ – DanielWainfleet May 26 '19 at 22:37
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In the last line of my previous comment that should be $\cup {\bar u:u\in U}$ instead of ${\bar u:u\in U}.$ – DanielWainfleet May 26 '19 at 22:44
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First of all, $\beta \Bbb N$ is separable (as it contains $\Bbb N$ as the countable dense subset). Moreover, $\beta \Bbb N\setminus \Bbb N$ is not first countable at any point (for example, see this answer).
For the sake of contradiction, assume $\beta\Bbb N\setminus\{p\}$ is paracompact. Since $\mathcal{U}=\{\beta \mathbb{N} \setminus \overline{U} : U \text { is a neighborhood of } p \}$ is an open cover of $\beta\Bbb N\setminus\{p\}$ ($\because\beta\Bbb N$ is Hausdorff), it has a locally finite open refinement $\mathcal U'$. After you've proven (1) and (2), you know $\mathcal U'$ is countable and $\displaystyle\bigcup\left\{\text{cl}_{\beta \Bbb N\setminus \{p\}}(U'):U'\in \mathcal U'\right\}=\beta \Bbb N\setminus\{p\}$ (Here $\overline{U'}$ and $\text{cl}_{\beta \Bbb N\setminus \{p\}} (U')$ denotes the closure of $U'$ in $\beta\Bbb N$ and in $\beta\Bbb N\setminus\{p\}$, resp.).
Then $\displaystyle\bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}\color{red}{\supseteq}\bigcup\left\{\text{cl}_{\beta \Bbb N\setminus \{p\}}(U'):U'\in \mathcal U'\right\}=\beta\Bbb N\setminus \{p\}$. We note that $p\notin\overline{U'}$ for every $U'\in\mathcal U'$ as we can find some neighborhood $U$ of $p$ such that $U'\subseteq \beta \Bbb N \setminus \overline{U}\Rightarrow U'\cap \overline{U}=\emptyset\Rightarrow U\cap \overline{U'}=\emptyset$. This implies $\displaystyle\bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}\color{red}{=}\beta \Bbb N\setminus \{p\}$. Taking complements of both sides, we get $\displaystyle\{p\}= \beta \Bbb N\setminus \bigcup\left\{\overline{U'}:U'\in \mathcal U'\right\}=\bigcap \left\{\left(\overline{U'}\right)^c:U'\in \mathcal U'\right\}$. Being a countable intersection of open set, $\{p\}$ has a countable local base (by the statement enclosed in square brackets; a complete proof can be found here), contrary to the fact that $\beta\Bbb N\setminus \Bbb N$ is not first countable at any point.
YuiTo Cheng
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Should be: contrary to the fact that $\beta \mathbb{N}\setminus \mathbb{N}$ is not first countable at any point. – Henno Brandsma May 26 '19 at 22:18
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The countability of $\mathcal{U}'$ is quite trivial: it's a point finite family and every member contains a member of $\mathbb{N}$. – Henno Brandsma May 26 '19 at 22:19