dense subspace of $\beta \Bbb N \times \beta \Bbb N$ Let $\beta \Bbb N$ be a Čech-Stone compactification of the discrete space $\Bbb N$ and fix a point $p\in \beta \Bbb N\setminus \Bbb N$. Put $X=\Bbb N\cup \{p\}$ and $Y=\beta \Bbb N \setminus \{p\}$. I have two questions,


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*$Z=(X\times Y)\cup \{(p,p)\}$ is dense in $\beta\Bbb N\times \beta\Bbb N$ ?

*$\{(p,p)\}$ is closed in $Z$?
 A: All are yes. 
Q1, $\mathbb N$ is dense in $\beta\mathbb N$, for $\beta\mathbb N$ is compactification of $\mathbb N$, and hence $\mathbb N \times \mathbb N$ is also dense in $\beta\mathbb N \times \beta\mathbb N$.
Q2, $\beta\mathbb N \times \beta\mathbb N$ is normal because it is Hausdorff compact, and hence $Z$ is at least Tychonoff (which we could see more). Note that the property Tychonoff is hereditary. And hence, $Z$ is $T_1$. Note that every point is closed in a $T_1$ space. So in the discrete space, every point is open and closed.
A: $\newcommand{\cl}{\operatorname{cl}}$Yes, and yes. $\Bbb N\times\Bbb N$ is dense in $\beta\Bbb N\times\beta\Bbb N$, so $Z$ certainly is, and $Z$ is $T_1$, so all singletons are closed.
Answering two questions from the comments: Let $U=\{\langle n,n\rangle:n\in\Bbb N\}$.


*

*Why is $U$ clopen in $X\times Y$? $\Bbb N$ is a dense set of isolated points in $\beta\Bbb N$, so every subset of $\Bbb N\times\Bbb N$ is open in $\beta\Bbb N\times\beta\Bbb N$ and hence in $X\times Y$. This also shows that $U$ has no limit points in $\Bbb N\times\Bbb N$. For each $n\in\Bbb N$ and $q\in\Bbb N\setminus(\Bbb N\cup\{p\})$, $(X\setminus\{n\})\times\{n\}$ is an open nbhd of $\langle p,n\rangle$ disjoint from $U$, and $\{n\}\times(Y\setminus\{n\})$ is an open nbhd of $\langle n,q\rangle$ disjoint from $U$. And since $p\ne q$, there is an $A\subseteq\Bbb N$ such that $A\in p\setminus q$; then $\Bbb N\setminus A\in q$, and $(\{p\}\cup A)\times\cl_Y(\Bbb N\setminus A)$ is an open nbhd of $\langle p,q\rangle$ disjoint from $U$. Every point of $(X\times Y)\setminus(\Bbb N\times\Bbb N)$ is of one of those three types, so $U$ has no limit points in $X\times Y$ and is therefore closed as well as open.

*Why is $\langle p,p\rangle\in\cl U\cap\cl\big((X\times Y)\setminus U\big)$, where the closure is taken in $\beta\Bbb N\times\beta\Bbb N$ (and hence in $X\times\beta\Bbb N$)? The point $\langle p,p\rangle$ has a local base consisting of all sets of the form $\cl_{\beta\Bbb N}A\times\cl_{\beta\Bbb N}B$ with $A,B\in p$. If $A,B\in p$, then $A\cap B\ne\varnothing$, so let $n\in A\cap B$; then $$\langle n,n\rangle\in U\cap(\cl_{\beta\Bbb N}A\times\cl_{\beta\Bbb N}B)\;,$$ so $\langle p,p\rangle\in\cl U$. Moreover, $A\cap B$ is infinite, so we can choose distinct $m,n\in A\cap B$ and observe that $$\langle m,n\rangle\in(\cl_{\beta\Bbb N}A\times\cl_{\beta\Bbb N}B)\cap\big((X\times Y)\setminus U\big)\;,$$ so that $\langle p,p\rangle\in\cl\big(X\times Y)\setminus U\big)$ as well.
