3-Point Compactification of $\mathbb{N} \times \mathbb{N}$

Question

Does there exist a compactification $X$ of $\mathbb{N} \times \mathbb{N}$ with the following properties?

• $X$ is compact

• $X$ is Hausdorff

• $\mathbb{N} \times \mathbb{N}$ is dense in $X$

• $X \setminus (\mathbb{N} \times \mathbb{N}) = \{\infty_1, \infty_2, \infty_3\}$

• If $A \subseteq \mathbb{N}$ is finite and $B \subseteq \mathbb{N}$ is infinite, then $\overline{A \times B} = (A \times B) \cup \{\infty_1\}$

• If $A \subseteq \mathbb{N}$ is infinite and $B \subseteq \mathbb{N}$ is finite, then $\overline{A \times B} = (A \times B) \cup \{\infty_2\}$

• If $A \subseteq \mathbb{N}$ is infinite and $B \subseteq \mathbb{N}$ is infinite, then $\overline{A \times B} = (A \times B) \cup \{\infty_1, \infty_2, \infty_3\}$

I was thinking of using the following construction: Let $N = \{1 - \frac1n \mid n \in \mathbb{N}\}$. Let $N' = N \cup \{1\}$. Take $X$ to be the quotient of $N' \times N'$ where we identify $N \times \{1\}$ to the single point $\infty_1$, $\{1\} \times N$ to the single point $\infty_2$ and $(1, 1)$ to $\infty_3$ and then embed $\mathbb{N} \times \mathbb{N}$ as $N \times N$, but I don't think the resulting space is Hausdorff.

The point of this is that we can informally express a sequence $(a_n)_{n \in \mathbb{N}}$ being Cauchy as "$d(a_n, a_m) \to 0$ for $n, m \to \infty$". If we have a compactification with the above properties, we can formally express this as "$d(a_n, a_m) \to 0$ for $(n, m) \to \infty_3$".

Answer 1

No, there can be no such compactification.

To see why, note first that if $U$ is a neighborhood of $\infty_1$, then there cannot be an infinite set $B$ with $(\{n\}\times B)\cap U=\emptyset$, otherwise taking $A=\{n\}$ we would have $\infty_1\notin \overline{A\times B}$. Thus any neighborhood of $\infty_1$ must include a set of the form $U_f:=\{(m,n)\mid n\geq f(m)\}$ for some function $f\colon \mathbb N\to \mathbb N$.

The same argument with the coordinates interchanged shows that any neighborhood of $\infty_2$ must include a set of the form $V_g:=\{(m,n)\mid m\geq g(n)\}$. Then by the Hausdorff condition there are $f$ and $g$ and some neighborhood $W$ of $\infty_3$ with $U_f$, $V_g$, and $W$ pairwise disjoint.

We now construct sets $A$ and $B$ as follows.

Let $(a_1,b_1)\in U_f$. Then proceed by induction, assuming at each stage $n$ we have picked $a_1,\dots, a_n$, and $b_1,\dots b_n$, so that if $A_n=\{a_1,\dots, a_n\}$, and $B_n=\{b_1,\dots,b_n\}$, then $A_n\times B_n\subseteq U_f\cup V_g$.

Having picked $A_n$ and $B_n$ in such a fashion, we pick some $a_{n+1}$ large enough so that $(a_{n+1},b_k)\in V_g$ for each $k\leq n$. Then we pick some $b_{n+1}$ large enough so that $(a_k,b_{n+1})\in U_f$ for each $k\leq n+1$. Thus $A_{n+1}\times B_{n+1}\subseteq U_f\cup V_g$ as well.

We continue for all $n$, and so obtain sets $A=\bigcup_n A_n$ and $B=\bigcup_n B_n$, such that $A\times B\subseteq U_f\cup V_g$. But then $(A\times B)\cap W=\emptyset$, so since $W$ is a neighborhood of $\infty_3$, $\infty_3\notin \overline{A\times B}$, contradicting the fact that $A$ and $B$ are infinite.