3-Point Compactification of $\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}$

Question

Does there exist a compactification $X$ of $\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}$ with the following properties?

• $X$ is compact

• $X$ is Hausdorff

• $\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}$ is dense in $X$

• $X \setminus (\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}) = \{\infty_1, \infty_2, \infty_3\}$

• If $A \subseteq \mathbb{R}_{\geq 0}$ is compact and $B \subseteq \mathbb{R}_{\geq 0}$ is closed but not compact, then $\overline{A \times B} = (A \times B) \cup \{\infty_1\}$

• If $A \subseteq \mathbb{R}_{\geq 0}$ is closed but not compact and $B \subseteq \mathbb{R}_{\geq 0}$ is compact, then $\overline{A \times B} = (A \times B) \cup \{\infty_2\}$

• If $A \subseteq \mathbb{R}_{\geq 0}$ is closed but not compact and $B \subseteq \mathbb{R}_{\geq 0}$ is closed but not compact, then $\overline{A \times B} = (A \times B) \cup \{\infty_1, \infty_2, \infty_3\}$

I was thinking of taking $X$ to be quotient of $[0,1] \times [0,1]$ where we identify $[0,1) \times \{1\}$ to the single point $\infty_1$, $\{1\} \times [0,1)$ to the single point $\infty_2$ and $(1, 1)$ to $\infty_3$ and then embed $\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}$ as $[0,1) \times [0,1)$, 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 is no such Hausdorff compactification. To see this, note that the complement of a compact subset of $\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}$ has precisely one non-relatively-compact connected component. As a result, if there are finitely many "points at infinity" (points added in the compactification), there must only be $1$.

To see why this is so, observe that if $\infty_1,\dots,\infty_n$ are the added points, then by Hausdorff we can find neighborhoods $U_k\ni\infty_k$ for $k=1,\dots,n$, where $U_k$ are disjoint. Then $(\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0})\backslash \bigcup_{k=1}^n U_k$ is compact, with a complement having at least $n$ non-relatively-compact components. We therefore have $n=1$.