Given two infinite sets $A$, $B$ of natural numbers, write $A\preceq B$ if $B\setminus A$ is a finite set. Define the equivalence relation $A\sim B$ if $A\preceq B$ and $B\preceq A$, and let $\partial\mathbb{N}$ be the set of equivalence classes of infinite sets under this equivalence relation. Write $[A]$ for the equivalence class of $A$.

Now define a topology on the disjoint union $\overline{\mathbb{N}}=\mathbb{N}\cup\partial\mathbb{N}$ as follows: A set $U\subseteq\overline{\mathbb{N}}$ is open if and only if, for every $[A]\in U\cap\partial\mathbb{N}$, $[B]\in U$ whenever $B\prec A$, and moreover $A'\subset U$ for some $A'\sim A$.

Is this a known topology? Does it have a name?

It is not hard to see that $\overline{\mathbb{N}}$ is compact (hence the question title): For any neighbourhood of $[\mathbb{N}]$ is the entire space minus a finite subset of $\mathbb{N}$. In particular, it is very much a non-Hausdorff space. Also, $\overline{\mathbb{N}}$ contains $\mathbb{N}$ as an open subset with the discrete topology on the latter. The subspace $\partial\mathbb{N}$ is indeed the boundary of $\mathbb{N}$ in this topology (hence my chosen notation), and it is an Alexandrov space.

I came up with this topology while thinking about sequences, subsequences, and their limits. It seems rather natural, so I don't think I am the first one to ever think of it.

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    $\begingroup$ Note that is not an Alexandrov space: $\bigcap_{n=0}^\infty \{n,n+1,\ldots\}\cup\partial\mathbb{N} = \partial\mathbb{N}$ is not open. $\endgroup$ – François G. Dorais Mar 16 '10 at 21:58
  • $\begingroup$ @François: I made the claim (Alexandrov space) only for the subspace $\partial\mathbb{N}$, not the full space $\overline{\mathbb{N}}$. $\endgroup$ – Harald Hanche-Olsen Mar 16 '10 at 23:04
  • $\begingroup$ Ah, I see. Sorry for the confusion. $\endgroup$ – François G. Dorais Mar 16 '10 at 23:18

Your set $\partial\mathbb{N}$ is also intensely studied in set theory and known as P(ω)/Fin. What you have done is mod out by the ideal of finite sets. People study more general properties P(X)/I, taking the quotient by many other ideals (or by an arbitrary ideal). P(X)/I is a Boolean algebra, and many forcing arguments can be viewed as forcing with this Boolean algebra. The topological properties are very much used in that forcing context, since the generic filters are exactly those containing elements from every ground model dense set. The finite sets become equivalent to the point [emptyset] in this algebra.

Perhaps Lusin was the first to study P(ω)/Fin seriously, and found the phenomenon of Lusin gaps. A gap in P(ω)/Fin is a cut in the order, where the left side increases and the right side decreases, and everything on the left is below everything on the right, with respect to almost-inclusion. Lusin found gaps of various types, including ones with uncountable cofinality.

A particularly interesting case is P(ω1)/I, where I is the ideal of non-stationary sets, and many set theoretic hypotheses, some engaging with large cardinals, interact with the topological properties of that situation.

A few quick examples:

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  • $\begingroup$ Thanks; that is a good start. But considering this space as a boundary of $\mathbb{N}$ as I did, is presumably less common? $\endgroup$ – Harald Hanche-Olsen Mar 16 '10 at 17:46
  • $\begingroup$ Harald, yes, most of the interest in descriptive set theory is focused on the boundary, or P(omega)/Fin. In the Boolean algebra quotient P(omega)/Fin, the finite sets get identified with 0. I'm not sure how much it changes things to put those individual points back in separately. $\endgroup$ – Joel David Hamkins Mar 16 '10 at 18:01

I was about to answer the same thing as Joel, but here is a more topological perspective. I will show that the lattice of open subsets of $\overline{\mathbb{N}}$ is very similar to the lattice of open subsets of the Stone-Čech compactification $\beta\mathbb{N}$. So, from the localic point of view, there is but a small difference between your space $\overline{\mathbb{N}}$ and $\beta\mathbb{N}$.

View $\beta\mathbb{N}$ as the set of ultrafilters on $\mathbb{N}$ and let $\mathbb{N}^*$ be the remainder $\beta\mathbb{N}\setminus\mathbb{N}$. (Prinicipal ultrafilters are identified with the corresponding point of $\mathbb{N}$, so $\mathbb{N}^*$ is the subspace of nonprincipal ultrafilters.) Recall that the clopen subsets of $\beta\mathbb{N}$ are precisely those of the form

$$\langle A \rangle = \{ \mathcal{U} \in \beta\mathbb{N} : A \in \mathcal{U} \}$$

for $A \subseteq \mathbb{N}$. Note that $A \preceq B$ iff $\langle A \rangle \cap\mathbb{N}^* \subseteq \langle B \rangle\cap\mathbb{N}^*$, so the points of $\partial\mathbb{N}$ can be identified with the clopen subsets $[A] = \langle A \rangle \cap \mathbb{N}^*$ of the remainder $\mathbb{N}^*$ (including the empty set). Given an open set $U \subseteq \beta\mathbb{N}$, the set

$$U' = (U \cap \mathbb{N}) \cup \{ [A] : [A] \subseteq U \}$$

is open in $\overline{\mathbb{N}}$. Conversely, given an open set $V \subseteq \overline{\mathbb{N}}$, your conditions ensure that

$$V' = (V \cap \mathbb{N}) \cup \bigcup \{ [A] : [A] \in V \}$$

is open in $\beta\mathbb{N}$. This correspondence is not perfect since $A \cup B \cup [A] \cup [B] = A \cup B \cup [A \cup B]$, but

$$A \cup B \cup \{[C] : C \preceq A \lor C \preceq B\} \quad\mbox{and}\quad A \cup B \cup \{[C] : C \preceq A \cup B\}$$

are not always the same. However, these are the only errors that occur, i.e. the translation is perfect for open subsets of $\overline{\mathbb{N}}$ whose part in $\partial\mathbb{N}$ is upward directed in the ${\preceq}$ ordering.

Although your space $\overline{\mathbb{N}}$ is interesting, this approximate translation suggests most of its applications could be transferred to work over the well studied space $\beta\mathbb{N}$ instead.

Here is yet another perspective which suggests that there may be more to $\overline{\mathbb{N}}$ after all. The soberification of $\partial\mathbb{N}$, which I will denote $\mathrm{Fil}_{\mathbb{N}}$, is the space of all nonprincipal filters on $\mathbb{N}$, with the topology generated by the basic open sets

$$[A] = \{ \mathcal{F} \in \mathrm{Fil}_{\mathbb{N}} : A \in \mathcal{F} \}.$$

The points of $\partial\mathbb{N}$ can be identified with the filters $\mathcal{F}_A = \{ B : A \preceq B \}$. Note that the space $\mathbb{N}^*$ is also a subspace of $\mathrm{Fil}_{\mathbb{N}}$, which explains the connection found above.

For the pointless topology aficcionados, the space $\mathrm{Fil}_{\mathbb{N}}$ is obtained by imposing the trivial Grothendieck topology on the preorder $(\mathcal{P}\mathbb{N},{\preceq})$ viewed as a category (or, equivalently, the quotient partial order $\mathcal{P}\mathbb{N}/\mathrm{fin}$ as suggested by Joel). The subspace $\mathbb{N}^*$ is similarly obtained by imposing the finite cover (aka coherent) Grothendieck topology on $(\mathcal{P}\mathbb{N},{\preceq})$.

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  • $\begingroup$ Very nice, François. But couldn't one also turn things around, and say that one can understand the clopen subsets of the Stone space by looking at Harald's space? After all, it might be more concrete to consider subsets of N with finite modifications than to think about (uncountable) sets of ultrafilters. $\endgroup$ – Joel David Hamkins Mar 16 '10 at 20:30
  • $\begingroup$ Thanks. There is some food for thought here. However, you lost me at the end where you write “… if you further required open sets of $\partial\mathbb{N}$ to be closed under finite unions”. Are you saying that my topology is in fact not a topology? $\endgroup$ – Harald Hanche-Olsen Mar 16 '10 at 20:53
  • $\begingroup$ @Harald: I didn't write that very well, I should have said joins of the ${\preceq}$ lattice be clear. I'll fix it in a second. $\endgroup$ – François G. Dorais Mar 16 '10 at 21:22
  • $\begingroup$ @Joel: That's the point of locales, which completely ignore points. In terms of locales, $\beta\mathbb{N}$ and $\overline{\mathbb{N}}$ are very similar and just as easy to understand, but the former is much better studied. $\endgroup$ – François G. Dorais Mar 16 '10 at 21:43

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