# Appert topology

In general topology, a branch of mathematics, the Appert topology, named for Appert (1934), is an example of a topology on the set Z+ = {1, 2, 3, …} of positive integers.[1] To give Z+ a topology means to say which subsets of Z+ are open in a manner that satisfies certain axioms:[2]

1. The union of open sets is an open set.
2. The finite intersection of open sets is an open set.
3. Z+ and the empty set ∅ are open sets.

In the Appert topology, the open sets are those that do not contain 1, and those that asymptotically contain almost every positive integer.

## Construction

Let S be a subset of Z+, and let N(n,S) denote the number of elements of S which are less than or equal to n:

${\displaystyle \mathrm {N} (n,S)=\#\{m\in S:m\leq n\}.}$

In Appert's topology, a set S is defined to be open if either it does not contain 1 or N(n,S)/n tends towards 1 as n tends towards infinity:[1]

${\displaystyle \lim _{n\to \infty }{\frac {{\text{N}}(n,S)}{n}}=1.}$

The empty set is an open set in this topology because ∅ is a set that does not contain 1, and the whole set Z+ is also open in this topology since

${\displaystyle {\text{N}}\!\left(n,{\mathbf {Z}}^{+}\right)=n\ ,}$

meaning that N(n,S)/n = 1 for all n.

## Related topologies

The Appert topology is closely related to the Fort space topology that arises from giving the set of integers greater than one the discrete topology, and then taking the point 1 as the point at infinity in a one point compactification of the space.[1] The Fort space is a refinement of the Appert topology.

## Properties

The closed subsets of Z+, equipped with the Appert topology, are the subsets S that either contain 1 or for which

${\displaystyle \lim _{n\to \infty }{\frac {\mathrm {N} (n,S)}{n}}=0.}$

As a result, Z+ is a completely normal space (and thus also Hausdorff), for suppose that A and B are disjoint closed sets. If AB did not contain 1, then A and B would also be open and thus completely separated. On the other hand, if A contains 1 then B is open and ${\displaystyle \scriptstyle \lim _{n\to \infty }\mathrm {N} (n,B)/n\,=\,0}$, so that Z+B is an open neighborhood of A disjoint from B.[1]

A subset of Z+ is compact in the Appert topology if and only if it is finite. In particular, Z+ is not locally compact, since there is no compact neighborhood of 1. Moreover, Z+ is not countably compact.[1]