Appert topology

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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[edit]

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:

 \mathrm{N}(n,S) = \#\{ m \in S : m \le 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]

\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

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

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

Related topologies[edit]

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[edit]

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

\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 \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]

Notes[edit]

  1. ^ a b c d e Steen & Seebach 1995, pp. 117–118
  2. ^ Steen & Seebach 1995, p. 3

References[edit]