Divisor topology

The sets $S\_n\; =\; \backslash \{x\; \backslash in\; X\; :\; x\backslash mathop|n\; \backslash \}$ for $n\; =\; 2,3,...$ form a basis for the divisor topologySteen & Seebach, example 57, p. 79-80 on $X$, where the notation $x\backslash mathop|n$ means $x$ is a divisor of $n$.

The open sets in this topology are the lower sets for the partial order defined by $x\backslash leq\; y$ if $x\backslash mathop|y$. The closed sets are the upper sets for this partial order.

All the properties below are proved in or follow directly from the definitions.

• The closure of a point $x\backslash in\; X$ is the set of all multiples of $x$.
• Given a point $x\backslash in\; X$, there is a smallest neighborhood of $x$, namely the basic open set $S\_x$ of divisors of $x$. So the divisor topology is an Alexandrov topology.
• $X$ is a T₀ space. Indeed, given two points $x$ and $y$ with
• $X$ is a not a T₁ space, as no point is closed. Consequently, $X$ is not Hausdorff.
• The isolated points of $X$ are the prime numbers.
• The set of prime numbers is dense in $X$. In fact, every dense open set must include every prime, and therefore $X$ is a Baire space.
• $X$ is second-countable.
• $X$ is ultraconnected, since the closures of the singletons $\backslash \{x\backslash \}$ and $\backslash \{y\backslash \}$ contain the product $xy$ as a common element.
• Hence $X$ is a normal space. But $X$ is not completely normal. For example, the singletons $\backslash \{6\backslash \}$ and $\backslash \{4\backslash \}$ are separated sets (6 is not a multiple of 4 and 4 is not a multiple of 6), but have no disjoint open neighborhoods, as their smallest respective open neighborhoods meet non-trivially in $S\_6\backslash cap\; S\_4=S\_2$.
• $X$ is not a regular space, as a basic neighborhood $S\_x$ is finite, but the closure of a point is infinite.
• $X$ is connected, locally connected, path connected and locally path connected.
• $X$ is a scattered space, as each nonempty subset has a first element, which is an isolated element of the set.
• The compact subsets of $X$ are the finite subsets, since any set $A\backslash subseteq\; X$ is covered by the collection of all basic open sets $S\_n$, which are each finite, and if $A$ is covered by only finitely many of them, it must itself be finite. In particular, $X$ is not compact.
• $X$ is locally compact in the sense that each point has a compact neighborhood ($S\_x$ is finite). But points don't have closed compact neighborhoods ($X$ is not locally relatively compact.)

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• {{Citation | last1=Steen | first1=Lynn Arthur | author1-link=Lynn Arthur Steen | last2=Seebach | first2=J. Arthur Jr. | author2-link=J. Arthur Seebach, Jr. | title=Counterexamples in Topology | orig-date=1978 | publisher=Springer-Verlag | location=Berlin, New York | edition=Dover Publications reprint of 1978 | isbn=978-0-486-68735-3 |mr=507446 | year=1995}}

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Category:Topological spaces