# Number of pos (just an attempt)

## Introduction

After this attempt I’ll try to advance the problem of counting pos seriously, in a new post: pos(n).

Positive integers are called natural numbers. When also integer   $0$   is included then they are called non-negative integers.

I’ll be concerned with the finite sets only. A partial order is abbreviated as po. Pos are in a bijective (1-1) correspondence with the $T_0$-topologies. We want to count the number of pos in arbitrarily fixed (finite) set, or–what is essentially the same–the $T_0$-topologies (in the same set).

## Covers

Let   $C(m\ n)$   be the number of coverings   $\gamma : V\rightarrow 2^W$   of an $n$-element set   $W$   by   $m$   sets   $\gamma(v)$,   i.e.   $\cup\gamma\ :=\ \cup_{v\in V}\ \gamma(v)\ =\ W$   (it’s a nice mood improver, i.e. a simple exercise that is easier than it looks). The sets   $\gamma(v)$   can be empty.

The following formula:

$C(m\ n)\ =\ (2^m-1)^n$

holds for arbitrary non-negative   $m\ n$;   here the exponentiation conventions honors equality   $C(0\ 0) = 1$.

## Pos or $T_0$-topologies. Signature.

A sequences of non-negative integers   $n_0\ n_1\ \ldots$   is called a signatures   $\Leftarrow:\Rightarrow$   if there exists a zero element   $n_k=0$;   and   $\forall_{i\ j}\ \ \left(\ \left(\left(i.

Next, we associate a unique signature with every finite poset. If two such sets are isomorphic then they will have the same signature.

Consider a finite poset (partially ordered set)   $\mathbf X := (X\ \preceq)$   (thus   $\preceq\subseteq X\times X$).   Then let   $M(\mathbf X)$   be the set of the minimal elements of   $\mathbf X$;   and   $\mathbf N(\mathbf X) := (N(\mathbf X)\ \preceq|N(\mathbf X)$   where   $N(\mathbf X) := X\setminus M(\mathbf X)$   and   $\preceq | N(\mathbf X)\ :=\,\ \preceq\ \cap\ \left(N\left(\mathbf X\right) \times N\left(\mathbf X\right)\right)$.

Let’s continue these definitions:   $\mathbf X_n := (X_n\ \preceq_n)$,   where:

• $X_0 := X\qquad \preceq_0\ :=\ \preceq \qquad M_0 := M(\mathbf X)$;
• $X_{n+1} := N(\mathbf X_n)\qquad \preceq_{n+1}\ :=\ \preceq | X_{n+1}\qquad M_{n+1} := M(X_{n+1})$.