# On $d$-completions of $T_0$-spaces

Author: Michael N. Shmatkov (Sobolev Institute of Mathematics, Novosibirsk State University)

### Abstract

A notion of $d$-space was initially introduced in [1]. Also the existence of a $d$-completion of an arbitrary $T_0$-space was proved.

Recall some necessary notions.

Let $X$ be topological space whose topology $T_X$ has $T_0$-property (i.e. $\forall \xi_0,\xi_1\in X(\xi_0\not=\xi_1\to \exists V\in T_X(\xi_0\in V\wedge\xi_1\notin V\vee \xi_0\notin V\wedge\xi_1\in V)$). Let us introduce the following relation on the elements of $X$ (called \textit{specialization relation}): $\xi_0\les\xi_1 \iff \forall U\in T_X(\xi_0\in U\to \xi_1\in U)$. A family $A$ of elements of $X$ is called \textit{directed}, if $\forall a_0,a_1\in A\exists a\in A(a_0\les a\wedge a_1\les a)$.

Topological space $X$ is called \textit{$d$-space}, if each directed family $\{\xi_i\mid i\in I\}$ of elements of $X$ has the exact upper bound $\xi$ that also have to be accumulation point of this family.

Yu.L. Ershov in [1] proved, that for any $T_0$-space $X$ and any its extension $Y$ being $d$-space the least $d$-space $X_0\subseteq Y$ containing $X$ is defined uniquely up to homeomorphism over $X$. This yields, in particular, the result of O.\,Wyler \cite{W:DCP} of existence of $d$-completions.

In order to prove the letter claim in [1] the special procedure of extension of an arbitrary $T_0$-space $X_0$ is described (denote this extension by $\hd{1}{X_0}$ and call it \textit{$d$-extension} of $X_0$). Then the sequence $X_\be=\hd{\be}{X_0}$ of topological spaces is constructed, where each space is $d$-extension of a previous one. This sequence occurs to stabilize, i.e. there exists an ordinal $\al$ such that $X_\al=X_{\al+1}$. For this $\al$ the space $X_\al$ is exactly $d$-completion of $X_0$.

Nevertheless, the question on possible length of the sequence mentioned was staying open.

The main aim of the present work is to prove that there are $T_0$-spaces with arbitrary long sequences of $d$-extensions. The following theorem is proved.

\textbf{Theorem.}\ {\itshape For any ordinal $\al$ there exists a topological space $X_\al$ such that $\hd{\al}{X_\al}$ is $d$-space, but the space $\hd{\be}{X_\al}$ is not $d$-space for every $\be<\al$. In other words, the (transfinite) sequence $X_\al=\hd{0}{X_\al}\subset\hd{1}{X_\al}\subset\dots\subset\hd{\be}{X_\al}\subset \dots\subset\hd{\al}{X_\al}$ of $d$-extensions of the space $X_\al$ to $d$-completion $\hd{\al}{X_\al}$ has the length $\al+1$. }

So, the process of constructing of $d$-completion of an arbitrary $T_0$-space may occur to be arbitrary long.

[1] O. Wyler, \textit{Dedekind complete posets and Scott topologies}, in: Continuous Lattices, Proc. Bremen, 1979. ed. B. Banaschewski, R.-E. Hoffman, Lect. Notes Math, 871 (1981), 384--389.

[2] Yu. L. Ershov, \textit{On $d$-spaces}, Theoretical computer science, 224 (1999), 59--72.