# An example of a regular but not linear-based topological space

Call a topological space $\langle X,\mathscr{O}\rangle$ regular iff it is both $T_0$ and $T_3$: for every point $x\notin A$, where $A$ is a closed subsets of $X$, there are open and disjoint sets $V$ and $U$ such that $x\in V$ and $A\subseteq U$.

A $\langle X,\mathscr{O}\rangle$ space is linear-based iff in every its point there is a local basis which is linearly ordered by $\subseteq$ relation.

Could you give me an example of a regular space which is not linear-based?

• Please provide the information about where in mathematics do arrive mentioned above by you linear-based topological spaces, why are them interesting? – Evgeny Kuznetsov Apr 26 '18 at 9:06
• Clearly all metric spaces are linear based. Thus not being linear based is an obstruction to metrizability. The example I gave is an example that the second countability assumption in Urysohns metrization theorem is essential. – HenrikRüping Apr 26 '18 at 10:10
• I am not aware of it. – HenrikRüping Apr 26 '18 at 10:27
• In my case the interest in such spaces stems from point-free topology, especially the part of it dealing with the so-called Boolean Contact Algebras. Point-free counterparts of linear bases (i.e. certain chains in BCAs) are good approximations of points and linear-based spaces are used for representation theorems for BCAs (at least in my studies of BCAs, there are different approaches too). As far as I know there is no book nor paper devoted to class of linear-based spaces. But check the so-called well-based spaces and radial spaces which are related to linear-based. – Mad Hatter Apr 26 '18 at 10:44
• @EvgenyKuznetsov This freely available paper by Dimiter Vakarelov may be a good starting point: link.springer.com/chapter/10.1007%2F978-0-387-69245-6_6 – Mad Hatter Apr 26 '18 at 14:11

Let us have a look at the latter. Let $X:= \prod_I \{0,1\}$, let $0\in X$ be the sequence which is zero everywhere and let $S_i\subset X$ be the subset consisting of all sequences whose $i$-th coordinate is zero.
By definition of the product topology, any open neighborhood of $0$ can be contained in only finitely many of the $S_i$'s.
Now suppose you have a linearly ordered basis $B$. Assign to each basis element $U$ the finite subset $f(U) = \{i\in I| S_i\supset U\}$.
Since $B$ is a basis, we know that every $i$ appears in some $f(U)$. It follows that $I$ is an union of nested, finite sets and hence countable. Contradiction!
• Henrik, the third paragraph should say "any basic open neighborhood of 0", for if you take for example $T_n$ to be the set of all sequences which have 0's at first $n$ terms, than $\bigcup_{n\in\omega}T_n$ is cotained in every $S_i$, for $i\in\omega$. Am I right? – Mad Hatter Nov 20 '18 at 22:03
• @MadHatter: The $T_i$'s form a decreasing sequence of open sets, Thus their union is just the first of the $T_i$'s, if we replaced union by intersection, we dont get something open. It also follows that if any basic open neighborhood of 0 is contained in only finitely many of the $S_i$'s, then any open nbhd of 0 can be contained in finitely many of the $S_i$'s (just pick a basic open nbh contained in it). – HenrikRüping Nov 21 '18 at 3:20