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?



An example of a regular but not linear-based topological space
An uncountable product of copies of the unit interval or of the discrete two element space. Indeed none of the points in this space has a linearly ordered local basis.


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!


