Given a point $x$ in a topological space $X$. I was wondering, whether one can always find a [local basis](http://en.wikipedia.org/wiki/Neighbourhood_system) at $x$, which is [totally ordered](http://en.wikipedia.org/wiki/Totally_ordered_set) (a chain) under inclusion. For example this is true for spaces, which have countable local basis: If $\{U\_i|i\in\mathbb{N}\}$ is a local basis at $X$, then $\{\bigcap\_{i=1}^nU\_i|n\in \mathbb{N}\}$ is totally ordered.


So my question is: Assume, there is a totally ordered local basis at $x\in X$. Does this already imply, that there is a countable local basis at $x\in X$?


The rest of this text contains the motivating example. So if you are motivated enough, you can stop reading.


For example the space $\prod\_{i\in I} \{0;1\}$ does not have a totally ordered local basis at $\{0\}^I$ for uncountable $I$. Assume $P$ is such a basis.
Let for any open neighborhood $U$ of $\{0\}^I$


$N\_U:=\{i\in I| 1\_i\notin U\}$


where $1\_i$ denotes the element of $\prod\_{i\in I} \{0;1\}$ with zeros everywhere except for the $i$-th entry. Note that $N\_U$ is always finite by definition of the product topology and that for $U\subset U'$ we get $N\_U\supset N\_{U'}$ Further note that $I=\bigcup\_{U\in P}N\_U$. 


Now construct an injection $f:\mathbb{N}\rightarrow I$ in the following way
Choose any $U\_0\in P$ and enumerate all elements of $N\_0$ in your favourite way. Then choose a $U\_1\in P$ with $N\_{U\_0}\subsetneq N\_{U\_1}$. Then enumerate all elements in $N\_{U\_1}\setminus N\_{U\_0}$. Continue this way to get an injection $\mathbb{N}\rightarrow I$.


As $I$ is uncountable, this map can't be surjective. Choose $i\in I\setminus Im(f)$ and any basis element $U\in P$ with $i\in N\_U$. So for any $n\in \mathbb{N}$, we get 


$i\notin N\_{U\_n} \Rightarrow U\_n\nsubseteq U\Rightarrow U\subseteq U\_n\Rightarrow N\_U \supset N\_{U\_n}$.


The second last implication holds, as the base is totally ordered. This holds for any $n$, so $N\_U \supset \bigcup\_nN\_{U\_n}$. But the left side is finite and the right side is (by constuction) infinite. Contradiction! So the cannot be a totally ordered local base.



chains and countability
The space $\omega\_1+1$ under the order topology, where $\omega\_1$ refers to the first uncountable ordinal, has a
linearly ordered local basis at the point $\omega\_1$ (and indeed at every point),
consisting of the intervals $(\alpha,\infty)$, but there is
no countable local basis at that point, because every
countable set of ordinals below $\omega\_1$ is bounded below
$\omega\_1$.


An essentially similar example arises from the long line, by placing a point at the top. This new point will have a linearly ordered local basis consisting of intervals, but no countable local basis. 


