Given a point $x$ in a topological space $X$. I was wondering, whether one can always find a local basis at $x$, which is totally ordered (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.
