Let $X$ be a locally connected topological space, which is covered by open sets $\{U\_{\alpha},\alpha\in A\}$ and let $C$ be an arc in $X$, i.e. a homeomorphic image of an interval.


Is it always possible to choose open sets $V\_1,...,V\_n$ that cover $C$ and such that


* Each $V\_i$ is contained in some $U\_{\alpha}$,
* $V\_i\bigcap V\_j=\varnothing$, whenever $|i-j|>1$ and
* $V\_i\bigcap V\_j$ is connected, whenever $|i-j|\le 1$?


If not always then what should be assumed about $X$?



Inscribing a "chain" into an open cover
The answer is ``yes'' if $X$ is Hausdorff. 


We can identify the arc $C$ with the unit interval $[0,1]$ and assume that $[0,1]$ is contained in $X$ and is covered by a family $\mathcal U$ of open subsets of $X$. By the compactness of $[0,1]$, there exists an increasing sequence $0=a\_0<a\_1<\dots<a\_n=1$ of real numbers such that for every positive $i\le n$ the interval $[a\_{i-1},a\_{n}]$ is contained in some set $U\_i\in\mathcal U$ of the cover. Using the local connectedness and the Hausdorff property of $X$, chose a sequence $(W\_i)\_{i=0}^n$ of pairwise disjoint open connected sets in $X$ such that $a\_i\in W\_i\subset U\_i\cap U\_{i-1}$ for all $i\in\{0,\dots,n\}$. Here we assume that $U\_{0}=X$.


For every $0<i\le n$ find a compact connected subset $K\_i$ in the open interval $(a\_{i-1},a\_{i})$ such that $[a\_{i-1},a\_i]\subset W\_{i-1}\cup K\_i\cup W\_i$ and observe that the sets $K\_i$, $0<i\le n$, are pairwise disjoint. Using the Hausdorff property of $X$ and the compactness of the pairwise disjoint sets $K\_i$, we can find a sequence $(O\_i)\_{i=1}^n$ of pairwise disjoint open sets of $X$ such that $K\_i\subset O\_i\subset U\_i$ for every $i\le n$. Moreover, since $X$ is locally connected, we can additionally assume that each open neighborhood $O\_i$ of the connected set $K\_i$ is connected. Finally for every positive $i\le n$ consider the open connected neighborhood $V\_i=W\_{i-1}\cup O\_i\cup W\_i$ of the segment $[a\_{i-1},a\_i]$ and observe that $V\_i\subset U\_i\in\mathcal U$. Moreover, for any $0\le i<j\le n$ if $j-i>1$, then the sets $V\_i,V\_j$ are disjoint and if $j-i=1$, then $V\_i\cap V\_j=W\_i$ is conneceted.


On the other hand, some separation axioms for $X$ are necessary: it may happen that $X$ contains a special point $x^\*\in X\setminus C$ contained in any non-empty open set. In this case the sets $V\_i,V\_j$ cannot be made disjoint.


Such pathological space can be constructing as the set $X=[0,1]\cup\{x^\*\}$ where $x^\*\notin[0,1]$, endowed with the topology consisting of the empty sets and sets $W\subset X$ that contain $x^\*$ and have open intersection $W\cap[0,1]$ in the Euclidean topology of $[0,1]$.


