Are the path components Borel sets?

Question

It is know that the connected components of a topological space are closed, and that the path connected components (pcc) need not to be closed, what I was wondering is if the pcc are always Borel sets, i.e. the pcc all belong to the $\sigma$-algebra generates by the topology.

Answer 1

No. For instance, let $I=\mathbb{R}\setminus\mathbb{Q}$ and let $A\subset I$ be any non-Borel set. Let $B\subset\mathbb{R}^2$ be union of the line segments connecting $(0,1)$ to each point of $A\times\{0\}$ and the line segments connecting $(0,-1)$ to each point of $(I\setminus A)\times\{0\}$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $I\times\{0\}$ would be Borel in $I\times\{0\}$. But that intersection is just $A\times\{0\}$, which is not Borel.