Similarly is the complement of any countable set in $\mathbb R^3$ simply connected?


Reading around I found plenty of articles discussing the path connectedness $\mathbb R^2 \setminus \mathbb Q^2$ and even an approach using cofiltered limits to approach that problem, but I am not read enough in that literature to see if this could be applied here.



Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
Yes, the complement of any countable set in $\mathbb{R}^3$ is simply connected, by the Baire category theorem.


Say your set is $X = \{x\_1, x\_2, ... \}$, and let $y$ be any point in $\mathbb{R}^3 \setminus X.$


Let $f:S^1 \rightarrow \mathbb{R}^3 \setminus X$, and consider the space of homotopies $h:S^1 \times [0,1] \rightarrow \mathbb{R}^3$, where $h(x, 0) = f(x)$ and $h(x, 1) = y$. With the natural topology the space of homotopies is a Baire space, and for each $n$, the set of homotopies that avoid the points ${x\_1, ... , x\_n}$ is open and dense. So the set of homotopies that miss all of $X$ is nonempty.


