Maybe not research level.


Let $Z\cong \mathbb{R}$ be the $z$-axis of $\mathbb{R}^3$. Clearly $\pi\_1(\mathbb{R}^3-Z)\cong \mathbb{Z}$. Now if $F\subset Z$ is a closed non-empty subset, then one easily sees that $\pi\_1(\mathbb{R}^3-(Z-F))=0$.


What is if $F$ is not closed (for example $F=\mathbb{Q}$)? 



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> Is for any nonempty subset $F\subseteq \mathbb{R}$ the space $\mathbb{R}^3-(Z-F)$ simply connected?
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Fundamental group of $\mathbb{R}^3-F$ where $F\subseteq \mathbb{R}\times \{0\} \times \{0\}$
Given a continuous pointed map $\gamma: (S^1,1) \to (\mathbb{R}^3 \setminus (Z-F), x)$, compactness of $S^1$ implies the intersection $\gamma(S^1) \cap Z$ is closed in $Z$. Thus, $\gamma$ represents an element of $\pi\_1(\mathbb{R}^3 \setminus (Z-F'), x)$ for some closed nonempty subset $F' \subset F \subset Z$, and as you mentioned, this group is trivial. In particular, there is a contracting homotopy that lies in this smaller space.


