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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 $F=\mathbb{Q}$)?

Is for any nonempty subset $F\subseteq \mathbb{R}$ the space $\mathbb{R}^3-(Z-F)$ simply connected?

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    $\begingroup$ I think so. Look at a loop representing the generator of $\pi_1(\R^3-Z)$. As soon as you remove at least one point from Z, then this loop will bound some embedded disk. $\endgroup$
    – ThiKu
    Commented Jul 30, 2014 at 14:17
  • $\begingroup$ The same should hold true for any loop in $R^3-Z$, just that in this case the disks won't be embedded anymore. $\endgroup$
    – ThiKu
    Commented Jul 30, 2014 at 14:18
  • $\begingroup$ I guess you can freely homotop an element in $[S^1,\mathbb R^3 -(Z-F)]$ parallel to $Z$ into an orthogonal plane of $Z$ based at a point of $F$. You can easily contract it in this plane to a point without problems. Hence the statement. $\endgroup$
    – Daniel Valenzuela
    Commented Jul 30, 2014 at 14:42

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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.

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