The complement of a union of uncountable many curves in $\mathbb{R^3}$ Let $C=\bigcup\limits_{i\in I}C_i$, where $I=[0,1)$ and $C_i$'s are curves in $\mathbb{R^3}$ such that there is only one intersection point for each pair $(C_i,C_j)$ if $i\not= j$ in $I$. Is it possible $\mathbb{R^3}\setminus C$ has a bounded component? So far I know some trivial examples for $C$, and $\mathbb{R^3}\setminus C$ has no bounded components. For example, all $C_i$'s are lines and they share one common point. I have no idea in general. Any suggestions? Thanks.
 A: Can't you enclose a sphere in the following way?
Consider standing at the North Pole, facing some direction. You can now start walking at any angle from $-90$ to $90$ degrees from this point and after having walked some circle, you'll be back on the North pole again. The further away from $0$ degrees you are, the smaller the circles will become. That is, if you start off at $0$ degrees, you move along the great circle through the south pole, if you start at an angle of $\pm45$ degrees, your southernmost point will be on the equator, etc.
By considering the complete collection of possible starting directions from the North Pole, you'll eventually be visiting each point on the sphere, so $\cup C_i$ is exactly the sphere. Moreover, if this point isn't the North Pole, it lies on the curve belonging to a unique starting direction, hence all distinct curves intersect at a single point.
There's still one problem with this approach: it gives a description of $C_i$ for $i\in(-90,90)$, which can easily be transformed in to a description with index set $(0,1)$, but not to $[0,1)$. In order to solve this, you might have to allow a degenerate curve, consisting of a single point.
