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Let $X$ be obtained by removing finitely many lines from $\mathbb{R}^3$. How do I prove $X$ is connected?

I think the way to go about this is to show $X$ is path-connected but I'm not quite sure.

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  • $\begingroup$ Good idea. Path-connected $\Rightarrow$ Connected. So, pick any two points in $X$ and try to find a path between them. $\endgroup$
    – Teddy38
    Oct 10 '17 at 6:42
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You can remove fewer than $2^{\aleph_0}$ lines from $\mathbb R^3$ and the remainder $X$ will still be path-connected.

Consider any two points $P,Q\in X.$ Consider the sphere having $PQ$ as a diameter. Each of the missing lines meets the sphere in at most two points; thus there are fewer than $2^{\aleph_0}$ missing points on the sphere. The $2^{\aleph_0}$ great circles through $P$ and $Q$ have no point in common besides $P$ and $Q,$ so we can choose one of them which has no missing points.

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  • $\begingroup$ nice ${}{}{}{}{}{}{}{}$ $\endgroup$
    – Asinomas
    Oct 10 '17 at 6:46
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Let $F$ be the union of all lines (we assume the number of lines is at most countable). Let $E=\mathbb R^3 \setminus F$ and let $a$ and $b$ be two points in $E$.

Pick two balls around $a$ and $b$ of the same size $\epsilon$ contained in $E$ ($E$ is open).

Consider the plane $P$ that is orthogonal to $a-b$ and passes through $a$. Consider the projection on $P$.

Notice that for every point in $P$ at distance at most $\epsilon$ from $a$ there is a straight line segment from the sphere of $a$ to the sphere of $b$ that is parallel to $a-b$. This line segment can be made into a path from $a$ to $b$ ( getting from $a$ or $b$ to any point on the respective sphere is obviously possible).

So now all we have to do is show that the projection of $F$ does not cover contain all of the points in $P$ of radius less than $\epsilon$ (these form a ball of dimension $2$). This last step can be done in a lot of ways (using measure theory, or noticing each straigh line contains at most two points at distance exactly $\epsilon$ from $a$).

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