# Removing countable no. of lines from $R^3$

What will happen to connectedness after removing countable no. of lines from $$R^3$$. Will it be connected or will it be disconnected?

I had read that removing finite number of points from $$R^2$$ is still connected. But I am unable to get any proof for this fact.

How do I prove/disprove it?

Edit

Choose two points in $$R^3$$, say $$a, b \in \mathbb{R} ^3$$ Then there are uncountably many spherical surfaces passing through those two points. Now any straight line and a sphere intersect in atmost two points. Hence there exists a sphere containing the two points but not any lines which has been deleted. So join those two points by a continous map, which exists as spheres are connected. Hence it is Path Connected, and hence connected. Hence proved.

• Someone please check it again, I have added my solution Dec 10 '19 at 12:14
• The statement beginning with "Hence there exists a sphere" is false. The point is that a sphere minus countably many points is path connected (since $\mathbb{R}^2$ minus countably many points is path connected, and $\mathbb{R}^2$ is just the sphere minus a point, or see here), hence there's a path from $a$ to $b$ in that holed sphere, hence in $\mathbb{R}^3$ minus some lines. Dec 10 '19 at 12:26

Proposition 1. If $$X$$ is a countable set of points in the plane $$M$$, then $$M\setminus X$$ is pathwise connected.

Proof. Consider two (distinct) points $$P_1,P_2\in M\setminus X$$. Choose a line $$L_1$$ in $$M$$ which goes through $$P_1$$ and does not go through any point of $$X$$. Choose a line $$L_2$$ in $$M$$ which goes through $$P_2$$ and does not go through any point of $$X$$ and is not parallel to $$L_1$$. The lines $$L_1$$ and $$L_2$$ meet in a point $$Q$$. The broken line $$P_1QP_2$$ provides a continuous path from $$P_1$$ to $$P_2$$ in $$M\setminus X$$.

Proposition 2. If $$X$$ is a subset of $$\mathbb R^3$$ which is the union of countably many lines $$L_1,L_2,\dots$$, then $$\mathbb R^3\setminus X$$ is pathwise connected.

Proof. Consider two (distinct) points $$P_1,P_2\in\mathbb R^3\setminus X$$. Choose a plane $$M$$ in $$\mathbb R^3$$ which contains the points $$P_1$$ and $$P_2$$ and does not contain any of the lines $$L_1,L_2,\dots$$. Thus $$M$$ meets each of those lines in at most one point, so the set $$M\cap X$$. is countable. Now $$M\setminus X$$ is a subset of $$\mathbb R^3\setminus X$$ containing $$P_1$$ and $$P_2$$, and is pathwise connected by Proposition 1.

• Is the set $\mathbb {R^2-Q\times Q}$ connected? Aug 11 '20 at 5:53
• @KishalaySarkar Yes, of course it is. Note that the set $S=(\mathbb R\times\mathbb R)\setminus(\mathbb Q\times\mathbb Q)$ is the union of all the vertical lines $x=c$ and all the horizontal lines $y=c$ where $c$ is an irrational number. Note that any two points of $S$ belong to a subset of $S$ which is either the union of two horizontal lines and one vertical line, or else one horizontal and two vertical lines. For instance, the points $(\sqrt2,0),(\sqrt3,1)\in S$ belong to the connected set $$\{(x,y):x=\sqrt2\}\cup\{(x,y):x=\sqrt3\}\cup\{(x,y):y=\sqrt5\}\subset S.$$
– bof
Aug 11 '20 at 6:50

Hence there exists a sphere containing the two points but not any lines which has been deleted.

There might not be such sphere. Let $$L$$ be a line segment connecting $$a$$ and $$b$$. Now let $$c$$ be any point on $$L$$ (other than $$a$$ and $$b$$) and let $$T$$ be any line passing through $$c$$ but not through $$a$$ and $$b$$. Note that $$T$$ intersects any sphere containing $$a$$ and $$b$$. The reason is because $$c$$ is an interior point of any such sphere.

So as you can see, simple "cardinality" argument is not enough. You have to consider geometry as well.

But you are on the right track. You don't need a sphere that misses all lines. In fact you can take any sphere that passes through $$a$$ and $$b$$, say $$S$$. Now since we have countable number of lines and each line intersects $$S$$ at at most $$2$$ points then we end up with $$S\backslash X$$ where $$X$$ is countable subset. And such subset is path connected because if $$X\neq\emptyset$$ then $$S\backslash X$$ is homeomorphic to $$\mathbb{R}^2\backslash(X\backslash\{*\})$$ (via stereographic projection). And as we all know $$\mathbb{R}^2$$ minus countable subset is path connected.