If n > 1 and $B \subset \mathbb R^n$ countable. Then $\mathbb R^n - B$ is connected ( James Dugundji)

Question

We can assume that $0 \in B$ , otherwise we move the origin.. We show that the origin and $ x \in \mathbb R^n - B$ are contained in a connected set lying in $\mathbb R^n - B$. Draw $\overrightarrow{ox}$ and l be any line segment intersecting $\overrightarrow{ox}$ at exactly one point . For each $z \in l$, let $l_z = \overrightarrow{ox} \cup \overrightarrow{zx}$ is a connected set . So Atleast one $l_z$ must lie in $ \mathbb R^n - B$.

How to prove the following points

1. If $z , z' \in l$, then $l_z \cap l_{z'} = \{0,x\}$

2. How to connlude that $\mathbb R^n - B$ is connected.

Please help me to see this question more clearly and help me to answer this.

Thank you.

Answer 1

The idea is the following: for every $x \neq y$ in $\mathbb{R}^n$, there are uncountably many paths (we can even chose them to be the combination of two line segments, as here) from $x$ to $y$ that have pairwise "disjoint" images (except that they all contain $x$ and $y$).

If $x \neq y$ are in $\mathbb{R}^n - B$, not all of these paths can intersect $B$ "in the middle", as $B$ is only countable and we have uncountably many paths that are "middle disjoint". So $\mathbb{R}^n - B$ is path connected, using a path that misses $B$.

Answer 2

Take two points $x$ and $y$ not in $B$. The cardinal of the set of the lines that pass through $x$ is uncountable; hence, there exists a line that passes through $x$ but not through $y$ or any point of $B$. The same thing can be said about $y$, with an extra condition: the chosen line can not be parallel to the first one.
Both lines describe a path from $x$ to $y$, and that proves that $\Bbb R^n-B$ is path connected, and hence, connected.

Answer 3

Suppose $x,y\in\mathbb R^2\setminus B$. There are uncountably many circles that pass through $x$ and $y$, and at most countably many of those circles contains a point of $B$, since no point besides $x$ and $y$ lies on more than one of those circle.