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


    
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*If $z , z' \in l$, then $l_z \cap l_{z'} =  \{0,x\}$
    
*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.
 A: 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 $B$. 
A: 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.
A: 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.
