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

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.

You want to show that $0$ and $x$ are connected by a path (piecewise straight-lined) which avoids $B$ entirely. You consider an uncountable familie of such paths, pairwise distinct (except at the start and end points, of course). Since the collection is uncountable, it requires uncountible many points from $B$ to stop each of these candidate paths. Since that is not possible, one of them must be satisfactory. — Jeppe Stig Nielsen, May 27 '14 at 14:29
Incidentally, this result is due to Cantor, who proved it near the end of a paper published in 1882. See here and here. — Dave L. Renfro, May 27 '14 at 14:55
@ajotatxe: if you have any alternate proof, please provide me. — user120386, May 28 '14 at 12:32
possible duplicate of Arcwise connected part of $\mathbb R^2$ — bof, May 29 '14 at 03:37
@wordsthatendinGRY The old question was less general, but Neon's answer points out that the same argument works for higher dimensions. — bof, May 29 '14 at 04:35

score 1 · Answer 1 · answered May 27 '14 at 14:36

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 $B$.

@ Henno Brandsma,Jeppe,Dave L,ajotatxe: thanks for your prompt reply. — user120386, May 28 '14 at 12:28

score 1 · Answer 2 · answered May 29 '14 at 02:21

1

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.

answered May 29 '14 at 02:21

ajotatxe

66,849

This is not an alternate proof. It's the same, although I've tried to write it more clearly. The "alternate proof" I mentioned was wrong. – ajotatxe May 29 '14 at 02:24

score 0 · Answer 3 · answered May 29 '14 at 04:41

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.

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

3 Answers3