Path connectedness in R^n

**vercammen** · October 16th 2012, 02:28 AM

Let $A$ be a countable subset of $\mathbb{R}^n$, where $n > 1$.
Prove that $\mathbb{R}^n \setminus A$ is path-connected.

Analogy of R^2. Notice that n = 1 does't work.

PICTURE ATTACHED!

Please Help!!!

**xxp9** · October 16th 2012, 08:14 AM

for any two points x, y not in A, pick any single parameter family of curves that connects x and y, having only x, y as their common points. There must be one curve c in the family such that c and A have no common points. Otherwise, suppose each curve intersects with A, A will have so many points that are not countable.

**vercammen** · October 16th 2012, 09:05 AM

thank you!

**johnsomeone** · October 16th 2012, 09:21 AM

Here's an explicit family of paths serving the purpose of xxp9's post:

$\text{PROP: Let } A \text{ be a countable subset of } \mathbb{R}^n, n>1. \text{ Then } A^c \text{ is path connected.}$

$\text{Proof: Assume not.}$

$\text{Then there exists }x, y \in A^c, x \ne y, \text{ such that every path connecting them in } \mathbb{R}^n \text{ intersects } A.$

$\text{Let }w \in \mathbb{R}^n \text{ such that } w \ne 0 \text{ and } <w, (y-x)> = 0.$

$\text{Note that such a } w \text{ exists if and only if } n > 1.$

$\text{For each }s \in \mathbb{R}, \text{ define } \gamma_s: [0, 1] \rightarrow \mathbb{R}^n \text{ by }\gamma_s(t) = x + t(y-x) + s(t-t^2)w.$

$\text{For all }s \in \mathbb{R}, \gamma_s(0) = x \text{ and } \gamma_s(1) = y, \text{ so these are paths from } x \text{ to } y.$

-------------

$\text{CLAIM: Distinct elements (paths) in } \{ \gamma_s \}_{s \in \mathbb{R}} \text{ only intersect at their endpoints } x \text{ and } y.$

$\text{proof:}$

$\text{By definition, }\gamma_{s_1}(t_1) = \gamma_{s_2}(t_2) \text{ if and only if }$

$x + t_1(y-x) + s_1(t_1-t_1^2)w = x + t_2(y-x) + s_2(t_2-t_2^2)w$

$\Leftrightarrow t_1(y-x) + s_1(t_1-t_1^2)w = t_2(y-x) + s_2(t_2-t_2^2)w$

$\Leftrightarrow (t_1-t_2)(y-x) + \{ s_1(t_1-t_1^2) - s_2(t_2-t_2^2) \}w = 0.$

$\text{Dotting with } y-x \text{ gives: }$

$0 = (t_1-t_2)\lVert y-x \rVert^2 + \{ s_1(t_1-t_1^2) - s_2(t_2-t_2^2) \}<w, y-x>$

$= (t_1-t_2)\lVert y-x \rVert^2 + \{ s_1(t_1-t_1^2) - s_2(t_2-t_2^2) \}(0) = (t_1-t_2)\lVert y-x \rVert^2,$

$\text{and since }y \ne x, \lVert y-x \rVert^2 \ne 0, \text { have that } 0 = (t_1-t_2)\lVert y-x \rVert^2 \Rightarrow t_1 = t_2.$

$\text{Thus }(t_1-t_2)(y-x) + \{ s_1(t_1-t_1^2) - s_2(t_2-t_2^2) \}w = 0 \Rightarrow t_1 = t_2, \text{ so have }$

$(t_1-t_2)(y-x) + \{ s_1(t_1-t_1^2) - s_2(t_2-t_2^2) \}w = 0$

$\Rightarrow (t_1-t_1)(y-x) + \{ s_1(t_1-t_1^2) - s_2(t_1-t_1^2) \}w = 0$

$\Rightarrow (0)(y-x) + (s_1-s_2)(t_1-t_1^2)w = 0 \Rightarrow (s_1-s_2)(t_1-t_1^2)w = 0$

$\Rightarrow (s_1-s_2)(t_1-t_1^2)\lVert w \rVert^2= <0, w> = 0 \overset{(w \ne 0)}{\Rightarrow} (s_1-s_2)(t_1-t_1^2) = 0.$

$\text{Thus } \gamma_{s_1}(t_1) = \gamma_{s_2}(t_2) \text{ implies } t_1 = t_2 \text{ and }$

$\text{either } s_1 = s_2 \text{ or } t_1 = t_2 \in \{ 0, 1 \}.$

$\text{If } s_1 \ne s_2, \text{ then } \gamma_{s_1}(t_1) = \gamma_{s_2}(t_2) \text{ implies } t_1 = t_2 \in \{ 0, 1 \},$

$\text{ and so } \gamma_{s_1}(t_1) = \gamma_{s_2}(t_2) \in \{ x, y \}.$

$\text{That proves that distinct elements of } \{ \gamma_s \}_{s \in \mathbb{R}} \text{ only intersect at their endpoints } x \text{ and } y.$

-------------

$\text{Since each of those paths intersects } A \text{ by assumption, and the endpoints } x \text{ and } y \text{ aren't in } A,$

$\text{have } \forall s \in \mathbb{R}, \gamma_s((0, 1)) \cap A \ne \emptyset, \text{ so } \exists \ a_s \in \gamma_s((0, 1)) \cap A.$

$\text{Since } \{ \gamma_s((0, 1)) \}_{s \in \mathbb{R}} \text{ are disjoint sets, } s_1 \ne s_2 \Rightarrow a_{s_1} \ne a_{s_2}, \text{ so each } a_s \text{ is distinct from the others.}$

$\text{Therefore } \{ a_s \}_{s \in \mathbb{R}} \text{ is an uncountable subset of } A.$

$\text{But that's impossible, since } A \text{ is by premise countable.}$

$\text{Thus the assumption that } A^c \text{ isn't path connected has produced a contradiction.}$

$\text{Therefore } A^c \text{ is path connected.}$

**vercammen** · October 16th 2012, 11:34 AM

Thank you!

Thread: Path connectedness in R^n

LinkBack

Thread Tools

Display

Path connectedness in R^n

Re: Path connectedness in R^n

Re: Path connectedness in R^n

Re: Path connectedness in R^n

Re: Path connectedness in R^n

Similar Math Help Forum Discussions

Path connectedness and connectedness

Simplest counterexample about path connectedness

Path-Connectedness proof

Local path-connectedness definition

path connectedness

Search Tags