0

If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected?

My attempt : I found the answer here

Here is the outline of the given answer

Here is a marginal variation:

Choose $x_0,x_1 \in A^c$. Let $d \neq 0$ be orthogonal to $x_1-x_0$. Define the collection of paths $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$. Since $\{\gamma_\alpha\}_{\alpha \in [0,1]}$ is uncountable, there exists $\alpha_0 \in [0,1]$ such that $\gamma_{\alpha_0}[0,1] \cap A = \emptyset$. Hence $A^c$ is path connected.

My doubt : why we let $d \neq 0$ be orthogonal to $x_1-x_0?$

My thinking : I draw the diagram in my mind by taking $x_0=(0,0) $ ,$x_1=(1,0)$ and $d=(0,1)$

I think here $d$ will be meaningless and orthogonal mean $x_1-x_0 \perp d =0$ i,e $(1,0)-(0,0) \perp (0,1)= (1,0)(0,1)=(0,0)$

Actually im not getting what is the motive behind $x_1-x_0 \perp d$ ?

Lastly i need some book /reference where i can find this kind of concept

jasmine
  • 15,021

1 Answers1

1

The motive behind $x_1- x_0 \perp d$ is $d$ is the perpendicular bisector of the segment from $x_0$ to $x_1$

construct a line from $x_0 $and $x_1$ with different slope i,e two line should not be parallel and not intersect with $ A$

Choose a line $L_1$ through $x_0$ and $L_2 $ line through $x_1$

Both $L_1$ and $L_2$ touch at some point of perpendicular bisector $d$

Now we have pair of lines intersecting at some point of $d$ i,e choose $r \in d$ and $r=(r_1,r_2)$

$\implies$ there is path $f$ from $x_0$ to $r$ and again path $g$ from $r$ to $x_1$

By using pasting lemma we get a path $h$ which is connecteding $x_0$ to $x_1$ via $r$

therefore $\mathbb{R}^2 \setminus A$ is path connected

jasmine
  • 15,021