R^2 - A, with A being a countable set, path-connected?

[curtdbz]

So I'm studying for an exam and one of the practice questions is to prove that the complement of R^2 to any countable set A is path connected.

I also assume this extends to any R^n larger than R^2 but that's not important. So this is intuitively obvious, since when you take out these discrete points from R^2 you can always find a path from X to Y anyway (with X and Y being points in R^2 not in A). But how do I go about proving this?

Is there some way to extend the proof that R^n - {0} is path connected? Or is there an easier way? Thank you for your help!

 

[JSuarez]

So this is intuitively obvious, since when you take out these discrete points from R^2 you can always find a path from X to Y anyway

Really? \mathbb Q^2 is both countable and dense in \mathbb R^2. Is it so obvious in this case?

Here's a hint: let x,y \in \mathbb R^2 and consider the set of all lines passing through them; do all these lines must contain points from A?

 

[curtdbz]

Lol, you're right, it's not as obvious as I thought. Usually nothing is that obvious to me in Topology anyway. Okay, so taking your hint:

Lets have two points in R^2, x and y. For x, let's take all the lines that intersect with it. Now visualizing it, it looks like a shining star centered at x (with rays going outwards). My claim is that there exists a line that does not intersect with A. Why does this line, call it L, exist?

Well since A is countable... it should imply A is uncountable if it DID intersect with L at some point. Thus contradiction. (This is where I need help).

Now since L exists for x, a similar argument shows line K exists for point y. Since these lines are of infinite length, they must intersect at some point (unless they're parallel), thus you can make a path from x to y using the lines.

Now obviously my "proof" is completely un-mathematically and full of holes. I'm unsure how to go about proving that L exists because if not it would mean A is uncountable. Also I don't know how to show that L and K aren't parallel.

If there were more than one L or K then the latter is easy. Thank you for your hint, and I hope this is the right approach. Could some one assist me a little further? Thanks so mcuh.

 

[JSuarez]

Ok, you are on the right track. For the main missing gap:

Well since A is countable... it should imply A is uncountable if it DID intersect with L at some point. Thus contradiction.

Consider the set of lines passing through x; they are completely specified by their slopes. Now, suppose that every line contains, at least, one point from A. As these intersect only at x, and x \notin A these points must be distinct. What are the implications for the cardinality of A?

If they are paralell, maybe a third line would help.

 

[curtdbz]

Hm.. Ohh! Yes, if they are completely specified by slope, which is a member of R (uncountable), then them intersecting at every point implies they can be paired up with R, thus uncountable! Genius! Thanks a lot JSuarez.

And, many of these unintersecting lines exists since you can take slopes ranging from 2..3 and show the same thing, and then another from 5..7 etc. Great!

 

[wofsy]

Hm.. Ohh! Yes, if they are completely specified by slope, which is a member of R (uncountable), then them intersecting at every point implies they can be paired up with R, thus uncountable! Genius! Thanks a lot JSuarez.

And, many of these unintersecting lines exists since you can take slopes ranging from 2..3 and show the same thing, and then another from 5..7 etc. Great!

So why does it mean that it is path connected?