January 10th, 2018, 08:26 AM  #1 
Newbie Joined: Dec 2017 From: vienna Posts: 8 Thanks: 1  topology
How can I show that for every two points in R^2_Q^2 they can be connected by a differentiable path?

January 10th, 2018, 10:38 AM  #2 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Take vertical and horizontal paths where one coordinate is irrational.

January 10th, 2018, 11:05 AM  #3 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Very interesting also is the statement: if $S$ is any countable subset of $\mathbb{R}^2$, then $\mathbb{R}^2\setminus S$ is path connected. As a hint for the proof: there are uncountably many lines through any given point of $\mathbb{R}^2$, so there are uncountably many lines with no points of $S$ on it. 
January 10th, 2018, 12:46 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,779 Thanks: 482  
January 10th, 2018, 01:20 PM  #5 
Newbie Joined: Dec 2017 From: vienna Posts: 8 Thanks: 1 
I think it's right what Micrm@ss said, but the issue is in the corners, but I think that I can always join two end points (2corners) by an arc as a path and as I've only countably many corner points, I can always count in a stupid way those arc paths, but I'd like to see a clever/systematic proof of this.
Last edited by skipjack; January 10th, 2018 at 05:34 PM. 
January 10th, 2018, 01:33 PM  #6  
Senior Member Joined: Aug 2012 Posts: 1,779 Thanks: 482  Quote:
I'm sure it can be done, but I'd like to see the proof. I'm sure you'd like to see the proof too! That's why you asked. Last edited by skipjack; January 10th, 2018 at 05:36 PM.  
January 10th, 2018, 06:15 PM  #7 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Replace the corner by the part of a circle. It's not hard to determine the rational points on a circle or how to determine when it has none (like you want).

January 10th, 2018, 08:05 PM  #8  
Senior Member Joined: Aug 2012 Posts: 1,779 Thanks: 482  Quote:
Quote:
I'm afraid I don't see this. I have no doubt that "it's not hard" as you say. Does your solution involve arcs? Or horizontal and vertical line segments? I can't tell if I'm missing something obvious or your proof is missing important details. ps  Perhaps you mean one horizontal and one vertical path. So what happens at the corner? Last edited by Maschke; January 10th, 2018 at 08:18 PM.  
January 11th, 2018, 09:15 AM  #9  
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92  Quote:
 
January 11th, 2018, 09:21 AM  #10 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Also, it is a common misconception that corners mean nondifferentiable, that is not the case. For example, see the path $t\rightarrow (t^2, t^3)$ (Neil's parabola). This is perfectly differentiable, but has a corner (a cusp) in the origin. What is violated is not the differentiability, but another property (nonzero Jacobian). So if you parametrize your path right, you can have corners. 

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