My Math Forum  

Go Back   My Math Forum > College Math Forum > Topology

Topology Topology Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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?
aminea95 is offline  
 
January 10th, 2018, 10:38 AM   #2
Senior Member
 
Joined: Oct 2009

Posts: 495
Thanks: 164

Take vertical and horizontal paths where one coordinate is irrational.
Micrm@ss is online now  
January 10th, 2018, 11:05 AM   #3
Senior Member
 
Joined: Oct 2009

Posts: 495
Thanks: 164

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.
Micrm@ss is online now  
January 10th, 2018, 12:46 PM   #4
Senior Member
 
Joined: Aug 2012

Posts: 2,010
Thanks: 574

Quote:
Originally Posted by Micrm@ss View Post
Take vertical and horizontal paths where one coordinate is irrational.
What happens at the corners? Is it differentiable as OP specified?
Maschke is offline  
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.
aminea95 is offline  
January 10th, 2018, 01:33 PM   #6
Senior Member
 
Joined: Aug 2012

Posts: 2,010
Thanks: 574

Quote:
Originally Posted by aminea95 View Post
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.
Yes and that differentiable arc must not pass through any rational points. And all the arcs together must be differentiable. In other words you need one arc end to end, it can't be piecewise. You can't have any corners.

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.
Maschke is offline  
January 10th, 2018, 06:15 PM   #7
Senior Member
 
Joined: Oct 2009

Posts: 495
Thanks: 164

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).
Micrm@ss is online now  
January 10th, 2018, 08:05 PM   #8
Senior Member
 
Joined: Aug 2012

Posts: 2,010
Thanks: 574

Quote:
Originally Posted by Micrm@ss View Post
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).
Which corner? Earlier you wrote

Quote:
Originally Posted by Micrm@ss View Post
Take vertical and horizontal paths where one coordinate is irrational.
So you have many horizontal and vertical paths, and many corners. And it's not at all clear what you mean. You can't start at a point and go "one point over" each time.

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.
Maschke is offline  
January 11th, 2018, 09:15 AM   #9
Senior Member
 
Joined: Oct 2009

Posts: 495
Thanks: 164

Quote:
Originally Posted by Maschke View Post
Which corner? Earlier you wrote



So you have many horizontal and vertical paths, and many corners. And it's not at all clear what you mean. You can't start at a point and go "one point over" each time.

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?
My posts are not meant to be a complete proof, merely hints for the OP.
Micrm@ss is online now  
January 11th, 2018, 09:21 AM   #10
Senior Member
 
Joined: Oct 2009

Posts: 495
Thanks: 164

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.
Micrm@ss is online now  
Reply

  My Math Forum > College Math Forum > Topology

Tags
topology



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Topology Sammy2121 Topology 1 January 11th, 2015 07:02 PM
Topology johnny_01 Topology 4 November 3rd, 2012 08:41 PM
Problem on product topology/standard topology on R^2. vercammen Topology 1 October 19th, 2012 11:06 AM
discrete topology, product topology genoatopologist Topology 0 December 6th, 2008 10:09 AM
discrete topology, product topology Erdos32212 Topology 0 December 2nd, 2008 01:04 PM





Copyright © 2018 My Math Forum. All rights reserved.