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 January 10th, 2018, 08:26 AM #1 Newbie   Joined: Dec 2017 From: vienna Posts: 12 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: 751 Thanks: 253 Take vertical and horizontal paths where one coordinate is irrational.
 January 10th, 2018, 11:05 AM #3 Senior Member   Joined: Oct 2009 Posts: 751 Thanks: 253 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
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Quote:
 Originally Posted by Micrm@ss Take vertical and horizontal paths where one coordinate is irrational.
What happens at the corners? Is it differentiable as OP specified?

 January 10th, 2018, 01:20 PM #5 Newbie   Joined: Dec 2017 From: vienna Posts: 12 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
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Quote:
 Originally Posted by aminea95 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.

 January 10th, 2018, 06:15 PM #7 Senior Member   Joined: Oct 2009 Posts: 751 Thanks: 253 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
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Quote:
 Originally Posted by Micrm@ss 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 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.

January 11th, 2018, 09:15 AM   #9
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Quote:
 Originally Posted by Maschke 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.

 January 11th, 2018, 09:21 AM #10 Senior Member   Joined: Oct 2009 Posts: 751 Thanks: 253 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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