A question involving connectedness

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Homework Help Overview

The problem involves determining the connectedness of the set S = R^2 \ Q^2, where points in S have at least one irrational coordinate. The original poster seeks to understand the nature of this set in the context of topology, specifically regarding its connectedness and path connectedness.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of connectedness and explore whether S can be separated into two nonempty open sets. There are attempts to visualize paths connecting points in S and to clarify the implications of irrational and rational coordinates. Some participants question the notation and the assumptions about the points in S.

Discussion Status

There is ongoing exploration of the concept of path connectedness, with some participants suggesting potential paths and others expressing confusion about constructing continuous functions. Hints have been provided regarding the nature of lines through points in S and their relationship to rational points, but no consensus has been reached on a definitive approach.

Contextual Notes

Participants note the complexity of constructing paths without intersecting rational points and the challenges posed by the need for continuous functions. There is an acknowledgment of various cases depending on the coordinates being rational or irrational.

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Homework Statement


Let S=R^2/Q^2. (Points (x,y) in S have at least one irrational coordinate. Is S connected?


Homework Equations


A set M is connected if it has no separation. A separation is 2 nonempty open sets A and B such that A union B is M and A intersection B is the empty set.

I know if X is connected then f(X) is connected


The Attempt at a Solution


I have no idea how to start this.
 
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raw said:

Homework Statement


Let S=R^2/Q^2. (Points (x,y) in S have at least one irrational coordinate. Is S connected?

Homework Equations


A set M is connected if it has no separation. A separation is 2 nonempty open sets A and B such that A union B is M and A intersection B is the empty set.

I know if X is connected then f(X) is connected

The Attempt at a Solution


I have no idea how to start this.

In fact, it's path connected with the subspace topology. Can you find such a path? Consider how to connect it to a point (x_1, y_1) where both are irrational.

EDIT: I think my hint is correct, although I am not sure because of your notation... S = R^2\Q^2 means that (x, y) in S cannot contain any rational point.
 
malicx said:
In fact, it's path connected with the subspace topology. Can you find such a path? Consider how to connect it to a point (x_1, y_1) where both are irrational.

EDIT: I think my hint is correct, although I am not sure because of your notation... S = R^2\Q^2 means that (x, y) in S cannot contain any rational point.

I'm not sure what you are asking. What I meant to say is that both x and y can't be irrational but x or y could be.
 
malicx said:
In fact, it's path connected with the subspace topology. Can you find such a path? Consider how to connect it to a point (x_1, y_1) where both are irrational.

EDIT: I think my hint is correct, although I am not sure because of your notation... S = R^2\Q^2 means that (x, y) in S cannot contain any rational point.

The hint is good. Points in R^2\Q^2 are not rational points, sure. That's the same thing as saying if (x,y) is in S it has at least one irrational coordinate. This is an easy question, but it's only easy if you know the trick. Here's an even more direct hint. The number of straight lines through any given point in R^2\Q^2 is uncountable. Can you show that? Now show the number of those lines containing a point in Q^2 is countable.
 
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Dick said:
The hint is good. Points in R^2\Q^2 are not rational points, sure. That's the same thing as saying if (x,y) is in S it has at least one irrational coordinate. This is an easy question, but it's only easy if you know the trick. Here's an even more direct hint. The number of straight lines through any given point in R^2\Q^2 is uncountable. Can you show that? Now show the number of those lines containing a point in Q^2 is countable.
Oh of course, if I would have thought about that for 5 seconds more I should have seen it (been a long night!).
 
Last edited:
malicx said:
In fact, it's path connected with the subspace topology. Can you find such a path? Consider how to connect it to a point (x_1, y_1) where both are irrational.

EDIT: I think my hint is correct, although I am not sure because of your notation... S = R^2\Q^2 means that (x, y) in S cannot contain any rational point.

OK so there are several cases and I can visualize it. If you have p=(x1,y1) and q=(x2,y2) then the first case is when both coordinates have two irrational coordinates. You simply draw a vertical line from p to (x1, y2) and then a horizontal one to q. Actually this case also includes when only one of the coordinates has two irrational components or when x1,y2 are irrational or y1, x2 are irrational with slight changes.

When only x1 and x2 are irrational we can find a rational number say c between y1 and y2 and draw a vertical line from (x1,y1) to (x1,c) and then a horizontal line from (x1,c) to (x2,c) and another vertical line to q.

I am just frustrated because I'm having trouble constructing the continuous functions. I want to do a composition but I am having trouble because I am turning a constant into a variable. Can someone help me out?
 
raw said:
OK so there are several cases and I can visualize it. If you have p=(x1,y1) and q=(x2,y2) then the first case is when both coordinates have two irrational coordinates. You simply draw a vertical line from p to (x1, y2) and then a horizontal one to q. Actually this case also includes when only one of the coordinates has two irrational components or when x1,y2 are irrational or y1, x2 are irrational with slight changes.

When only x1 and x2 are irrational we can find a rational number say c between y1 and y2 and draw a vertical line from (x1,y1) to (x1,c) and then a horizontal line from (x1,c) to (x2,c) and another vertical line to q.

I am just frustrated because I'm having trouble constructing the continuous functions. I want to do a composition but I am having trouble because I am turning a constant into a variable. Can someone help me out?

I think you are kind of missing the point I was trying to make in post 3. If (x1,y1) is in R^2/Q^2, consider all of the lines that pass through (x1,y1). Do you see why there are an uncountable number of such lines? Now, do you see why only a countable number of them intersect Q^2? If a line doesn't intersect Q^2 then it must lie entirely in R^2/Q^2, right?
 
Dick said:
I think you are kind of missing the point I was trying to make in post 3. If (x1,y1) is in R^2/Q^2, consider all of the lines that pass through (x1,y1). Do you see why there are an uncountable number of such lines? Now, do you see why only a countable number of them intersect Q^2? If a line doesn't intersect Q^2 then it must lie entirely in R^2/Q^2, right?
Sorry, I didn't really look at thread after I responded to malicx. I just went and thought about what he said. Thank you for the help and I will think about your hint for a bit (I'm quite slow when it comes to this stuff). I am just wondering whether my approach would be ok too? I wanted to see if I could go through with it.
 
raw said:
Sorry, I didn't really look at thread after I responded to malicx. I just went and thought about what he said. Thank you for the help and I will think about your hint for a bit (I'm quite slow when it comes to this stuff). I am just wondering whether my approach would be ok too? I wanted to see if I could go through with it.

Well, you can try and go through with it with only horizontal and vertical lines. But I don't think it's going to work very well. Let's call symbols with an 'i' in them irrational and with an 'r' in them rational. How to you get from (i1,r1) to (i2,r1) without hitting a rational point?
 
  • #10
Dick said:
Well, you can try and go through with it with only horizontal and vertical lines. But I don't think it's going to work very well. Let's call symbols with an 'i' in them irrational and with an 'r' in them rational. How to you get from (i1,r1) to (i2,r1) without hitting a rational point?

Oh crap I didn't think about that.
 
  • #11
What if you went to some (i1,c) where c is irrational and then to (i2,c) and back to (i1,r1)?
 
  • #12
raw said:
What if you went to some (i1,c) where c is irrational and then to (i2,c) and back to (i1,r1)?

Actually, I think that does work. I was thinking of the more general case of R^2\{any countable set}. This is easier.
 
  • #13
Dick said:
Actually, I think that does work. I was thinking of the more general case of R^2\{any countable set}. This is easier.
This sounds bad but I'm still confused about how to construct the function. I'm running into issues when I take the composition of 2 straight lines with constants turning into variables.
 
  • #14
raw said:
This sounds bad but I'm still confused about how to construct the function. I'm running into issues when I take the composition of 2 straight lines with constants turning into variables.

If you want to connect (x1,y1) with (x2,y2), then f(t)=(1-t)*(x1,y1)+t*(x2,y2) for t in [0,1] does the job, right? g(t)=(1-(t-1))*(x2,y2)+(t-1)*(x3,y3) connects (x2,y2) with (x3,y3) for t in [1,2]. Now define the path h(t) piecewise. h(t)=f(t) for t in [0,1] and h(t)=g(t) for t in [1,2]. h(t) connects (x1,y1) with (x3,y3) for t in [0,2]. You don't take a 'composition', you make a piecewise definition.
 

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