Connected Sets

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  • #1
HallsofIvy
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Since things are a bit quiet here, I thought I would throw out a puzzle I came up with several years ago, after reading an article on connected sets:

Find two sets, P and Q, satisfying:

1) Both P and Q are completely contained in the (closed) rectangle in R2 with vertices at (1, 1), (1, -1), (-1, -1), and (-1, 1).

2) P contains the diametrically opposite points (1, 1) and (-1, -1) while Q contains(1, -1) and (-1, 1).

3) P and Q are both connected sets.

4) P and Q are disjoint.

The solution involves the difference between "connected" and "path-wise connected".
 

Answers and Replies

  • #2
NateTG
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P is the set of points [itex](x,y)[/itex] such that [itex]x[/itex] is rational and [itex]-1\leq x\leq 1[/itex], and [itex]y[/itex] is irrational and [itex]-1\leq y \leq 1[/itex] along with (1,1) and (-1,-1).

Q is the set of points [itex](x,y)[/itex] such that [itex]x[/itex] is irrational and [itex]-1\leq x\leq 1[/itex], and [itex]y[/itex] is rational and [itex]-1\leq y \leq 1[/itex] along with (1,1) and (-1,-1).
 
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  • #3
NateTG
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AKG said:
Make Q the straight segment. Make P have two disjoint parts, each the mirror image of the other. Any half of P will start at its corner, and approach Q like the topologist's sine curve approaches the y-axis.
P can be partitioned into the closed sets that correspond to [itex]x+y > 0[/itex] and [itex]x+y < 0[/itex]. Since solutions to [itex]x+y=0[/itex] are not in P, both of those sets are closed (in P), and they're obviously disjoint.
 
  • #4
matt grime
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Nate, your idea doesn't work. P and Q are not connected. You can disconnect P say be splitting it into two sets, those points in P to the left of the vertical line x=any irrational in (-1,1) and those to the right.
 
  • #5
NateTG
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Editing to make things nicer...

Consider the following sets:
[tex]P[/itex]
is a union of the following:
A vertical line segment at [itex]x=-1, y \in (-1,.5][/itex]
A vertical line segment at [itex]x=1, y \in [-.5,1)[/itex]
Horizontal line segments [itex]x \in [-1,.5], y = \sqrt{2} * n - k (\rm{for some} k,n \in \mathbb{Z} \rm{and} y \in [-1,.5][/itex]
and
Horizontal line segments [itex]x \in [-.5,.-1], y = \sqrt{2} * n - k (\rm{for some} k,n \in \mathbb{Z} \rm{and} y \in [-.5,.1][/itex]

This could be described as two interleaved infernal combs.
And [itex]Q[/itex] is [itex]P[/itex]'s complement.
Clearly each comb is a path-connected subset, so the only possible partition into non-empty closed sets is to split this into the combs, but each comb contains part of the other in its closure. Hence [itex]P[/itex] is connected.

(I'm open to suggestions on how to improve this section.)
Now, let's assume that [itex]Q[/itex] can be partitioned into disjoint non-empty closed sets A, and B. Since any horizontal intersecting [itex]Q[/itex] forms a connected sets, the projections of A and B onto the horizontal line must be disjoint. Since line segements are connected, at least one of two cannot be a closed set. Without loss of generality, that set is A. Since the projection of A onto the y axis is not closed there is some [itex]y[/itex] that is a limit point of the projection, but not in the projection, but that limit point clearly corresponds to a line segment of limit points of A that are not in A, but are in [itex]A[/itex] - contradicting that A is closed.
 
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  • #6
NateTG
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If anyone cares, I thought of a better example (in the sense that it's a bit easier to prove connectedness) on my way home last night, but it uses the same idea.
 
  • #7
HallsofIvy
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NateTG, I'm going to have to think about that for a while. Here' my answer:

Let A be the straight line from (-1,-1) to (0, -1/2). Let B be the set
{(x,y)| 0< x<= 1/pi, y= 0.8 sin(1/x)+ .1} (0.8 and 0.1 are chosen to lift that slightly above the x-axis but stay within the square), and let C be the straight line from (1/pi,0.1) to (1,1). Let P= A union B union C.

Let X be the straight line from (-1,1) to (0, 1/2). Let Y be the set {(x,y)| 0< x< 1/pi, y= 0.8 sin (1/x)- .1}, and let Z be the straight line from (1/pi,-0.1) to (1,-1). Let Q= X union Y union Z

Each of A, B, C, X, Y, Z is connected B union C and Y union Z are clearly connected since B,C and Y,Z have a point in common. The fact that
P= A union B union C is connected is clear from the fact that the closure of B includes the entire line (0, y) with y from -0.7 to 0.9 and includes (0, 0.5). The fact that Q= X union Y union Z is connected is clear from the fact that the closure of Y includes the entire line (0, y) with y from -0.9 to 0.7 and includes (0, 0.5).
 
  • #8
NateTG
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HallsofIvy said:
NateTG, I'm going to have to think about that for a while. Here' my answer.
Nice.

I had thought about using
[tex] (1-|x|) \sin (\frac{1}{x}) \pm {x}[/itex]
but for whatever reason didn't think of simply splitting the y axis into, for example, positive and negative sections.
 

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