2 edited body; added 12 characters in body; added 34 characters in body

Hi all, This is really interesting qn. But it only involved involves basic homotopy theory, not anything as subtle as the jordan curve theorem.

Proof:

Let the paths be parameterized as v(t), and w(s). t,s in I := [0,1]

assume the paths never intersect. Then the map f : I x I -> S^1, given by f(s,t) = (v(t)-w(s))/|v(t)-w(s)| is well defined.

Think of f I x I as being a homotopy between the paths,

$a1(t) = { (0,2t) : 0< t <1/2 { (2t-1,1) : 1/2 < t<1$

and

$a2(t) = { (2t,0) : 0 < t<1/2 { (1,2t-1) : 1/2 < t < 1$

(i.e. the two boundary components of IxI, as paths from (0,0) -> (1,1) )

Then, we see that f(a1(t)) is a path that starts at the northpole north pole of the circle, and ends at the circlesouth pole, and traverses clockwise, whereas f(a2(t)) starts and ends the same, but traverses counterclockwise. but f Now f(I x I) provides a homotopy between these paths, however they are not homotopic as paths in the circle, this give a contradiction. and hence the paths must intersect.

1

Hi all, This is really interesting qn. But it only involved basic homotopy theory, not anything as subtle as the jordan curve theorem.

Proof:

Let the paths be parameterized as v(t), and w(s). t,s in I := [0,1]

assume the paths never intersect. Then the map f : I x I -> S^1, given by f(s,t) = (v(t)-w(s))/|v(t)-w(s)| is well defined.

Think of f as being a homotopy between the paths,

$a1(t) = { (0,2t) : 0< t <1/2 { (2t-1,1) : 1/2 < t<1$

and

$a2(t) = { (2t,0) : 0 < t<1/2 { (1,2t-1) : 1/2 < t < 1$

(i.e. the two boundary components of IxI, as paths from (0,0) -> (1,1) )

Then, we see that f(a1(t)) is a path that starts at the northpole of the circle, and ends at the circle, and traverses clockwise, whereas f(a2(t)) traverses counterclockwise. but f provides a homotopy between these paths, however they are not homotopic as paths in the circle, this give a contradiction. and hence the paths must intersect.