Visualizing Point at Infinity - Euclidean Plane

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Can anyone provide a visual of a Point at Infinity (or the ideal point)? I'm trying to visualize it and I apparently always end up interpreting every point on the Euclidean plane as an ideal point.
 
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Gear300 said:
Can anyone provide a visual of a Point at Infinity (or the ideal point)? I'm trying to visualize it and I apparently always end up interpreting every point on the Euclidean plane as an ideal point.
The easiest example to visualize is the infinite points of the extended real line. The real line has the same shape1 as the open interval (0, 1), and the extended real line has the same shape as the closed interval [0, 1]. In this picture, the points [itex]+\infty[/itex] and [itex]-\infty[/itex] of the extended real line correspond to 1 and 0.

If you curl [0, 1] into a circle (so that 0 and 1 become the same point), you get the projective real line.

The projective complexes have a similar picture: you can wrap the complexes up into a sphere missing a point. Under this picture, complex projective infinity would correspond to that missing point.


1: topologically, at least. There is a homeomorphism between the two spaces. For example, the map [itex]f(x)= (\arctan(x) + \pi / 2) / \pi[/itex]
 
Hurkyl said:
The easiest example to visualize is the infinite points of the extended real line. The real line has the same shape1 as the open interval (0, 1), and the extended real line has the same shape as the closed interval [0, 1]. In this picture, the points [itex]+\infty[/itex] and [itex]-\infty[/itex] of the extended real line correspond to 1 and 0.

If you curl [0, 1] into a circle (so that 0 and 1 become the same point), you get the projective real line.

The projective complexes have a similar picture: you can wrap the complexes up into a sphere missing a point. Under this picture, complex projective infinity would correspond to that missing point.


1: topologically, at least. There is a homeomorphism between the two spaces. For example, the map [itex]f(x)= (\arctan(x) + \pi / 2) / \pi[/itex]

I'm somewhat getting what you're saying...the vision you provided goes beyond how I was thinking...thanks
 
Well, you are certainly wrong if you "end up interpreting every point on the Euclidean plane". None of the points on the Euclidean plane is an "ideal point"!

One way to visualize it is this: Imagine a sphere sitting on the Euclidean plane so its "south pole" at (0,0). Draw a line from it "north pole" to a point on the plane. The point on the sphere where that line intersects the sphere is "identified" with the point in the plane. That identifies, in a one-to-one manner, every point on the plane with every point on the sphere except one- the "north pole". It is the "north pole" that now corresponds to the "point at infinity" or the "ideal point".

Here's another way. Draw a circle, of radius R, with center at (0,0). Identify points within the circle as "regular points", points on the circle as "ideal points". Of course that does not give a "one-to-one" correspondence so we let R go to infinity. In the limit, the point "inside the circle" are the points of the plane and the points "on the circle" are the "points at infinity". Notice that this gives an infinite number of "ideal points" or "points at infinity" not just one as the first construction does. These two models are not "topologically equivalent".

Yet a third method (the "projective plane"): Imagine not points but lines in the plane. "Points" are identified with the intersection of two lines. Two parallel points determine a "point at infinity". Of course, if I start with L1 and L2 not parallel, lines parallel to L1 will determine a different "point at infinity" that lines parallel to L2 so here we also have an infinite number of "ideal points" or "points at infinity". This model is topologically equivalent to the second of the two above.