A nice, artistic ray-traced picture of a block of glass with the Borromian rings missing (name that movie)
John Robinson makes beautiful real-life Borromean rings, thanks to his admirable sponsors:
Paul Bourke has several brightly colored diagrams and pictures
Borromean rings in VRML
A cat looking at borromean rings
For a grand finale, here's
a page with 3 rings, 4 rings, 6 rings, mpeg animation, stereoscopic views,
raytraced pictures, ...
(probably originally by Mike Friedman)
Suppose you have a link in R3 made of circles such that any two circles have linking number 0. Then the link is trivial. Why? Consider the 4-D unit ball B4. First put R3 stereographically onto the ball's surface S3 (this preserves circles). Each circle now bounds a (flat) 2-disk in B4. When two circles have linking number 0, the corresponding disks do not intersect (Can you picture this well enough to believe this claim? I have trouble visualizing this). Now consider the family of concentric 3-spheres centered at the origin. The intersections of the spheres with the family of disjoint disks gives a homotopy of the link, except that at certain strategic moments, certain of the circles vanish to a point and disappear. Anyway this homotopy shows that the link is the trivial link.
(this is a 3D-ization by me of the above proof)
Given any arrangement of circles in 3-D, you can steadily shrink the areas of the circles without any of the circles ever touching each other, until each circle is reduced to a point (therefore nothing "went through" it), whereupon you can remove that circle from the others (therefore it wasn't linked to anything in any way).
(This can easily be made fully rigorous, if we specify that the circles should shrink so that each remains in a constant plane with a constant center, and the amount of area lost by each circle should equal the time T that has passed so far. If the disks bounded by two rings don't intersect, then the rings clearly won't touch as they shrink, so the only interesting case is when the disks do intersect. If the disks are in the same plane, then one is inside the other, and its radius is shrinking faster, so they won't meet. If the disks are not in the same plane, then, looking at the line of intersection, one can easily check that a shrinking circle's two points on this line are shrinking just as if they were themselves the diameter of a shrinking circle, so this case can be transformed in this way to a corollary of the case where the circles are in the same plane.)
(by Conway and myself)
The disc bounded by any ring (A) must intersect another (B) (since otherwise it could be shrunk to a point without hitting the others), and therefore must intersect it twice. I'll say A swallows B. Then plainly B can't also swallow A, so B swallows C and C swallows A. The planes of the A and B rings intersect in a line L -- if this is parallel to the plane of the C ring, then looking along it we see a picture:
in which we can see that one "end" of any ring can be "shrunk" until it is no longer "swallowed" (eg, the lower end of the C ring), so that the object plainly isn't Borromean.
The same argument works if L hits the plane of C in a point outside
the disc of C - we just project from this point (which is just the intersection
of the planes of all three rings. The only case left is therefore
that in which the three discs have a common point. Call it P.
The power of the point P (the product of the two distances to a given circle
on any line through P and the circle) will always be larger with respect
to the swallower than with respect to the swallowee, since P is inside
both discs. Chasing this inequality around the three circles gives
a contradiction, so it is impossible for three (or more) discs to share
a common point if each one swallows the next. So such a case in fact
cannot occur, so we are done.
I conjecture that the converse is also true:
That is, given any three simple unknotted closed curves in 3-D, they can always be arranged in a Borromean arrangement unless they are all circles.
For example, try making a Borromean arrangement with: