The True/False problem 2 of the
final exam was a tougher cookie.
You must allow teachers to entertain themselves when creatively writing new exam problems.
We have seen various distance problems
before in TF problems:
d(P,R ) ≤ d(P,Q) + d(Q,R) for three points. This is the
standard triangle inequality.
We have seen in practice exams the problem that there is
no inequality relating
d(K,M) and d(K,L) + d(L,M) , if L,K,M are lines.
The reason is that L can intersect two parallel lines M,L
leading to one inequality. But then there is also the case of
three parallel lines, where we have a different inequality.
The same holds also if K,L,M are planes in three dimensions.
Also then, there is no inequality. Again, there examples where
one side of the inequality holds and examples where the
other side holds.
We have also seen that if U,V,W are three spheres, then
there is in general no inequality between d(U,W) and
d(U,V) and d(V,W).
I mentioned in personal communications to students during
office hours that in general one can not draw any triangular inequality
conclusion for higher dimensional objects. Take three cylinders
for example. Or take three hyperboloids. In most cases, just there
is no inequality like that. I also mentioned (maybe a bit misleading)
that if you see an inequality you have never seen before, then that
the chance is big that it is simply wrong.
When writing the exam, I thought about this a bit more and also looked at
the case of three spheres, where one sphere degenerates to a point.
This became TF problem 2 in the final exam. A bit surprisingly, now
there is a triangle inequality! It surprised me quite a bit too.
The statement actually holds in any dimension.
Here is the Summer 2020 theorem:
Theorem: if U,V are two spheres and P is a point, then
d(U,V) ≤ d(U,P) + d(P,V)
Proof. One can analyze this in a planar situation as one can intersect
the space with the 2 dimensional plane containing the centers of the spheres
and P. Now one has two circles and a point and the claim is that the distance
between the two circles is smaller or equal than the sum of the distance between
the two distances from P to the circles. Make a picture and denote the closest
point on the circle U from P with A and the closest point from P to V with B.
Now d(A,B) ≤ d(P,A) + d(P,B) by the triangle inequality. We also
have d(U,V) ≤ d(A,B) because by definition, the distance between U and V
is the smallest distance d(X,Y) where X is U and Y is on V. This in particular
holds for X=A, Y=B. From these two inequalities d(A,B) ≤ d(P,A) + d(P,B)
and d(U,V) ≤ d(A,B) the inequality d(U,V) ≤ d(U,P) + d(P,V) holds
by the transitivity property of the distance.
Now to make the proof tight, we also have to look at the situation where one
sphere is contained in the other. Also this can be verified in a similar way.
QED.
The reason why this theorem is surprising is that if P becomes a sphere too
(even with a tiny radius), the result becomes false!
I leave it to you to see whether the Summer 2020 theorem can be generalized
to ellipsoids or other shapes. Remember the
Polya lecture we have
seen early in the course and discussed. Asking new questions and exploring
such situations in variations is what mathematical research is about.
It is essential in all such explorations to play around and keep an open mind.
Here are some further explorations:
For which quadrics U,V does the result hold?
Does the result hold for Packman regions U,V?
Does the result hold for cubes?
Does the result hold if U is a circle and V is a cube?
Can you formulate a rather general theorem,
for a class of regions U,V, where U,V even do not have
to be of the same type?
Can you find cases of classes of regions U,V, where
the inequality fails?
Maybe you can even generalize the above theorem in full
generality: let (X,d) be a metric space (look up what this means)
and let U,V be two compact sets in X (look up what compact means).
Define the distance between two sets as the minimal distance of
points d(x,y) with x in one set and y in the other set.
Now if P is a point, then d(U,V) ≤ d(U,P) + d(P,V) holds.
And here is an even more challenging problem: Can you figure out
how to measure distances between sets in a metric space so that
the distance always satisfies the triangle inequality?
There is one which is used in fractal geometry.