Exotic functions (elementary)

I've recently made a couple of posts to another
group (not usenet) that I thought might be of
interest to some people in sci.math, and so
I've merged those posts together to form this one.

http://mathforum.org/kb/thread.jspa?threadID=1463266

Incidentally, neither of the properties being considered
(unbounded in every interval, dense graph) can occur for
a Baire one function. However, the first example I give
is a Baire two function, and I'm pretty sure it's possible
(use condensation of singularities?) to obtain a Baire two
function whose graph is dense in the plane.


TABLE OF CONTENTS

1. A FUNCTION UNBOUNDED IN EVERY INTERVAL

2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE


-----------------------


1. A FUNCTION UNBOUNDED IN EVERY INTERVAL

I had a little free time, so I thought I'd
discuss a function, related to the ruler function

<http://mathforum.org/kb/thread.jspa?messageID=4594375>,

whose behavior illustrates a possibility that some
might consider impossible for a function to have.

We define f from the reals to the reals as follows.

f(x) = 0 if x is irrational or x=0.

f(p/q) = q if x is rational and x = p/q in lowest
terms and x has the same sign as p (i.e. q > 0).

Then f has the property of being unbounded in
every interval. To see this, consider an arbitrary
interval J. Then, for each integer n > 0 (no matter
how large), there exist infinitely many reduced-form
fractions p/q in J such that p > n. This is easy
to see if you consider "grids" on the number line
formed by marking the points

...., -3/q, -2/q, -1/q, 0, 1/q, 2/q, 3/q, ...,

first for q = M+1, then for q = M+2, then for
q = M+3, and so on. If the interval J is very
short, the first few of these grids might not
produce any points in J, but once q gets large
enough, the grids will start producing points
in J. Of course, only those p/q's where p and q
are relatively prime will matter, but this is
easily taken care of by just considering those
integers q > M such that q is a prime number.

Note that the value of f is bounded at each point.
[All I'm saying is that for each real number r,
|f(r)| < oo, which is a no-brainer.] However, for
each point r, the limiting behavior of f at x=r
is "unbounded". By this, I mean that f is unbounded
in every neighborhood of x=r.

Incidentally, f is unbounded in only one way,
namely from above. However, if we replace
"f(p/q) = q" with "f(p/q) = [(-1)^p] * q",
then I believe we'll get a function whose graph
is unbounded, both from above and from below,
in every interval.

As badly behaved as this function is (it's worse
than being discontinuous at each point, because
a function can be discontinuous at each point and
still be bounded on the entire real line), its
graph is relatively sparse in the plane. For
example, the graph of this function is certainly
a subset of the union of all horizontal lines
having an integer y-intercept, and even this
larger set misses a lot of regions with positive
area in the plane (e.g. the interior of the circle
with center (1, 1/2) of radius 1/2).


-----------------------


2. A FUNCTION WHOSE GRAPH IS DENSE IN THE PLANE

Now I'll show that we can define a function g,
much worse than this, so that the graph of g is
dense in the plane. This means that the interior
of *every* circle in the plane will contain points
belonging to the graph of g.

Let D_1, D_2, D_3, ... be an infinite sequence of
sets of nonzero real numbers such that each pair
of the sets has empty intersection and each of the
sets is a dense subset of the real numbers. See below
for several ways to obtain such a sequence of sets.

Write the rational numbers as a sequence r_1, r_2,
r_3, ... We can do this because the rational numbers
form a countable set. Indeed, there are many explicit
methods for carrying this out -- google phrases such
as "rationals are countable", "list the rationals",
"countability of the rationals", etc.

For each n = 1, 2, 3, ..., let L_n be the horizontal
"line" L_n = { (x, r_n) : x belongs to D_n }.

That is, L_n is the set of all points in the plane
whose x-coordinates are numbers chosen from D_n and
whose y-coordinates are the (fixed) number r_n.

The function g, and here I'm identifying the function
with its graph, consists of all the points in any
of the sets L_n, along with every point (if any) of
the form (x',0), where x' is a real number that doesn't
belong to any of the sets D_1, D_2, D_3, ...

g is function -- It is easy to see that g satisfies
the vertical line test by using the fact that the sets
D_1, D_2, ... are pairwise disjoint and the fact that
none of these sets contains 0. Also, the domain of g
is the set of all real numbers. This is because each
real number is either in one of the "D-sets" (in which
case g contains an ordered pair whose x-coordinate is
that real number) or it isn't in any of the "D-sets"
(in which case the value of g at that real number
is 0).

g is dense in the plane -- It suffices to show that
the (possibly) smaller set consisting of all the
horizontal "lines" L_1, L_2, ... is dense in the
plane. To see this, we need to show that each
rectangle [a,b] x [c,d] in the plane contains
in its interior at least one point belonging to
these "L-sets". Because the rational numbers are
dense, there exists a rational number, call it
r_k (i.e. let k be the index for one such rational
number in our listing of the rational numbers as
r_1, r_2, ...), such that c < r_k < d. Because the
set D_k is dense, there exists a number s in D_k
such that a < s < b. Then (s, r_k) belongs to L_k,
and hence to the union of the horizontal "lines",
and (s, r_k) lies in the interior of the rectangle
[a,b] x [c,d].


HOW TO OBTAIN THE SETS D_1, D_2, D_3, ...

One way is to let D_1 be the set of all nonzero
rational numbers whose reduced-fraction form
has a denominator that is some power of 2,
D_2 be the set of all nonzero rational numbers
whose reduced-fraction form has a denominator
that is some power of 3, ..., D_n be the
set of all nonzero rational numbers whose
reduced-fraction form has a denominator that
is some power of the n'th prime number, ...

A 2'nd way is to let D_1 be the set of all
nonzero rational numbers whose reduced-fraction
form has a prime denominator, D_2 be the set
of all nonzero rational numbers whose reduced-
fraction form has a denominator that is a product
of two prime numbers, ..., D_n be the set of
all nonzero rational numbers whose reduced-fraction
form has a denominator that is a product of n
prime numbers, ...

A 3'rd way is to let
D_1 = { t + sqrt(2): t is a rational number}
D_2 = { t + sqrt(3): t is a rational number}
..
..
..
D_n = { t + sqrt(p_n): t is a rational number},
where p_n is the n'th prime number,
..
..
..

A 4'th way is to let
D_1 = { t*sqrt(2): t is a nonzero rational number}
D_2 = { t*sqrt(3): t is a nonzero rational number}
..
..
..
D_n = { t*sqrt(p_n): t is a nonzero rational number},
where p_n is the n'th prime number,
..
..
..

The 4'th way is probably the easiest to verify the
properties for. To show D_n is dense, let a,b be
two real numbers with a < b. Choose t' to be a
nonzero rational number such that t' lies between
a / sqrt(p_n) and b / sqrt(p_n). [We can do this
because the rational numbers, and hence the nonzero
rational numbers as well, are dense in the reals.]
Then t'*sqrt(p_n) lies between a and b. To show that
the sets are pairwise disjoint, suppose t*sqrt(p)
belongs to one of the sets and t'*sqrt(p') belongs
to another one of the sets. Now it might be the case
that t = t', but since the two numbers are taken
from different "D-sets", we must have p not equal
to p'. Could these two numbers be equal? No, because
if t*sqrt(p) = t'*sqrt(p'), we'd have t/t' = sqrt(p'/p),
a contradiction, since t/t' is a rational number
and sqrt(p'/p) is an irrational number (because
the ratio of two different prime numbers can't be
a perfect square).


-----------------------


Dave L. Renfro

.

Relevant Pages

  • Re: Dense vs. Continuous
    ... mentioned that the rationals were "dense". ... function on the reals is a continuous function. ... If I were to graph the same function and said the domain ...
    (sci.math)
  • Re: counter example in analysis
    ... Since I referred to rationals, ... different not just from the irrationals but from the real ones, ... On the other hand, uncountable implies dense. ... If I consider the reals one by one, then they are not dense but countable. ...
    (sci.math)
  • Re: counter example in analysis
    ... Since I referred to rationals, ... infinite sauce-like continuum of locations. ... to be overall dense, and not even this corresponds to the original. ... If I consider the reals one by one, then they are not dense but countable. ...
    (sci.math)
  • Re: DSP riddle
    ... the rational numbers lie dense in the real numbers (every ... The mathematician who did it was called Cantor ("triangulation of Cantor" is the name of the proof that there are reals which are not rationals) and he became mad after that ...
    (comp.dsp)
  • Re: Cardnality of integers > Cardnality of integers
    ... In the diagonalization of the reals, every infinite sequence of digits ... let me form the Ulam spiral with these rationals. ... let me consider any 'square area' of the plane. ...
    (sci.math)