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Wednesday, July 13, 2005

Weird Functions; why continuity matters so much

This summer I have been working through a delightful book by Gelbaum and Olmsted called Counterexamples in Analysis. No, I don't own stock in the Dover book company, nor do I know either of the authors.

I can recommend this book to anyone who is either learning elementary analysis, interested in learning the "whys" behind calculus, or is teaching calculus, real analysis or complex variables.

I'll share a few cool examples that I was not familiar with prior to reading this book.

• A function that is continuous precisely at the irrational numbers whose discontinuities are removable.

A reminder: if the limit as "x" approaches "a" of a function "f" exists but is not equal to the value of the function at "a", we say that the discontinuity is "removable". For example, the function f(x) = (x^2 -1)/(x + 1) has a removable discontinuity at x = 1 because f(1) doesn't exist (zero in the denominator) but the limit of f as x approaches 1 is clearly zero.

When we bring this concept up in calculus, we show a graph with a single "hole" where (a, f(a)) should be.

Now recall that a rational number is a number that can be put in the form p/q where p and q are integers and p, q have no common factors other than 1 or -1. If we insist that, say, q be positive, then this form is unique.

Now define f(x) to be "1/q" if x = p/q (i.e., x is rational) and let f(x) = 0 if x is irrational (e. g. f(pi) = 0).

Then note that f is continuous at every irrational number; it is a wonderful delta-epsilon argument to see this. Also note that if r is any rational number and we redefine f(r)=0 and leave f alone elsewhere, f becomes continuous at r. Hence each discontinuity is removable.

• A function that is bounded at every point but is unbounded on any interval (closed or open; we don't allow for empty or single point intervals)

Now let g(x) = q if x is rational and x = p/q and let g(x) = 0 if x is irrational.

Note that any interval whatsoever contains rational numbers of arbitrarily high magnitude (to see this, divide the real number line into, say, 1/10'ths, 1/100'ths, 1/1000's and so on, the way you would a ruler). Hence g is unbounded on any given interval, even though g(x) is finite for any given x. Different River said...

Don't leave me hanging! What's that epsilon-delta argument that f(x) is continuous at every irrational number? (I think I understand why it's discontinuous at every rational number. Forgive me if this is obvious, but it's been 14 years since I took Analysis.)

5:54 PM ollie said...

Ok D. R., let's step though this.

First, imagine the real line turned into an infinite ruler. The first set of "marks" are the whole numbers; this is where f = 1.
The second set of marks are the "odd number"/2; here f = 1/2
The third set of marks are the fractions (3k +1)/3, (3k+2)/3 . here f = 1/3 and so on.

Now let y be an irrational number and let "e" (for epsilon) be given.
Note that y is not equal to any of the marks. Now let n be the least integer so that n > 1/e.

Now let "d" (for delta) be the minimum of {distance between y, the irrational number in question, and all of the marks corresponding to the whole numbers, numbers over 2, numbers over 3, ....numbers over n}

It makes sense to talk about this distance as 1, 2, 3, ....n is a finite set.

Now let z be ANY number that is within "d" of y. If z is irrational, f(z)=0 therefore
|f(z) - f(y)|=0 < e
IF z is RATIONAL, then by the way we chose d, the denominator of z (in lowest terms) is greater than n, hence f(z)< 1/n, therefore
|f(z) - f(y)| < |1/n - 0| = 1/n < e
because n was chosen to be bigger than 1/e.

(note: I probably won't give this problem to average calculus students. :-) )

1:45 PM Different River said...

Wow. That is insanely clever. Thanks!

8:25 PM