Re: Sequence of continuous functions unbounded on irrationals


david petry wrote:
David C. Ullrich wrote:
On 3 Oct 2006 16:38:44 -0700, "david petry"
<david_lawrence_petry@xxxxxxxxx> wrote:


TCL wrote:
I am trying to find a sequence of nonnegative continuous functions f_n(x)
such that f_n(a) is unbounded for every irrational number a and bounded
for every rational number a.
Can't think of an easy example.

Let {a_k} be an enumeration of the rational numbers,
and let P_m(x) = product n=1..m (x-a_n)
and let F_m(x) = C_m | P_m (x) | where C_m is a well chosen rapidly
growing sequence of positive numbers. Then {F_m(x)} is the sequence you
want.

That's certainly _plausible_. I got that far all by myself, in fact.
Now how do you _prove_ that if C_m is large enough then F_m(x) is
unbounded for every irrational x?

Here's one way.

Define 'f' so that for all n,m |a_n - a_m| > f( max(n,m) ) > 0

That's not right. What I really meant was to define 'f' so that it is
a bound on how close two elements of the set {a_1, ..., a_m} can be to
each other.



Choose C_m so that F_m(x) > m for all x except possibly on an
exceptional set consisting of intervals of length f(m+100)/3
surrounding a_k for k <= m

Then a given irrational number 'r' may lie within the exceptional
inteveral of a rational number a_p for a great many of the F_m, but
eventually it will lie outside the exceptional interval for some F_n,
and for the next 100 functions in the sequence, it will not lie in the
exceptional interval of any rational number.

.

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