Re: Sequence of continuous functions unbounded on irrationals

In article
<waderameyxiii-E1FDE4.13483403102006@xxxxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:

In article <Ue0Ug.4072$753.1487@trnddc05>,
"TCL" <tlim1@xxxxxxxxxxx> 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.

For n = 1, 2, ... let E_n = {m/n : m an integer}. Define f_1 to
be the constant function 1. Assuming f_1, ..., f_n have been
defined, let f_(n+1) = f_n on E_1 U ... U E_n, f_(n+1) = n+1 on
E_(n+1) \ (E_1 U ... U E_n), then interpolate linearly. Then f_1,
f_2, ... is an increasing sequence of continuous piecewise-linear
functions on R.

If x is rational, let m be the smallest integer such that x is in
E_m. Then f_m(x) = m, and by construction all later f_n's are
also m at x. So the f_n's are bounded on the rationals.

If x is irrational, let N be a positive integer. Then all
rationals sufficiently close to x must lie outside of E_1 U ... U
E_N. It follows that f_n > N




contains two consecutive points of E_n, but no points of E_1 U
... U E_N.

Sorry, hit the Post button by mistake. Work in progress, maybe.
.

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