| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1888.1 | See | UTROP1::BEL_M | Michel Bel@UTO - Telecommie | Fri Aug 19 1994 17:59 | 1 |
| C is not equal to Aleph 0 ( Easy way out)
|
| 1888.2 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Fri Aug 19 1994 20:35 | 35 |
| One non-continuous example is
{ 0 if [x] is even
f(x) = {
{ sqrt(2) if [x] is odd
where [x] is the greatest integer <= x.
A proof similar to 1887.1 shows no such continuous function
can exist. Let f:R->R have the property that for all real x,
f(x) is rational if and only if f(x+1) is irrational. Either
f(0) is rational, or it is irrational and f(1) is rational.
So there are reals x such that f(x) is rational, and we can
choose one of them, call it c, so that f(c) is rational. Then
f(c+1) is irrational and f(c+2) is rational. Let a = f(c+2) - f(c)
and note that a is rational.
Let g(x) = f(x) - f(c) - (1/2)a(x - c). If f is continuous
then g is continuous. g(c) = 0 and g(c+2) = 0. g(c+1) =
f(c+1) - f(c) - (1/2)a = f(c+1) - (f(c) + (1/2)a) is the
difference between an irrational and a rational, and so g(c+1)
is irrational and therefore g(c+1) is nonzero.
Let h(x) = g(x) - g(x+1). If g is continuous then h is continuous.
h(c) = g(c) - g(c+1) = -g(c+1) and h(c+1) = g(c+1) - g(c+2) = g(c+1).
Since g(c+1) is nonzero, we have h is a function which takes on
opposite signs at h(c) and h(c+1). If h is continuous there is a
value d with c < d < c+1 such that h(d) = 0. So g(d) - g(d+1) = 0 or
(f(d) - f(c) - (1/2)a(d-c)) - (f(d+1) - f(c) - (1/2)a(d+1-c)) = 0
or f(d) - f(d+1) = -(1/2)a, which is impossible as f(d) - f(d+1)
is irrational but -(1/2)a is rational. So h cannot be continuous.
which means g and f must not be continuous.
Dan
|
| 1888.3 | re 1.: Too shortcut for me! | EVTSG8::ESANU | | Mon Aug 22 1994 07:34 | 2 |
| Can you elaborate your solution? I don't quite c how you get
the solution from C <> Aleph 0...
|
| 1888.4 | A shorter proof than 1888.2 | EVTSG8::ESANU | | Mon Aug 22 1994 07:52 | 11 |
| re 2.:
Suppose there is such a continuous function f.
Define h(x) = f(x+1) - f(x)
Then h is continuous and h(x) is irrational for all real x, hence h is constant,
more precisely, there is an irrational a such that h(x) = a for all real x.
We have f(x+1) - f(x) = a
Hence f(x+2) - f(x) = 2a
Choose x real such that f(x) is rational, thence f(x+2) is rational. It results
that 2a is rational, hence a is rational - contradiction.
|
| 1888.5 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Tue Aug 23 1994 13:22 | 7 |
| Yes, that works and is simpler. If h is continuous then
{ h(x) | x is in R } is connected, and so if it contains at
least two points then it contains a rational between them. So
if { h(x) | x is in R } contains only irrationals then it must
have exactly one element.
Dan
|