Question

Prove that f is a constant function

Answer 1

Did you get part a) yet? It has to be used in the solution to b)

Answer 2

@joemath314159: Yes, just use induction on m right?

Answer 3

Yes, thats correct.

Basically, you can use part a) to show that the value of the function at rational points is constant. Then to extend this property to the irrationals as well, you should use the fact that the rational numbers are dense in R.

Answer 4

If you need help with the details, just let us know :)

Answer 5

What happens if \[
f(r) =\begin{cases}
1&r\in \mathbb{Q} \\
0 &r\in \mathbb{R}\setminus \mathbb{Q}
\end{cases}
\]

Answer 6

Oh never mind, it has to be continuous...

Answer 7

If it has to be continuous then perhaps you could use \[
f'(r) = 0
\]

Answer 8

It might not be differentiable. Differentiability implies continuity, but not the other way around.

Answer 9

Ah thanks guys, I'm trying real hard to get it. If I'm not mistaken then the Density Theorem states that there exists a rational between any two real numbers but how to make use of it? Also should we take continuity into consideration, as it's what our professor gave us as a hint?

Answer 10

A corollary of the Density Theorem as you are stating is that for any real number x, there exists a sequence of rational numbers r_n such that:\[\lim_{n\rightarrow \infty}r_n = x\] The continuity of f is going to give us this fact:\[\lim_{n\rightarrow \infty}f( x_n)=f\left( \lim_{n\rightarrow \infty}x_n\right)\]This is called the sequential criterion for continuous functions. Are these things you have seen before? or maybe have seen in your analysis book?

Answer 11

Yes, I've seen it before, just give me some time to think of it :D

Answer 12

great! i just wanted to make sure I wasn't getting ahead of the material in your class.

Answer 13

i think you are on the right track, from the little I was able to read of your post :)

Answer 14

Oops I pressed the wrong button, I'll type it again in a minute!

Answer 15

By part (i), f(x)=c fpr some constant c and all rationals x. Suppose f is not a constant function, then there exists a real number r s.t \[f(r) \neq c\]. By the Density Theorem, there exists a sequence (r_n) of rationals converging to r. Now since f is continuous on R, \[f(r)=f(\lim_{n \rightarrow \infty}r_n)=\lim_{n \rightarrow \infty}f(r_n)=\lim_{n \rightarrow \infty}c=c\], which is a contradiction!. Hence f is a constant function.

Answer 16

perfect :)

Answer 17

Oh thanks for your kind help!