Find all continuous functions $f:\mathbf R\rightarrow \mathbf R$ such that $f(x)-f(y)$ is rational for rational $x-y$. Can someone please help me with the solution ? Any Hint will be appreciated.
Asked
Active
Viewed 54 times
2
-
1the title is different from the body of the post. did you want the cardinality of the set or did you want some sort of characterization? – Tim kinsella Jan 13 '18 at 19:24
-
@Tim Ugh! Yes, I assumed the question was just about the number of such functions. – Andrés E. Caicedo Jan 13 '18 at 19:27
1 Answers
0
There are continuum many (i.e., $|\mathbb R|$) such functions. First of all, there are only $|\mathbb R|$ many continuous functions, so this is an upper bound. On the other hand, for any real $r$, $f(x)=x+r$ satisfies the requrements, so there are at least $|\mathbb R|$ many such functions.
Andrés E. Caicedo
- 79,201
-
(The usual argument showing that there are only $|\mathbb R|$ many continuous functions $f!:\mathbb R\to\mathbb R$ goes by noting that any such function is determined by $f|\upharpoonright \mathbb Q$, and there are only $|\mathbb R|^{|\mathbb Q|}=|\mathbb R|$ functions from $\mathbb Q$ to $\mathbb R$). – Andrés E. Caicedo Jan 13 '18 at 19:24