I was wondering if there exists a function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies the property that for every $a<b$, $Image(f|_{(a, b)})=\mathbb{R}$. I tried to show that there isn't but falied. Any ideas?
Asked
Active
Viewed 129 times
2 Answers
9
Conway's base 13 function does exactly what you want: it maps every non-empty open interval to the whole of $\mathbb R$. See this Wikipedia article for details.
Note that this function is discontinuous everywhere. If you require a continuous $f$, then it can't be done.
TonyK
- 64,559
-
How about somewhere continuous? – MathematicsStudent1122 Jun 28 '16 at 13:12
-
1@MathematicsStudent1122 if it was at a point continuous then in a neighborhood of that point all the values would be near the value there – clark Jun 28 '16 at 13:14
-
Thank you, buy wow, I didn't manage to follow – 35T41 Jun 28 '16 at 14:17
5
For another counterexample unfortunately I cannot recall the reference, I have seen this a while back.
Let $ \mathbb{R} /\ \mathbb{Q}$ so $[x] = [y]$ if and only if $x-y\in\mathbb{Q}$.
because $\mathbb{R} /\ \mathbb{Q}$ has the same cardinatlity as $\mathbb{R}$ we get a surjection
$h:\mathbb{R} /\ \mathbb{Q}\rightarrow \mathbb{R}$.
Now and define $f (x) = h([x])$. The important algebraic property of $\mathbb{R} /\ \mathbb{Q}$ is that in any interval $(a,b)$ there is a full set of representatives, since for any $x$ we can find $r$ a rational number such that $x-r\in (a,b)$.
And hence $f( (a,b)) = \mathbb{R}$. for any interval.
clark
- 15,327