Does there exist continuous onto function from $[0,1)$ to $(0,1)$.
I have made some conclusions. Such function can not be one in a neighborhood of zero.
Please help.
Asked
Active
Viewed 80 times
0
Mohsen Shahriari
- 6,718
Monotonic
- 145
-
Welcome to MSE! Please add some context to the problem and what have you tried? – Arjun Jul 30 '20 at 17:57
-
Perhaps the simple $f(x)=x$? – K.defaoite Jul 30 '20 at 17:58
-
See if f is one one in a nbd of zero.. Then by approaching 1, I can approach either 0 or 1. So range can not be (0,1) – Monotonic Jul 30 '20 at 17:58
-
3Consider $f(x) = 1/2 + 1/2x\sin(\frac{1}{1-x})$ on the domain [0,1). – John Lathrop Jul 30 '20 at 18:01
-
@john_lathrop, it's range is (0,1)? – Monotonic Jul 30 '20 at 18:05
-
I've clarified in an answer below – John Lathrop Jul 30 '20 at 18:21
-
How did you think of this? Can you give another function? – Monotonic Jul 30 '20 at 18:29
-
1This is a problem that seems to be testing knowledge of the extreme value theorem. We need to take advantage of the open bound of our domain, otherwise the EVT would apply (therefore making the statement false). Any function that attains $0$ and $1$ repeatedly as $x\rightarrow \infty$ can be substituted instead of $\sin$, so that $\frac{1}{1-x}\rightarrow \infty$ as $x\rightarrow 1$. – John Lathrop Jul 30 '20 at 18:40
-
Great..can you suggest me book where I can get more such questions of this type. – Monotonic Jul 30 '20 at 18:50
-
Any real analysis textbook that includes chapter problems should have some on functions on open/closed sets and testing knowledge of theorems. See this other post for a variety of analysis textbooks. – John Lathrop Jul 30 '20 at 18:57
1 Answers
5
Consider the function $f:[0,1)\rightarrow(0,1)$ defined as $f(x) = 1/2 + 1/2*x*\sin(\frac{1}{1-x})$. I've attached an image of a graph of $f(x)$ to hopefully aid understanding of what's going on with $f$:
$f$ is enclosed in the envelope bound by $y_{upper} = \frac{x+1}{2}$ and $y_{lower} = \frac{-x+1}{2}$ and is therefore will not attain $y=0$ or $y=1$ on $[0,1)$.
However for any $y$ value arbitrarily close to 1, there will be some $x$ near 1 s.t. $\frac{1}{1-x} = \pi/2 + 2\pi k$ for $k\in\mathbb{Z}$, so that $f(x)=1/2+1/2*x = \frac{x+1}{2}$, which approaches 1 as $x\rightarrow1$.
Similarly, for any desired $y$ value close to 0, there will be $x$ near 1 s.t. $\frac{1}{1-x} = 3\pi/2 + 2\pi k$, so that $f(x) = 1/2 + 1/2*x*(-1) = \frac{-x+1}{2}$, which approaches 0 as $x\rightarrow1$.
John Lathrop
- 331