I know that bijection from $[a,b]$ to $(c,d)$ can't be continuous, but I'm wondering if such a function could exist if it was discontinuous at just countable points, and if this isn't possible, why?
Asked
Active
Viewed 75 times
2 Answers
8
The function $f:[0,1]\to [0,1)$ given by $$ f(x) = \cases{\frac x2 & if $x = \frac{1}{2^n}$ for some $n\in \Bbb N$\\ x & otherwise} $$ is the standard example of a bijection $[0,1]\to [0,1)$ and it is discontinuous only at a countable number of points. It can also without too much work be altered to fit $[a, b]\to(c, d)$ without losing that property.
Arthur
- 199,419
-
Could you elaborate a bit on how to alter it into a $[0,1]\to (0,1)$ function? – YoTengoUnLCD Dec 24 '15 at 06:21
-
@YoTengoUnLCD I think the easiest is to use this technique on $[-1,1]$, doing as above for $x=\frac1{2^n}$, and at the same time also for $x=-\frac1{2^n}$. – Arthur Dec 24 '15 at 08:29
3
Such functions exist. For simplicity, I'll map the interval $[-1,1]$ to $(-1,1)$, but the technique works in the general case.
For $n \geq 1$, let $a_n = \frac{1}{n}$ and $b_n = -\frac{1}{n}$. Define $f\colon\thinspace [-1,1]\rightarrow (-1,1)$ as follows:
-For values in the sequence $a_n$, $f(a_n) = a_{n+1}$.
-For values in the sequence $b_n$, $f(b_n) = b_{n+1}$.
-For all other $x$, $f(x) = x$.
One can easily verify that $f$ is a bijection from $[-1,1]$ to $(-1, 1)$. And it is continuous everywhere except on the sequences $\{a_n\}$ and $\{b_n\}$, which are countably many points.
MartianInvader
- 6,394