1
$\begingroup$

Does a bijective map from $(-\pi/2,\pi/2)\to (0,1)$ exist?

My first guess was using the sine function but it doesn't really comply with the bijective map since it goes from $(-\pi/2,\pi/2)\to (-1,1)$. Am I missing something with the sine function or is there another way I can achieve this bijection?

$\endgroup$
  • $\begingroup$ Notice that the diameter of both intervals are the same, so it's really easy to come up with a bijective function in this case (as qalpha has already done below). $\endgroup$ – Irregular User May 29 '16 at 20:16
8
$\begingroup$

Consider the function $f(x) = \frac{x}{\pi} + \frac{1}{2}$.

$\endgroup$
  • $\begingroup$ $f(-\pi/2)=0$ and the function is increasing… $\endgroup$ – egreg May 29 '16 at 20:24
  • $\begingroup$ @egreg Yes, and the question is asking for a bijection from $(\frac{-\pi}{2}, \frac{\pi}{2})$ to $(0, 1)$. What's the problem? $\endgroup$ – qaphla May 29 '16 at 20:30
  • 1
    $\begingroup$ Sorry, I was misled by the text of the question and didn't read carefully the title. $\endgroup$ – egreg May 29 '16 at 20:34
2
$\begingroup$

As @leibnewtz said, for any two intervals $(a,b), (c,d)$ there is a function as you asked. Consider

$$f(x)=\frac{x-a}{b-a}(d-c)+c.$$

In this case,

$$f(x) = \frac{x-(-\pi/2)}{\pi/2-(-\pi/2)}(1-0)+0 = \frac{ x + \pi/2}{\pi}=\frac{x}{\pi}+\frac{1}{2}$$

as @qaphla has pointed out.

Note that $f$ is continuous, $f(a)=c, f(b)=d$ and $f'(x)=\frac{d-c}{b-a}>0$, hence bijective.

$\endgroup$
2
$\begingroup$

Take any bijection $f(x)$ from $(0,1)$ to $(0,1)$. Take a linear change of variable like $l_{-}(y) = \frac{1-2y}{2\pi} $ or $l_{+}(y) = \frac{1+2y}{2\pi} $. Then $f(l_\pm(y))$ is a bijection from $(-\frac{\pi}{2},\frac{\pi}{2})$ to $(0,1)$, because $l_{-}$ or $l_{+}$ maps $(-\frac{\pi}{2},\frac{\pi}{2})$ to $(0,1)$ first, and is bijective.

Now for $f(x)$ you can take for instance $x\to x^p$ or $x\to \sin^p(2x/\pi)$, with $p>0$, and you have at least a bijection with a sine function.

$\endgroup$
0
$\begingroup$

Any interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to any other interval $(c,d)$, which in particular implies that there is a bijection $f:(a,b) \to (c,d)$.

$\endgroup$
  • 3
    $\begingroup$ How do you prove the homeomorphism? By constructing a bicontinuous bijection. So this is no answer. $\endgroup$ – egreg May 29 '16 at 20:25
  • $\begingroup$ The bicontinuous bijection may be constructed without reference to specific values of $a$, $b$, $c$, and $d$. The question was if a bijection exists, and not for a specific example of such a bijection $\endgroup$ – leibnewtz May 29 '16 at 20:28
  • 2
    $\begingroup$ Let $f(x)=\frac{x-a}{b-a}(d-c)+c$. $\endgroup$ – B. Rivas May 29 '16 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.