1
$\begingroup$

I'm looking for a bijective function $f : (-1,1) \to \mathbb{R}$. I'm having trouble finding an example.

$\endgroup$
6
  • 7
    $\begingroup$ $(-1,1)\to(-1,1)$, $x \mapsto x$? $\endgroup$ – azif00 Nov 9 '20 at 23:30
  • $\begingroup$ What is the codomain of the envisaged function with domain $(-1,1)$? Without both pieces of data trying to think of a bijection, let alone any function, doesn't make any sense unless we automatically consider the codomain to be $(-1,1)$ - in which case the identity on this interval will do just fine. $\endgroup$ – Taylor Rendon Nov 9 '20 at 23:36
  • $\begingroup$ Hi @TaylorRendon. The problem only says that the codomain belongs to the real numbers. $\endgroup$ – CHM93 Nov 9 '20 at 23:43
  • $\begingroup$ In your opinion could the intent of the problem be that your function be a bijection from $(-1, 1)$ to all of $\mathbb{R}.$ That is, could the problem be that every real number must be in the range of $f$? $\endgroup$ – user2661923 Nov 10 '20 at 0:40
  • 1
    $\begingroup$ See also: Is there a bijective map from $(0,1)$ to $\mathbb{R}$? $\endgroup$ – Martin Sleziak Nov 10 '20 at 6:34
8
$\begingroup$

$f:: (-1,1) \rightarrow \mathbb{R}$

$f: x \mapsto \tan(\frac{\pi}{2} x)$

$\endgroup$
1
  • 5
    $\begingroup$ +1 for nailing it and -0.49 for making me pull out my Calculus book to confirm. I rounded up. $\endgroup$ – user2661923 Nov 10 '20 at 1:05
7
$\begingroup$

For a different example, $$f(x) = \frac{1}{x - 1} + \frac{1}{x + 1}$$ has this property:

  1. It is injective: if $x,y\in(-1,1)$ and $f(x) = f(y)$, then we have $$\frac{1}{x-1}+\frac{1}{x+1} = \frac{1}{y-1} + \frac{1}{y+1}.$$ Multiplying by $(x-1)(x+1)(y-1)(y+1)$ gives $$(x+1)(y-1)(y+1) + (x-1)(y-1)(y+1) = (x-1)(x+1)(y+1) + (x-1)(x+1)(y-1).$$ Now, we can simplify that down to $$x(y - 1)(y + 1) = y(x - 1)(x + 1).$$ At this point, we look to be stuck with expanding the brackets, so we'll do that, obtaining $$xy^2 - x = yx^2 - y,$$ or $$xy(y - x) = x - y.$$ If $x = y$, we're done, so suppose not (so that we can divide by $x - y$). Then we have $xy = -1,$ so $|xy| = 1$. But we know that $|x| < 1$ and $|y| < 1$, so $|xy| = |x||y| < 1$, a contradiction. Thus, $f$ is injective.

  2. It is surjective: for this, we shall simply show that it is unbounded above and below, and appeal to a version of the intermediate value theorem. First, note that, for $x = 1 - \varepsilon$ and $\varepsilon < 1$, we have $$f(x) = \frac{1}{-\varepsilon} + \frac{1}{2 + \varepsilon} = \frac{2}{-\varepsilon(2 + \varepsilon)}\leq \frac{-2}{3\varepsilon} \to -\infty$$ as $\varepsilon \to 0$ from above. Similarly, for $x = -1 + \varepsilon$, we have $f(x) \geq \frac{2}{3\varepsilon} \to +\infty$ as $\varepsilon \to 0$ from above.

Thus, for any $y \in \mathbb{R}$, there are some $x_0, x_1$ such that $f(x_0) < y < f(x_1)$ (and, indeed, we see that $x_0 > x_1$). Applying the intermediate value theorem (after noting that $f$ is continuous on $(-1, 1)$ as a sum of rational functions whose denominators have no zeroes in that interval) to $[x_1, x_0]$ then gives us some $x_2 \in [x_1, x_0] \subset (-1, 1)$ such that $f(x_2) = y$, and so $f$ is surjective.

$\endgroup$
2
  • $\begingroup$ I did not work this out, but I suppose one can also solve $f(x)=y$ to show bijectivity. It’s a quadratic equation. $\endgroup$ – Carsten S Nov 10 '20 at 7:52
  • 1
    $\begingroup$ @CarstenS Indeed: you'll find that for every $y$ (other than $0$), there are exactly two $x \in \mathbb{R}$ such that $f(x) = y$, exactly one of which lies in $(-1,1)$. $\endgroup$ – user3482749 Nov 10 '20 at 13:15
2
$\begingroup$

Under the assumption that the bijection must be from
$(-1, 1) \to \mathbb{R}$:

$x \geq 0:~~$ Let

$$f(x) = - \log|x - 1|.$$

$x < 0:~~$ Let

$$f(x) = + \log| ~|x| - 1|.$$

$\endgroup$
0
$\begingroup$

HINT: If the intent of this problem is for the codomain of $f$ to be all of $\mathbb{R}$,

Define the map $f : (-1,1) \to \mathbb{R}$ by $f(x) := \frac{x}{1-x^{2}}$.

$\endgroup$
0

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.