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

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    $\begingroup$ $(-1,1)\to(-1,1)$, $x \mapsto x$? $\endgroup$
    – azif00
    Nov 9, 2020 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. $\endgroup$ Nov 9, 2020 at 23:36
  • $\begingroup$ Hi @TaylorRendon. The problem only says that the codomain belongs to the real numbers. $\endgroup$
    – CHM93
    Nov 9, 2020 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 $f$? $\endgroup$ Nov 10, 2020 at 0:40
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    $\begingroup$ See also: Is there a bijective map from $(0,1)$ to $\mathbb{R}$? $\endgroup$ Nov 10, 2020 at 6:34

4 Answers 4

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

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

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    $\begingroup$ +1 for nailing it and -0.49 for making me pull out my Calculus book to confirm. I rounded up. $\endgroup$ Nov 10, 2020 at 1:05
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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.

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  • $\begingroup$ 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, 2020 at 7:52
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    $\begingroup$ @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 $(-1,1)$. $\endgroup$ Nov 10, 2020 at 13:15
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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

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

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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 $f(x) := \frac{x}{1-x^{2}}$.

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