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I tried $\tan(x)$, and $\log(x)$, but seems it does not work, so I wonder is there a bijection or not?

qwerty
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3 Answers3

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Consider that $$ [0,\infty) = \{0\} \cup (0,1) \cup \{1\} \cup (1,2) \cup \{2\} \cup \cdots $$ and $$ (0,1) = (0,1/2) \cup \{1/2\} \cup (1/2,3/4) \cup \{3/4\} \cup (3/4, 7/8) \cup \{7/8\} \cup \cdots $$

So we can make a bijection between then so that $0$ maps to $1/2$, $1$ maps to $3/4$, $2$ maps to $7/8$, etc., and the interval $(0,1)$ maps to the interval $(0,1/2)$ and the interval $(1,2)$ maps to the interval $(1/2,3/4)$, etc.

You cannot find a bijection with a single continuous function, because of topological reasons which are more advanced than the construction just described. In particular, $[0,\infty) \setminus \{0\}$ is connected but $(0,1)$ minus any point is disconnected, and the image of a connected set under a continuous map is connected.

Carl Mummert
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$$ f(x) = \begin{cases} \dfrac x {x+1} & \text{if } x\notin\{0,1,2,3,\ldots\}, \\[8pt] \dfrac{x+1}{x+2} & \text{if } x\in\{0,1,2,3,\ldots\}. \end{cases} $$ $f : [0,\infty) \to (0,1)$ is one-to-one and onto.

Michael Hardy
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  • How would you prove that f is bijective? –  Nov 24 '17 at 00:25
  • @user391135 The idea is that if you just have $g(x)=\dfrac x{x+1},$ that's a bijection from $[0,\infty)$ to $[0,1),$ but then you modify this function as follows: $$ \begin{array}c f(g^{-1}(0)) = 1/2 \ {} \ f(g^{-1}(1/2)) = 2/3 \ {} \ f(g^{-1}(2/3)) = 3/4 \ {} \ f(g^{-1}(3/4)) = 4/5 \ {} \ \vdots \end{array} $$ Think about that pattern in order to see why this is bijective. $\qquad$ – Michael Hardy May 04 '20 at 19:06
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Yes. You can find an injection from $(0,1)$ to $[0,\infty)$ by using $f(x) = 1/x$. You can find an injection from $[0,\infty)$ to $(0,1)$ by $\frac{1}{10} \arctan(x)+\frac{1}{10}$.

Since you have injections from each set to the other, there exists a bijection between them (via cantor-schroder-bernstein).

Batman
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  • No. I'm showing that $f(x) = 1/x$ is an injection not a bijection (i.e. we do not require the function to be onto), and similiarly for the use of arctan. By invoking cantor-schroder-bernstein, if you have an injection both ways, there exists a bijection. – Batman May 03 '16 at 02:09