Prove that the intervals $(0,1)$ and $(0,\infty)$ have the same cardinality I have to proof that the intervals $(0,1)$ and $(0,\infty)$ have the same cardinality. I find some similar example with $(0,1)$ and $\mathbb{R}$ but I still have no idea to solve it.
 A: Define $f : (0, 1) \to (0, \infty)$ by $$f(x) = -\frac{x^2}{x-1}.$$ We can use methods from calculus to verify that this is a bijection. Since $f$ is a bijection between $(0, 1)$ and $(0, \infty)$, these two sets have the same cardinality.
A: $x \mapsto \frac{1}{x}$ is a bijection between $(0,1)$ and $(1, + \infty)$. Then use a translation to get $(0,+\infty)$.
A: Yet another way to do it:
The function
$y = \tanh(x) \tag{1}$
maps $(0, \infty) \to (0, 1)$.  Furthermore,
$y'(x) = \cosh^{-2}(x) > 0 \tag{2}$
for $x \in (0, \infty)$.  Thus $\tanh(x)$ is a bijection, as is $\tanh^{-1}(x)$ which maps $(0, 1) \to (0, \infty)$.
A: Another explicit bijection: $x \mapsto \tan{(\pi x/2)}$ is continuous, increasing on $(0,1)$, tends to $0$ as $x \to 0$, and $\infty$ as $x \to 1$, so it is a bijection $(0,1) \to (0,\infty)$
A: A bijection is a function that, for each distinct input has a unique and distinct output, and for each distinct output has a unique and distinct input.
To prove that 2 sets have the same cardinality, you can simple prove that there is a bijective transformation from one to the other.
For $(0, 1)$ to $(0, +\infty)$, there are an infinite number bijective functions.
For example:
$$x \mapsto -ln(x)$$
A: What about $f(x) = \frac{2}{\pi}arctan(x)$ is a bijection between $(0, \infty)$ and $(0, 1)$. So they have the same cardinality since we can construct a bijection between the two intervals.
