To prove that there is a bijection between the set of real numbers $\mathbb{R}$ and the interval $[0, 1]$, we need to establish a one-to-one and onto (bijective) function between these two sets. Here’s a structured approach to understand and establish such a bijection:

1. Bijection Concept

A bijective function $f: \mathbb{R} \rightarrow [0,1]$ must satisfy:

• Injectivity (One-to-One): Each element in $\mathbb{R}$ maps to a unique element in $[0,1]$.
• Surjectivity (Onto): Every element in $[0,1]$ is mapped by some element in $\mathbb{R}$.

2. Establishing Bijection

Cantor's Diagonal Argument and Cardinality:

• The set $\mathbb{R}$ and the interval $[0,1]$ both have the same cardinality, $\mathfrak{c}$ (the cardinality of the continuum).
• By Cantor's diagonal argument, we know the set of real numbers $\mathbb{R}$ is uncountable, and so is the interval $[0,1]$. Therefore, there exists a bijection between them because any two sets with the same cardinality have a bijective correspondence.

Constructing a Specific Bijection: While the existence of a bijection is guaranteed by the cardinality argument, constructing an explicit bijection can be more challenging. Here's one approach using a transformation technique:

Example Bijection: We can use the tangent function to map $\mathbb{R}$ to $(0,1)$: $f(x) = \frac{1}{2} \left(1 + \frac{2}{\pi} \arctan(x) \right)$

Properties:

• Injectivity: The arctangent function is continuous and strictly increasing on $\mathbb{R}$, and maps $\mathbb{R}$ bijectively onto $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.
• Surjectivity: The transformation $\frac{2}{\pi} \arctan(x)$ maps $\mathbb{R}$ onto $(-1,1)$, and then scaling and translating it maps it to $(0,1)$.

3. Ensuring Bijection

To ensure that the function $f(x)$ is indeed a bijection:

• Proof of Injectivity: Assume $f(x_1) = f(x_2)$, then $\arctan(x_1) = \arctan(x_2)$, implying $x_1 = x_2$.
• Proof of Surjectivity: For every $y \in (0,1)$, we need to find an $x \in \mathbb{R}$ such that $f(x) = y$. Given $y$, we can solve: $y = \frac{1}{2} \left(1 + \frac{2}{\pi} \arctan(x) \right)$ $2y - 1 = \frac{2}{\pi} \arctan(x)$ $\arctan(x) = \frac{\pi}{2} (2y - 1)$ $x = \tan \left(\frac{\pi}{2} (2y - 1)\right)$ Thus, for each $y \in (0,1)$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$.

Conclusion

The function $f(x) = \frac{1}{2} \left(1 + \frac{2}{\pi} \arctan(x) \right)$ serves as a bijection between $\mathbb{R}$ and the interval $(0,1)$.

Next Steps

Would you like further details on this bijection, or do you have any other questions? Here are some related questions you might be interested in:

1. How can we prove that $\mathbb{R}$ and $[0,1]$ have the same cardinality using Cantor's diagonal argument?
2. Can you provide another example of a bijective function between $\mathbb{R}$ and $[0,1]$?
3. How does the concept of bijection apply in other areas of mathematics?
4. What are some applications of bijective functions in real-world problems?
5. How can we visualize the transformation of $\mathbb{R}$ to $[0,1]$?

Tip: When studying bijections, always check both injectivity and surjectivity to confirm the function is bijective.