So I am trying to connect some very common mapping into a new set that is to consider to be one unit, one as in whole. From the cardinality perspective and bijection I believe is sufficient to prove that mapping each element from R into space [0,1] is possible being room for all the values.

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.

Understand how to prove a bijection between the set of real numbers and the interval [0,1] using Cantor's diagonal argument and a specific example function involving the arctangent. Explore the concept of bijection in set theory and its applications in advanced mathematics.

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