Prove that the interval (−1,1) in R is homeomorphic to R

To prove that the interval \((-1, 1)\) in \(\mathbb{R}\) is homeomorphic to \(\mathbb{R}\), we need to show that there exists a continuous, bijective function with a continuous inverse between \((-1, 1)\) and \(\mathbb{R}\). A common approach is to construct an explicit homeomorphism between these two spaces.

### Proof:

1. **Define the Function:**
   Consider the function \( f : (-1, 1) \to \mathbb{R} \) given by:

   \[
   f(x) = \tan\left(\frac{\pi}{2} x\right).
   \]

   This function maps the interval \((-1, 1)\) onto \(\mathbb{R}\).

2. **Check Continuity of \(f\):**
   - The function \( f(x) = \tan\left(\frac{\pi}{2} x\right) \) is continuous on \((-1, 1)\) because the tangent function is continuous wherever it is defined.
   - Since \(\frac{\pi}{2} x\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) as \(x\) ranges from \(-1\) to \(1\), \(f\) continuously maps the interval \((-1, 1)\) to all of \(\mathbb{R}\).

3. **Check Bijectivity:**
   - **Injectivity (one-to-one):** Suppose \( f(x_1) = f(x_2) \). Then, \(\tan\left(\frac{\pi}{2} x_1\right) = \tan\left(\frac{\pi}{2} x_2\right)\). Since the tangent function is injective on the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), we have \(\frac{\pi}{2} x_1 = \frac{\pi}{2} x_2\), implying \(x_1 = x_2\).
   - **Surjectivity (onto):** For any \(y \in \mathbb{R}\), there exists an \(x \in (-1, 1)\) such that \(f(x) = y\). Specifically, for any \(y\), we can find \(x\) by solving \(x = \frac{2}{\pi} \arctan(y)\), which clearly lies in \((-1, 1)\).

4. **Check Continuity of the Inverse:**
   - The inverse function is \( f^{-1}(y) = \frac{2}{\pi} \arctan(y) \).
   - Since the arctangent function is continuous on \(\mathbb{R}\), \(f^{-1}\) is also continuous.

5. **Conclusion:**
   - Since \(f\) is a continuous bijection with a continuous inverse, \(f\) is a homeomorphism between \((-1, 1)\) and \(\mathbb{R}\).

Thus, the interval \((-1, 1)\) is homeomorphic to \(\mathbb{R}\).

### Questions for Further Exploration:
1. Can you find a different function that also establishes a homeomorphism between \((-1, 1)\) and \(\mathbb{R}\)?
2. How would the proof change if we were considering a closed interval \([-1, 1]\) instead?
3. What is the geometric interpretation of the function \( f(x) = \tan\left(\frac{\pi}{2} x\right) \)?
4. How does this result connect to other homeomorphisms in topology?
5. Can you prove that other intervals, such as \((0, 1)\), are homeomorphic to \(\mathbb{R}\)?

**Tip:** Understanding homeomorphisms is crucial in topology as they allow us to classify spaces that are "topologically the same," even if they look different geometrically.

Prove that the interval (−1,1) in R is homeomorphic to R.

Explore the proof that the interval (-1,1) in the real number line is homeomorphic to R. The solution involves constructing a continuous bijection with a continuous inverse using the function f(x) = tan(π/2 * x). Suitable for university-level topology students.

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