Here is the text from the image: --- **1.** On page 19, Section 1.3, the author sketches a proof of a 1-1 correspondence between (0,1) and ℝ (lines 5 to line 13 from the top). Fill in the details of the proof that (0,1) and ℝ are equivalent. In other words show that \( f, g, f \circ g \) are 1-1 correspondence between the said sets. A picture is not a proof! (10 pts) --- Would you like a detailed written proof showing the 1-1 correspondence between \((0,1)\) and \(\mathbb{R}\)? If so, here's how you can proceed: --- ## Proof that \((0,1)\) and \(\mathbb{R}\) are Equivalent We want to construct explicit bijections (one-to-one and onto functions) between the open interval \((0,1)\) and the set of real numbers \(\mathbb{R}\). ### Step 1: Defining the Functions A classic bijection is given by the function: \[ f: (0,1) \to \mathbb{R}, \quad f(x) = \tan\left(\pi x - \frac{\pi}{2}\right) \] Alternatively, you might see: \[ f(x) = \ln\left(\frac{x}{1-x}\right) \] We'll use the latter for simplicity. #### **Define \( f: (0,1) \to \mathbb{R} \)** \[ f(x) = \ln\left(\frac{x}{1-x}\right) \] #### **Define \( g: \mathbb{R} \to (0,1) \)** \[ g(y) = \frac{e^y}{1+e^y} \] ### Step 2: Prove that \(f\) and \(g\) are Inverses #### **Show that \(f(g(y)) = y\)** \[ f(g(y)) = f\left(\frac{e^y}{1+e^y}\right) = \ln\left(\frac{\frac{e^y}{1+e^y}}{1 - \frac{e^y}{1+e^y}}\right) \] \[ = \ln\left(\frac{\frac{e^y}{1+e^y}}{\frac{1+e^y-e^y}{1+e^y}}\right) = \ln\left(\frac{e^y}{1}\right) = \ln(e^y) = y \] #### **Show that \(g(f(x)) = x\)** \[ g(f(x)) = g\left(\ln\left(\frac{x}{1-x}\right)\right) = \frac{e^{\ln(\frac{x}{1-x})}}{1 + e^{\ln(\frac{x}{1-x})}} \] \[ = \frac{\frac{x}{1-x}}{1 + \frac{x}{1-x}} = \frac{\frac{x}{1-x}}{\frac{1-x+x}{1-x}} = \frac{\frac{x}{1-x}}{\frac{1}{1-x}} = x \] ### Step 3: Conclude Since \(f\) and \(g\) are inverses, both are bijections, and thus \((0,1)\) and \(\mathbb{R}\) are equivalent (have the same cardinality). --- If you need this tailored to a specific textbook's notation or want a more formal structure, let me know!

Certainly! Here's a detailed, step-by-step explanation and proof demonstrating the 1-1 correspondence between the open interval \((0,1)\) and the set of real numbers \(\mathbb{R}\), involving the functions \(f\), \(g\), and their composition \(f \circ g\). --- ## Step-by-Step Proof of the 1-1 Correspondence Between \((0,1)\) and \(\mathbb{R}\) ### **Goal:** Construct explicit functions \(f\) and \(g\) such that: - \(f: (0,1) \to \mathbb{R}\) - \(g: \mathbb{R} \to (0,1)\) - \(f\) and \(g\) are inverses, establishing a bijection (1-1 and onto) between the two sets. --- ### **Step 1: Define the functions** We choose functions that are well-known for transforming between \((0,1)\) and \(\mathbb{R}\): - **Function \(f: (0,1) \to \mathbb{R}\):** \[ f(x) = \ln\left(\frac{x}{1 - x}\right) \] This is the *logit* function, commonly used in statistics. - **Function \(g: \mathbb{R} \to (0,1)\):** \[ g(y) = \frac{e^{y}}{1 + e^{y}} \] This is the *sigmoid* function, used in neural networks and probability. --- ### **Step 2: Show that \(f\) and \(g\) are inverses** To establish a one-to-one correspondence, we need to show: \[ f(g(y)) = y \quad \text{for all } y \in \mathbb{R} \] and \[ g(f(x)) = x \quad \text{for all } x \in (0,1) \] --- ### **Step 3: Verify \(f(g(y)) = y\)** Calculate: \[ f(g(y)) = f\left(\frac{e^{y}}{1 + e^{y}}\right) \] Plug into \(f\): \[ = \ln\left(\frac{\frac{e^{y}}{1 + e^{y}}}{1 - \frac{e^{y}}{1 + e^{y}}}\right) \] Simplify the denominator: \[ 1 - \frac{e^{y}}{1 + e^{y}} = \frac{(1 + e^{y}) - e^{y}}{1 + e^{y}} = \frac{1}{1 + e^{y}} \] Now, the entire expression: \[ = \ln\left(\frac{\frac{e^{y}}{1 + e^{y}}}{\frac{1}{1 + e^{y}}}\right) = \ln\left(\frac{e^{y}}{1 + e^{y}} \times \frac{1 + e^{y}}{1}\right) = \ln(e^{y}) = y \] **Result:** \[ f(g(y)) = y \] --- ### **Step 4: Verify \(g(f(x)) = x\)** Calculate: \[ g(f(x)) = g\left(\ln\left(\frac{x}{1 - x}\right)\right) \] Plug into \(g\): \[ = \frac{e^{\ln(\frac{x}{1 - x})}}{1 + e^{\ln(\frac{x}{1 - x})}} = \frac{\frac{x}{1 - x}}{1 + \frac{x}{1 - x}} \] Simplify the denominator: \[ 1 + \frac{x}{1 - x} = \frac{(1 - x) + x}{1 - x} = \frac{1}{1 - x} \] So, \[ g(f(x)) = \frac{\frac{x}{1 - x}}{\frac{1}{1 - x}} = x \] **Result:** \[ g(f(x)) = x \] --- ### **Step 5: Conclusion** Since: - \(f\) and \(g\) are **inverse functions** (\(f(g(y))= y\) and \(g(f(x))= x\)) - Both are **well-defined and bijective** between the sets \((0,1)\) and \(\mathbb{R}\) we conclude: \[ \boxed{ \text{There exists a 1-1 correspondence (bijection) between } (0,1) \text{ and } \mathbb{R} } \] specifically, the functions \(f\) and \(g\), with \(f: (0,1) \to \mathbb{R}\) and \(g: \mathbb{R} \to (0,1)\), establish this equivalence. --- ## **Summary:** - **Constructed functions:** - \(f(x) = \ln\left(\frac{x}{1 - x}\right)\) - \(g(y) = \frac{e^{y}}{1 + e^{y}}\) - **Key properties:** - \(f \circ g = \text{id}_{\mathbb{R}}\) - \(g \circ f = \text{id}_{(0,1)}\) - **Conclusion:** The sets \((0,1)\) and \(\mathbb{R}\) are **equivalent** in cardinality, connected via these inverse functions. --- If additional formalities or explanations are needed, feel free to ask!