Can you write a proof showing that the punctured Euclidean plane \(\mathbb{R}^2\setminus\{(0,0)\}\) is homeomorphic to the open annulus of internal radius 1 and external radius 2? You should include a formula for the homeomorphism.
Certainly! Let’s construct a proof showing that the punctured Euclidean plane \(\mathbb{R}^2 \setminus \{(0,0)\}\) is homeomorphic to the open annulus \(\{(x,y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2\}\).
Let \(A = \mathbb{R}^2 \setminus \{(0,0)\}\) be the punctured Euclidean plane, and let \(B = \{ (x, y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2 \}\) be the open annulus of inner radius 1 and outer radius 2.
We want to construct a homeomorphism between \(A\) and \(B\). Define the function \(f : A \to B\) by the formula:
\[ f(x, y) = \left( \frac{3}{2} \cdot \frac{(x, y)}{\sqrt{x^2 + y^2}} - \frac{1}{2} \cdot \frac{(x, y)}{\sqrt{x^2 + y^2}} \right). \]
Simplifying:
\[ f(x, y) = \frac{1}{2} \]
You did not write a correct formula for the function \(f(x,y)\). I would suggest using polar coordinates instead of Cartesian coordinates in the Euclidean plane.
You’re right—using polar coordinates will make the definition of the homeomorphism clearer. Let’s revise the proof and use polar coordinates to define the mapping.
Let \(A = \mathbb{R}^2 \setminus \{(0,0)\}\) be the punctured Euclidean plane, and let \(B = \{ (x, y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2 \}\) be the open annulus of inner radius 1 and outer radius 2.
We want to construct a homeomorphism \(f : A \to B\).
Recall that any point \((x, y)\) in the Euclidean plane can be expressed using polar coordinates \((r, \theta)\), where:
In polar coordinates:
We now define a function \(f : A \to B\) using polar coordinates. The idea is to map the radial component \(r\) of \(A\) to the radial component of \(B\) while preserving the angular component \(\theta\). One simple way to do this is by using a function that scales \(r\) into the interval \((1, 2)\).
Define \(f\) as: \[ f(r, \theta) = \left( \frac{r + 1}{r}, \theta \right), \] where \(r > 0\) and \(\theta \in [0, 2\pi)\).
We need to show that \(f\) is a bijective, continuous function with a continuous inverse.
Injectivity: Suppose \(f(r_1, \theta_1) = f(r_2, \theta_2)\). This means: \[ \left( \frac{r_1 + 1}{r_1}, \theta_1 \right) = \left( \frac{r_2 + 1}{r_2}, \theta_2 \right). \] Thus, \(\frac{r_1 + 1}{r_1} = \frac{r_2 + 1}{r_2}\) and \(\theta_1 = \theta_2\). It is straightforward to show that \(\frac{r_1 + 1}{r_1} = \frac{r_2 + 1}{r_2}\) implies \(r_1 = r_2\). Hence, \(f\) is injective.
Surjectivity: For any point \((r', \theta') \in B\), we need to find a point \((r, \theta) \in A\) such that \(f(r, \theta) = (r', \theta')\). Let \(r\) satisfy \(\frac{r + 1}{r} = r'\). Solving this equation gives: \[ r' r = r + 1 \quad \Rightarrow \quad r = \frac{1}{r' - 1}. \] Since \(1 < r' < 2\), it follows that \(r > 0\). Thus, \(f(r, \theta') = (r', \theta')\), so \(f\) is surjective.
Continuity: Since \(f\) is defined using elementary operations on \(r\) and \(\theta\), it is continuous in polar coordinates. When expressed in Cartesian coordinates, \(f\) is still continuous.
Inverse Function: The inverse \(f^{-1} : B \to A\) is given by: \[ f^{-1}(r', \theta') = \left( \frac{1}{r' - 1}, \theta' \right). \] This is continuous as \(r' > 1\). Therefore, \(f\) is a homeomorphism.
The function \(f : \mathbb{R}^2 \setminus \{(0,0)\} \to \{ (x, y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2 \}\) defined by: \[ f(x, y) = \left( \frac{r + 1}{r} \cos \theta, \frac{r + 1}{r} \sin \theta \right), \] where \((r, \theta)\) are the polar coordinates of \((x, y)\), is a homeomorphism. This shows that the punctured Euclidean plane is homeomorphic to the open annulus of internal radius 1 and external radius 2.
Unfortunately this version of the function \(f\) does not map the punctured plane onto the required annulus. As \(r\) tends to \(0\), \((r+1)/r\) tends to infinity and not to \(2\). Hence you are mapping the punctured plane onto the plane with unit disc removed. You need a different function of \(r\). By the way, a rational function of \(r\), similar to what you suggested, would work.
You’re absolutely right! I appreciate your feedback, and I can see where the issue arises. If we want to map the punctured Euclidean plane \(\mathbb{R}^2 \setminus \{(0,0)\}\) onto an open annulus with radii \(1\) and \(2\), we need to ensure that our function maps \(r \to 0\) to \(r \approx 2\), and \(r \to \infty\) to \(r \approx 1\).
A rational function of \(r\) will indeed be effective in achieving this. Let’s define a new homeomorphism with the appropriate properties.
Let \(A = \mathbb{R}^2 \setminus \{(0,0)\}\) be the punctured Euclidean plane, and let \(B = \{ (x, y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2 \}\) be the open annulus of internal radius \(1\) and external radius \(2\).
We want to construct a homeomorphism \(f : A \to B\).
In the punctured Euclidean plane \(A\), any point \((x, y)\) can be expressed in polar coordinates as \((r, \theta)\), where:
Thus, \(A = \{ (r, \theta) : r > 0, \ \theta \in [0, 2\pi) \}\).
In the annulus \(B\), a point can be expressed similarly in polar coordinates as \((r, \theta)\), where:
We need a function that maps \(r \in (0, \infty)\) bijectively to \(r' \in (1, 2)\). Consider the following rational function of \(r\): \[ f(r) = \frac{r + 2}{r + 1}. \]
Now, define \(f : A \to B\) as: \[ f(r, \theta) = \left( \frac{r + 2}{r + 1}, \theta \right). \]
In Cartesian coordinates, this becomes: \[ f(x, y) = \left( \frac{\sqrt{x^2 + y^2} + 2}{\sqrt{x^2 + y^2} + 1} \cdot \frac{x}{\sqrt{x^2 + y^2}}, \frac{\sqrt{x^2 + y^2} + 2}{\sqrt{x^2 + y^2} + 1} \cdot \frac{y}{\sqrt{x^2 + y^2}} \right). \]
Injectivity: Suppose \(f(r_1, \theta_1) = f(r_2, \theta_2)\). This means: \[ \left( \frac{r_1 + 2}{r_1 + 1}, \theta_1 \right) = \left( \frac{r_2 + 2}{r_2 + 1}, \theta_2 \right). \] Thus, \(\frac{r_1 + 2}{r_1 + 1} = \frac{r_2 + 2}{r_2 + 1}\) and \(\theta_1 = \theta_2\). Solving the equation \(\frac{r_1 + 2}{r_1 + 1} = \frac{r_2 + 2}{r_2 + 1}\) gives \(r_1 = r_2\). Hence, \(f\) is injective.
Surjectivity: For any point \((r', \theta') \in B\), we want to find a point \((r, \theta) \in A\) such that \(f(r, \theta) = (r', \theta')\). Consider: \[ \frac{r + 2}{r + 1} = r' \quad \Rightarrow \quad r' (r + 1) = r + 2. \] Solving for \(r\): \[ r' r + r' = r + 2 \quad \Rightarrow \quad r (r' - 1) = 2 - r' \quad \Rightarrow \quad r = \frac{2 - r'}{r' - 1}. \] Since \(1 < r' < 2\), \(2 - r' > 0\) and \(r' - 1 > 0\), so \(r > 0\). Thus, \(f(r, \theta') = (r', \theta')\), which proves that \(f\) is surjective.
Continuity: The function \(f(r, \theta) = \left( \frac{r + 2}{r + 1}, \theta \right)\) is continuous in polar coordinates. Since \(f\) is defined using continuous elementary operations on \(r\) and \(\theta\), it is continuous in Cartesian coordinates as well.
Inverse Function: The inverse \(f^{-1} : B \to A\) is given by: \[ f^{-1}(r', \theta') = \left( \frac{2 - r'}{r' - 1}, \theta' \right). \] This inverse is also continuous, since \(1 < r' < 2\).
Thus, \(f\) is a homeomorphism.
The function \(f : \mathbb{R}^2 \setminus \{(0,0)\} \to \{ (x, y) \in \mathbb{R}^2 : 1 < \sqrt{x^2 + y^2} < 2 \}\) defined by: \[ f(r, \theta) = \left( \frac{r + 2}{r + 1}, \theta \right) \] is a homeomorphism, proving that the punctured Euclidean plane is homeomorphic to the open annulus of internal radius 1 and external radius 2.