YB:

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.

AI:

Certainly! Let’s construct a proof showing that the punctured Euclidean plane ${\mathbb{R}}^2\setminus \{(0,0)\}$ is homeomorphic to the open annulus $\{(x,y)\in {\mathbb{R}}^2:1<\sqrt{x^2+y^2}<2\}$.

Proof:

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:

$$f(x,y)=(\frac{3}{2}\cdot \frac{(x,y)}{\sqrt{x^2+y^2}}-\frac{1}{2}\cdot \frac{(x,y)}{\sqrt{x^2+y^2}}).$$

Simplifying:

$$f(x,y)=\frac{1}{2}$$

YB:

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.

AI:

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.

Proof:

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 $f:A\to B$.

Step 1: Using Polar Coordinates

Recall that any point $(x,y)$ in the Euclidean plane can be expressed using polar coordinates $(r,\theta )$, where:

• $r=\sqrt{x^2+y^2}$ (the distance from the origin)
• $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ (the angle in radians with respect to the positive $x$-axis)

In polar coordinates:

• The punctured plane $A$ can be described as $A=\{(r,\theta ):r>0,\theta \in [0,2\pi )\}$.
• The annulus $B$ in polar coordinates is $B=\{(r,\theta ):1<r<2,\theta \in [0,2\pi )\}$.

Step 2: Defining the Homeomorphism

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: $$f(r,\theta )=(\frac{r+1}{r},\theta ),$$ where $r>0$ and $\theta \in [0,2\pi )$.

• When $r$ is very large ($r\to \mathrm{\infty }$), the radial component $\frac{r+1}{r}\approx 1$, so $f$ maps points far from the origin to points close to the inner radius of the annulus.
• When $r=1$, $\frac{r+1}{r}=2$, which maps to the outer radius of the annulus.

Step 3: Verifying the Homeomorphism

We need to show that $f$ is a bijective, continuous function with a continuous inverse.

1. Injectivity: Suppose $f(r_1,{\theta }_1)=f(r_2,{\theta }_2)$. This means: $$(\frac{r_1+1}{r_1},{\theta }_1)=(\frac{r_2+1}{r_2},{\theta }_2).$$ 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.

2. Surjectivity: For any point $(r^{\prime },{\theta }^{\prime })\in B$, we need to find a point $(r,\theta )\in A$ such that $f(r,\theta )=(r^{\prime },{\theta }^{\prime })$. Let $r$ satisfy $\frac{r+1}{r}=r^{\prime }$. Solving this equation gives: $$r^{\prime }r=r+1\Rightarrow r=\frac{1}{r^{\prime }-1}.$$ Since $1<r^{\prime }<2$, it follows that $r>0$. Thus, $f(r,{\theta }^{\prime })=(r^{\prime },{\theta }^{\prime })$, so $f$ is surjective.

3. 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.

4. Inverse Function: The inverse $f^{-1}:B\to A$ is given by: $$f^{-1}(r^{\prime },{\theta }^{\prime })=(\frac{1}{r^{\prime }-1},{\theta }^{\prime }).$$ This is continuous as $r^{\prime }>1$. Therefore, $f$ is a homeomorphism.

Conclusion

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: $$f(x,y)=(\frac{r+1}{r}\cos \theta ,\frac{r+1}{r}\sin \theta ),$$ 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.

YB:

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.

AI:

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 $r\to \mathrm{\infty }$ 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.

Revised Proof:

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 $f:A\to B$.

Step 1: Using Polar Coordinates

• In the punctured Euclidean plane $A$, any point $(x,y)$ can be expressed in polar coordinates as $(r,\theta )$, where:

  • $r=\sqrt{x^2+y^2}$ (the radial distance from the origin)
  • $\theta ={\tan }^{-1}(y/x)$ (the angle with respect to the positive $x$-axis)

  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:

  • $1<r<2$
  • $\theta \in [0,2\pi )$

Step 2: Constructing the Homeomorphism

We need a function that maps $r\in (0,\mathrm{\infty })$ bijectively to $r^{\prime }\in (1,2)$. Consider the following rational function of $r$: $$f(r)=\frac{r+2}{r+1}.$$

• Properties:
  • When $r\to 0$, $f(r)\to \frac{2}{1}=2$.
  • When $r\to \mathrm{\infty }$, $f(r)\to \frac{\mathrm{\infty }+2}{\mathrm{\infty }+1}\approx 1$.
  • For $r>0$, $f(r)$ satisfies $1<f(r)<2$.

Now, define $f:A\to B$ as: $$f(r,\theta )=(\frac{r+2}{r+1},\theta ).$$

In Cartesian coordinates, this becomes: $$f(x,y)=(\frac{\sqrt{x^2+y^2}+2}{\sqrt{x^2+y^2}+1}\cdot \frac{x}{\sqrt{x^2+y^2}},\frac{\sqrt{x^2+y^2}+2}{\sqrt{x^2+y^2}+1}\cdot \frac{y}{\sqrt{x^2+y^2}}).$$

Step 3: Verifying that $f$ is a Homeomorphism

1. Injectivity: Suppose $f(r_1,{\theta }_1)=f(r_2,{\theta }_2)$. This means: $$(\frac{r_1+2}{r_1+1},{\theta }_1)=(\frac{r_2+2}{r_2+1},{\theta }_2).$$ 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.

2. Surjectivity: For any point $(r^{\prime },{\theta }^{\prime })\in B$, we want to find a point $(r,\theta )\in A$ such that $f(r,\theta )=(r^{\prime },{\theta }^{\prime })$. Consider: $$\frac{r+2}{r+1}=r^{\prime }\Rightarrow r^{\prime }(r+1)=r+2.$$ Solving for $r$: $$r^{\prime }r+r^{\prime }=r+2\Rightarrow r(r^{\prime }-1)=2-r^{\prime }\Rightarrow r=\frac{2-r^{\prime }}{r^{\prime }-1}.$$ Since $1<r^{\prime }<2$, $2-r^{\prime }>0$ and $r^{\prime }-1>0$, so $r>0$. Thus, $f(r,{\theta }^{\prime })=(r^{\prime },{\theta }^{\prime })$, which proves that $f$ is surjective.

3. Continuity: The function $f(r,\theta )=(\frac{r+2}{r+1},\theta )$ 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.

4. Inverse Function: The inverse $f^{-1}:B\to A$ is given by: $$f^{-1}(r^{\prime },{\theta }^{\prime })=(\frac{2-r^{\prime }}{r^{\prime }-1},{\theta }^{\prime }).$$ This inverse is also continuous, since $1<r^{\prime }<2$.

Thus, $f$ is a homeomorphism.

Conclusion

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: $$f(r,\theta )=(\frac{r+2}{r+1},\theta )$$ is a homeomorphism, proving that the punctured Euclidean plane is homeomorphic to the open annulus of internal radius 1 and external radius 2.