Can Homeomorphism exists between One point compactification of Real Line and Unit Circle?

Circle is a wonderful closed shape, and Real Line is endless and seems very long and unbounded. Consider adding one imaginary point at the end. How are they the same? How are they homeomorphic? This wonderful question was asked by one of my favourite professors in a topology semester exam. I felt this was a great question because that day I did not know how to write it clearly. So today I decided to explore a more detailed answer.

Homeomorphism is one of the great topological properties of mathematical objects. I first heard and understood this concept when I started studying manifolds, because its definition contains the idea of homeomorphism. For this question, we have two ways to prove that a homeomorphism can exist.

Question:

Define a homeomorphism $f$ between $\mathbb{R}\cup \{\infty\}$ and $S^1 = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 =1\}$ . Prove that f is homeomorphism. Use this homeomorphism to characterize all the open sets $U \subseteq \mathbb{R} \cup \{\infty\}$ such that $\infty \in U$ .

Diagram showing two horizontal lines in blue and red above and a black circle below, representing geometrical relationships.

Simply, the question is to define the homeomorphism between the unit circle and the one point compactification of real line. That means the function we define must be continuous, it should be bijective, and its inverse function should also be continuous. And we need to characterize the open subsets of one point compactification of real line contains $\infty$ .

We can think of two different representations for the unit circle. One is the Cartesian coordinate representation

$S^1 = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 =1\}$

and another is the polar coordinate representation

$S^1 = \{(1,\theta) : \theta \in (-\pi,\pi)\}$

From intuition, we can say that if we take one point like $(0,1)$ , then the circle seems like a curved line. If we stretch it, it becomes identical to the real line, and we need to map the removed point to infinity. So they look the same. But in mathematics, we need a clear proof to say anything about mathematical objects.

First way: Cartesian Perspective

The representation was already given as

$S^1 = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 =1\}$

Take the North Pole of the circle, $N = (0,1)$

Now we should find a bijective map between the real line and $S^1 \setminus \{N\}$ . We can use stereographic projection for this. For every real number $t \in \mathbb{R}$ , let

$f(t) = \left( \frac{2t}{1+t^2},\frac{t^2-1}{1+t^2}\right)$

This gives a point on the unit circle except the North Pole.

Now define the map

$f : \mathbb{R} \to S^1 \setminus \{N\}$ ,

This function is continuous, and it is bijective.

Now we define the inverse map. For any point $(x,y) \in S^1 \setminus \{N\}$ ,

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

This inverse is also continuous.

Now we extend this map

$\lim_{t \to \infty} f(t) = N(0,1) = f(\infty)$

So we get a function

$f : \mathbb{R} \cup \{\infty\} \to S^1$

This function is continuous and bijective and we have inverse which is also continuous. Hence this is a homeomorphism.

Second way: Polar Perspective

Now consider the polar coordinate representation

$S^1 = \{(1,\theta) : \theta \in (-\pi,\pi]\}$

Every point on the unit circle can be written as

$(\cos \theta, \sin \theta)$

where $\theta \in (-\pi,\pi]$ .

Now we define a map between $\mathbb{R}$ and angles using

$\theta = 2\arctan (t)$

So define

$g(t) = (\cos(2\arctan t),\sin(2 \arctan t))$

This maps $\mathbb{R}$ to $S^1 \setminus \{(-1,0)\}$ .

As $t \to \infty$ , we get

$\theta \to \pi$

and this corresponds to the missing point when we compactify. So again we assign

$g(\infty) =(-1,0)$

Thus we extend

$g : \mathbb{R} \cup\{\infty\} \to S^1$

This map is continuous, bijective, and its inverse is also continuous.

So finally, we proved that

$\mathbb{R}\cup\{\infty\} \cong S^1$

This means the one point compactification of real line is homeomorphic to the unit circle.

What looked like a straight infinite line and a closed curved circle are actually the same from a topological point of view. The only trick was adding one point at infinity.

This is the beauty of topology. It does not care about shape, length, or curvature. It only cares about structure. Read our post on Topology and Homeomorphism: Understanding Topology: The Bridge Between Math and Geometry

Characterization of open sets containing Infinity

Now we use this homeomorphism to characterize all the open sets $U$ in $\mathbb{R} \cup {\infty}$ such that $\infty \in U$ .

We already defined a homeomorphism

$f : \mathbb{R} \cup {\infty} \to S^1$

with

$f(\infty) = N = (0,1)$

Since $f$ is a homeomorphism, a set $U \subset \mathbb{R} \cup {\infty}$ is open if and only if $f(U)$ is open in $S^1$ .

Now assume $\infty \in U$ . Then
$N \in f(U)$

So $f(U)$ is an open set in $S^1$ containing the North Pole.

From the geometry of the circle, an open set containing $N = (0,1)$ looks like a small arc around the North Pole. If we remove this small arc, what remains is a closed arc away from $N$ .

Now we translate this back using the inverse map $f^{-1}$ . That removed arc around $N$ corresponds to a bounded region in $\mathbb{R}$ . More precisely, it corresponds to a compact subset of $\mathbb{R}$ .

So the original set $U$ must contain everything outside that compact region, along with the point at infinity.

Hence, every open set $U \subset \mathbb{R} \cup {\infty}$ with $\infty \in U$ is of the form

$U = (\mathbb{R} \setminus K) \cup {\infty}$

where $K \subset \mathbb{R}$ is compact.

In $\mathbb{R}$ , compact sets are exactly closed and bounded sets, so intuitively this means that $U$ contains all sufficiently large positive and negative real numbers, together with $\infty$ .

So neighborhoods of infinity are exactly those sets that contain everything outside some bounded region. This matches our intuition: points going far away in either direction of the real line are all approaching the same point $\infty$ .

Thus, using the homeomorphism with the circle, we get a complete topological characterization of open sets containing infinity in the one-point compactification of $\mathbb{R}$ .

Can Homeomorphism exists between One point compactification of Real Line and Unit Circle?

First way: Cartesian Perspective

Second way: Polar Perspective

Characterization of open sets containing Infinity

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Can Homeomorphism exists between One point compactification of Real Line and Unit Circle?

First way: Cartesian Perspective

Second way: Polar Perspective

Characterization of open sets containing Infinity

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