Hausdorff's Theorem on Homeomorphisms

The theorem states that if $ f: X \to Y $ is a homeomorphism and $ X $ is a Hausdorff space, then $ Y $ will also be a Hausdorff space.

This means that being a Hausdorff space is a topological property preserved under homeomorphisms.

In other words, if there is a bijective and continuous mapping between $ X $ and $ Y $, with the inverse also continuous, and $ X $ has the property of being a Hausdorff space (i.e., for any two distinct points, there exist disjoint open neighborhoods containing each point), then $ Y $ will share this property.

The result leverages the fact that homeomorphisms preserve topological properties, and since the Hausdorff property is topological, $ Y $ inherits this from $ X $.

Note: In many sources, this is simply presented as a consequence of the fact that the Hausdorff property is a topological property, which is preserved under homeomorphisms.

A Practical Example
The Proof

A Practical Example

Consider two topological spaces:

$ X = \mathbb{R} $, the real line with the standard topology.
$ Y = (0, 1) $, the open interval with the topology induced by the real line.

Let's define a function $ f: \mathbb{R} \to (0, 1) $ that maps the real line onto the open interval $ (0, 1) $.

A well-known example of such a function is the sigmoid function, which compresses the values of $ \mathbb{R} $ into the interval $ (0, 1) $:

$$ f(x) = \frac{1}{1 + e^{-x}} $$

This function takes real values and transforms them into numbers in the range (0, 1), excluding the endpoints.

Let's verify that $ f $ is a homeomorphism:

Continuity: The function $ f(x) = \frac{1}{1 + e^{-x}} $ is continuous because it is composed of continuous operations (exponentials and algebraic functions). It maps each real number to a point in the open interval $ (0, 1) $.
Injectivity: The function is injective, meaning each $ x \in \mathbb{R} $ is mapped to a unique value $ f(x) $ in the interval $ (0, 1) $. The sigmoid function is strictly increasing, so it never takes the same value for different inputs.
Continuity of the inverse: The inverse of the sigmoid function is given by: $$ f^{-1}(y) = \ln\left(\frac{y}{1 - y}\right) $$ This function is also continuous, ensuring that $ f $ is a homeomorphism.

We know that $ \mathbb{R} $, with the standard topology, is a Hausdorff space.

By the Hausdorff Theorem on Homeomorphisms, since $ f $ is a homeomorphism, $ (0, 1) $ must also be a Hausdorff space.

In conclusion, by the theorem, the open interval $ (0, 1) $ is a Hausdorff space because it is homeomorphic to $ \mathbb{R} $, which already possesses the Hausdorff property.

This example illustrates how a continuous, bijective mapping between $ \mathbb{R} $ and an open interval like $ (0, 1) $ preserves the Hausdorff topological property.

The Proof

We start with the following assumptions:

$ f : X \to Y $ is a homeomorphism, meaning $ f $ is a bijective (injective and surjective) continuous function with a continuous inverse.
The space $ X $ is a Hausdorff space.

We need to prove that $ Y $ is also a Hausdorff space. That is, for any two distinct points $ y_1, y_2 \in Y $, there must exist disjoint open neighborhoods containing them.

Since $ f $ is surjective, there exist points $ x_1, x_2 \in X $ such that $ f(x_1) = y_1 $ and $ f(x_2) = y_2 $.

For two distinct points $ x_1 \ne x_2 $, there exist disjoint open neighborhoods $ U_1 $ and $ U_2 $ in $ X $, such that $ x_1 \in U_1 $ and $ x_2 \in U_2 $, since $ X $ is Hausdorff.

Because $ f $ is injective, $ f(x_1)=y_1 $ and $ f(x_2)=y_2 $ imply that $ y_1 \neq y_2 $.

Now, since $ f $ is continuous (as it is a homeomorphism by assumption), the sets $ f(U_1) $ and $ f(U_2) $ are open in $ Y $, as the continuity of $ f $ ensures that the image of an open set in $ X $ is open in $ Y $.

Thus, the images of the open sets $ U_1 $ and $ U_2 $ in $ X $ are open sets $ f(U_1) $ and $ f(U_2) $ in $ Y $.

Furthermore, the sets $ f(U_1) $ and $ f(U_2) $ contain $ y_1 $ and $ y_2 $, respectively, because $ f(x_1) = y_1 $ and $ f(x_2) = y_2 $.

Finally, the sets $ f(U_1) $ and $ f(U_2) $ are disjoint. If there were a point $ z \in f(U_1) \cap f(U_2) $, then $ z = f(x_1') = f(x_2') $ for some $ x_1' \in U_1 $ and $ x_2' \in U_2 $, which would contradict the fact that $ U_1 $ and $ U_2 $ are disjoint in $ X $, and that $ f $ is injective.

Therefore, we have found two disjoint open neighborhoods $ f(U_1) $ and $ f(U_2) $ in $ Y $, containing $ y_1 $ and $ y_2 $, respectively.

Thus, we conclude that the space $ Y $ is Hausdorff.

We have demonstrated that if $ f : X \to Y $ is a homeomorphism and $ X $ is Hausdorff, then $ Y $ is also Hausdorff.

And so on.