Question

Please don't try this if you are not strongest in

Topology.
Show that the Tychonoff plank T is C*-embedded in its one point compactification T*.

Answer 1

To show that the Tychonoff plank T is C*-embedded in its one-point compactification T*, we need to demonstrate that the inclusion map from T to T* is C*-embedded.

First, let's define the Tychonoff plank T and its one-point compactification T*:

The Tychonoff plank T is the product space [0,1] × [0,1) endowed with the product topology.

The one-point compactification T* of T is obtained by adding a single point at infinity to T, resulting in space [0,1] × [0,1] with the usual topology.

Now, we need to show that the inclusion map from T to T* is C*-embedding.

C*-embedding is a continuous injection that preserves topological and algebraic structures. In our case, we need to demonstrate that the inclusion map is a continuous injection and preserves the relevant topological and algebraic structures.

1. Injection: The inclusion map from T to T* is defined as the identity map on T, i.e., it maps each point in T to the corresponding point in T*. Since T* is obtained by adding a single point at infinity, there is no overlap between T and the added point. Therefore, the inclusion map is an injection.
2. Continuity: To show that the inclusion map is continuous, we need to consider the open sets in T*. We need to show that their pre-images under the inclusion map are open sets in T

Let U be an open set of T*. We have two cases to consider:

• If U does not contain the added point at infinity, then its pre-image under the inclusion map is simply U ∩ T, which is open in T since the product topology is induced by the usual topology on [0,1] × [0,1). Therefore, the inclusion map is continuous on this subset of T*.
• If U contains the added point at infinity, then its pre-image under the inclusion map is the entire T since every point in T maps to the corresponding point in T*. The entire T is open to T since T is a topological space. Therefore, the inclusion map is continuous on this subset of T* as well.

Hence, in both cases, the inclusion map is continuous.

3. Preservation of algebraic structures: Since both T and T* are product spaces, they inherit the algebraic structures of their respective factors. The inclusion map from T to T* preserves these algebraic structures as it maps each point in T to the corresponding point in T*.

Therefore, we have shown that the inclusion map from T to T* is a continuous injection and preserves the relevant topological and algebraic structures. Hence, T is C*-embedded in its one-point compactification T*.