Let be a completely regular space. The space is the space of all real-valued continuous functions defined on endowed with the pointwise convergence topology. In this post, we show that can be represented as the product of a subspace of with the real line . We prove the following theorem. See here for an application of this theorem.
Let be a completely regular space. Let . Let be defined by:
Then is homeomorphic to .
The above theorem can be found in  (see Theorem I.5.4 on p. 37). In , the homeomorphism is stated without proof. For the sake of completeness, we provide a detailed proof of Theorem 1.
Proof of Theorem 1
Define by for any . The map is a homeomorphism.
The map is one-to-one
First, we show that it is a one-to-one map. Let where . Assume that . Then . So assume that . Then the functions and are different, which means .
The map is onto
Now we show maps onto . Let . Let . Note that . Then . We have .
Note. Showing the continuity of and is a matter of working with the basic open sets in the function space carefully (e.g. making the necessary shifting). Some authors just skip the details and declare them continuous, e.g. . Readers are welcome to work out enough of the details to see the key idea.
The map is continuous
Show that is continuous. Let . Let be an open set in such that and,
where are arbitrary points in and is some large positive integer. Define the following:
Then define the open set as follows:
Clearly . We need to show . Let . Then . We need to show that and . Note that . For each , . So we have the following:
Subtracting the above two inequalities, we have the following:
The above inequality shows that for each , . Hence . It is clear that . This completes the proof that the map is continuous.
The inverse is continuous
We now show that is continuous. Let . Note that . Let be an open set in such that and
where are arbitrary points of and is some large positive integer. Now define an open subset of such that and
We need to show that . Let . We then have the following inequalities.
Adding the above two inequalities, we obtain:
The above implies that , . It is clear that . Thus . This completes the proof that is continuous.
- Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.