The evaluation map is a useful tool for embedding a space into a product space. In this post we demonstrate that any Tychonoff space can be embedded into a cube where is the unit interval and is some cardinal. Any regular space with a countable base (second-countable space) can also be embedded into the Hilbert cube (Urysohn’s metrization theorem). The evaluation map also plays an important role in the theory of Cech-Stone compactification.
The Evaluation Map
Let be a space. Let be a product space. For each , we use the notation to denote a point in the product space . Suppose we have a family of continuous functions where for each . Define a mapping as follows:
For each , is the point .
This mapping is called the evaluation map of the family of continuous functions . If the family is understood, we may skip the subscript and use to denote the evaluation map.
The family of continuous functions is said to separate points if for any two distinct points , there is a function such that . The family of continuous functions is said to separate points from closed sets if for each point and for each closed set with , there is a function such that .
Theorem 1. Given an evaluation map as defined above, the following conditions hold.
- The mapping is continuous.
- If the family of continuous functions separates points, then is a one-to-one map.
- If the family of continuous functions separates points from closed sets, then is a homeomorphism from into the product space .
In this post, basic open sets in the product space are of the form where is finite, for each , is an open set in and .
Proof of 1. We show that is continuous at each . Let . Let and let where is a basic open set. Consider . It is easy to verify that and .
Proof of 2. Let be distinct points. There is such that . Clearly, .
Proof of 3. Note that by condition 2 in this theorem, the map is one-to-one. It suffices to show that is an open map. Let be open. We show that is open in . To this end, let . Then . Since separates points from closed sets, there is some such that . Let . Then . We show that . For each , we have . If , then , a contradiction. So we have and this means that . It follows that .
A space is a Tychonoff space (also known as completely regular space) if for each and for each closed set where , there is a continuous function such that and for all . The following is a corollary to theorem 1.
Corollary 1. Any Tychonoff space can be embedded in a cube .
Proof. Let be the family of all continuous functions from the Tychonoff space into the unit interval . By the definition of Tychonoff space, separates points from closed sets. By theorem 1, the evaluation map is a homeomorphism from into the cube where .
We now turn our attention to regular second countable space. Having a countable base has many strong properties, one of which is that it can be embedded into the Hilbert Cube . Before we prove this, observe that any regular space with a countable base is a regular Lindelof space. Furthermore, the property of having a countable base is hereditary. Thus a regular space with a countable base is hereditarily Lindelof (hence perfectly normal). The Vendenisoff Theorem states that in a perfectly normal space, every closed set is a zero-set (i.e. every open set is a cozero-set). So we make use of this theorem to obtain continuous functions that separate points from closed sets. There is a proof of The Vendenisoff Theorem in this blog. A set is a zero-set in the space if there is a continuous function such that . A set is a cozero-set if is a zero-set. We are now ready to prove one part of the Urysohn’s metrization theorem.
Urysohn’s metrization theorem. The following conditions are equivalent.
- The space is a regular space with a countable base.
- The space can be embedded into the Hilbert cube .
- The space is a separable metric space.
We prove the direction . Let be a countable base for the regular space . Based on the preceding discussion, is perfectly normal. By the Vendenisoff Theorem, for each , is a zero-set. Thus for each , there is a continuous function such that and . Let . It is easy to verify that separates points from closed sets. Thus the evaluation map is a homeomorphism from into .