The Evaluation Map

The evaluation map is a useful tool for embedding a space X into a product space. In this post we demonstrate that any Tychonoff space X can be embedded into a cube I^{\mathcal{K}} where I is the unit interval [0,1] and \mathcal{K} is some cardinal. Any regular space with a countable base (second-countable space) can also be embedded into the Hilbert cube I^{\omega} (Urysohn’s metrization theorem). The evaluation map also plays an important role in the theory of Cech-Stone compactification.

The Evaluation Map
Let X be a space. Let \displaystyle Y=\Pi_{\alpha \in A}Y_\alpha be a product space. For each y \in Y, we use the notation y=\langle y_\alpha \rangle_{\alpha \in A} to denote a point in the product space Y. Suppose we have a family of continuous functions \mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace where f_\alpha:X \rightarrow Y_\alpha for each \alpha. Define a mapping E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha as follows:

    For each x \in X, E_{\mathcal{F}}(x) is the point \langle f_\alpha(x) \rangle_{\alpha \in A} \in Y.

This mapping is called the evaluation map of the family of continuous functions \mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace. If the family \mathcal{F} is understood, we may skip the subscript and use E to denote the evaluation map.

The family of continuous functions \mathcal{F} is said to separate points if for any two distinct points x,y \in X, there is a function f \in \mathcal{F} such that f(x) \neq f(y). The family of continuous functions \mathcal{F} is said to separate points from closed sets if for each point x \in X and for each closed set C \subset X with x \notin C, there is a function f \in \mathcal{F} such that f(x) \notin \overline{f(C)}.

Theorem 1. Given an evaluation map E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha as defined above, the following conditions hold.

  1. The mapping E_{\mathcal{F}} is continuous.
  2. If the family of continuous functions \mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace separates points, then E_{\mathcal{F}} is a one-to-one map.
  3. If the family of continuous functions \mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace separates points from closed sets, then E_{\mathcal{F}} is a homeomorphism from X into the product space \displaystyle Y=\Pi_{\alpha \in A}Y_\alpha.

In this post, basic open sets in the product space \displaystyle Y=\Pi_{\alpha \in A}Y_\alpha are of the form \bigcap_{\alpha \in W} [\alpha,V_\alpha] where W \subset A is finite, for each \alpha \in W, V_\alpha is an open set in Y_\alpha and [\alpha,V_\alpha]=\lbrace{y \in Y:y_\alpha \in V_\alpha}\rbrace.

Proof of 1. We show that E_{\mathcal{F}} is continuous at each x \in X. Let x \in X. Let h=\langle f_\alpha(x) \rangle_{\alpha \in A} and let h \in V \cap E_{\mathcal{F}}(X) where V=\bigcap_{\alpha \in W} [\alpha,V_\alpha] is a basic open set. Consider U=\bigcap_{\alpha \in W} f_\alpha^{-1}(V_\alpha). It is easy to verify that x \in U and E_{\mathcal{F}}(U) \subset V\cap E_{\mathcal{F}}(X).

Proof of 2. Let x,y \in X be distinct points. There is \alpha \in A  such that f_\alpha(x) \neq f_\alpha(y). Clearly, E_{\mathcal{F}}(x)= \langle f_\beta(x) \rangle_{\beta \in A} \neq E_{\mathcal{F}}(y)=\langle f_\beta(y) \rangle_{\beta \in A}.

Proof of 3. Note that by condition 2 in this theorem, the map E_{\mathcal{F}} is one-to-one. It suffices to show that E_{\mathcal{F}} is an open map. Let U \subset X be open. We show that E_{\mathcal{F}}(U) is open in E_{\mathcal{F}}(X). To this end, let \langle f_\alpha(x) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U). Then x \in U. Since \mathcal{F} separates points from closed sets, there is some \beta such that f_\beta(x) \notin \overline{f_\beta(X-U)}. Let V_\beta=Y_\beta-\overline{f_\beta(X-U)}. Then \langle f_\alpha(x) \rangle_{\alpha \in A} \in [\beta,V_\beta] \cap E_{\mathcal{F}}(X)=W_\beta. We show that W_\beta \subset E_{\mathcal{F}}(U). For each \langle f_\alpha(y) \rangle_{\alpha \in A} \in W_\beta, we have f_\beta(y) \notin \overline{f_\beta(X-U)}. If y \notin U, then f_\beta(y) \in f_\beta(X-U), a contradiction. So we have y \in U and this means that \langle f_\alpha(y) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U). It follows that W_\beta \subset E_{\mathcal{F}}(U).

Some Applications

A space X is a Tychonoff space (also known as completely regular space) if for each x \in X and for each closed set C \subset X where x \notin C, there is a continuous function f:X \rightarrow I such that f(x)=1 and f(y)=0 for all y \in C. The following is a corollary to theorem 1.

Corollary 1. Any Tychonoff space can be embedded in a cube I^{\mathcal{K}}.

Proof. Let \mathcal{F} be the family of all continuous functions from the Tychonoff space X into the unit interval I. By the definition of Tychonoff space, \mathcal{F} separates points from closed sets. By theorem 1, the evaluation map E_{\mathcal{F}} is a homeomorphism from X into the cube I^{\mathcal{K}} where \mathcal{K}=\lvert \mathcal{F} \lvert.

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 I^{\omega}=I^{\aleph_0}. 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 Z \subset X is a zero-set in the space X if there is a continuous function f:X \rightarrow I such that f^{-1}(0)=Z. A set W \subset X is a cozero-set if X-W 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.

  1. The space X is a regular space with a countable base.
  2. The space X can be embedded into the Hilbert cube I^{\aleph_0}.
  3. The space X is a separable metric space.

We prove the direction 1 \Rightarrow 2. Let \lbrace{B_0,B_1,B_2,...}\rbrace be a countable base for the regular space X. Based on the preceding discussion, X is perfectly normal. By the Vendenisoff Theorem, for each nX-B_n is a zero-set. Thus for each n, there is a continuous function f_n:X \rightarrow I such that f_n^{-1}(0)=X-B_n and f_n^{-1}((0,1])=B_n. Let \mathcal{F}=\lbrace{f_0,f_1,f_2,...}\rbrace. It is easy to verify that \mathcal{F} separates points from closed sets. Thus the evaluation map E_{\mathcal{F}} is a homeomorphism from X into I^{\aleph_0}.

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One thought on “The Evaluation Map

  1. Pingback: Extracting more information from Dowker’s theorem | Dan Ma's Topology Blog

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