# Embedding Completely Regular Spaces into a Cube

This is a continuation of a discussion on completely regular spaces (continuing from these two posts: Completely Regular Spaces and Pseudocompact Spaces, Completely Regular Spaces and Function Spaces). This post gives another reason (one of the most important ones) why the class of completely regular spaces occupies a central place in general topology, which is that completely regular spaces are precisely the spaces that can be embedded in a cube (the product of copies of the unit interval). This theorem was proved by Tychonoff in 1930 . The tool that makes this theorem possible also allowed Stone and Cech in 1937 to construct for any completely regular space $X$, a compact Hausdorff space $\beta X$ that contains $X$ as a dense set ($\beta X$ is called the Stone-Cech compactification of $X$). In this post we discuss the role played by complete regularity in this construction. We discuss the following theorem.

Theorem 1
Let $X$ be a space. Then $X$ is completely regular if and only if $X$ is homeomorphic to a subspace of a cube.

Original articles are [1], [4] and [5]. The Stone-Cech compactification and the related concepts are classic topics that are covered in standard texts (see [2] and [6]).

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Completely Regular Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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The Evaluation Map

The evaluation map can be defined in a more general setting. We define it here just to deal with the task at hand, namely to discuss Theorem 1.

Let $X$ be a completely regular space. Let $I=[0,1]$ be the unit interval in the real line $\mathbb{R}$. A cube is of the form $I^\mathcal{K}$, i.e., the product of $\mathcal{K}$ many copies of $I$ where $\mathcal{K}$ is some cardinal. Let $C(X,I)$ be the set of all continuous real-valued functions defined on the space $X$. Consider the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$. This is the cube where $X$ is embedded as a subspace. We can represent each point in the cube $\prod \limits_{f \in C(X,I)} I_f$ as a function $H:C(X,I) \rightarrow I$ or as a sequence $< H_f >_{f \in C(X,I)}$ such that each term (or coordinate) $H_f \in I=[0,1]$.

The key to embed $X$ into a cube is through the evaluation map, which is a map $E$ from $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$. Thus we have:

$\displaystyle (1) \ \ \ \ \ \ E:X \longrightarrow \prod \limits_{f \in C(X,I)} I_f$

We now define the map $E$. For each $x \in X$, $E(x)=< H_f >_{f \in C(X,I)}$ such that $H_f=f(x)$ for each $f \in C(X,I)$. In other words, $E(x)$ is the point in the product space $\prod \limits_{f \in C(X,I)} I_f$ whose $f^{th}$ coordinate is $f(x)$.

We show that because $X$ is completely regular, the evaluation map $E$ is a homeomorphism. We show the following:

• The map $E$ is continuous.
• The map $E$ is one-to-one.
• The map $E^{-1}$ is continuous.

The continuity of the map $E$ follows from the fact that each $f \in C(X,I)$ is continuous. The map $E$ being one-to-one follows from the fact that for each pair $x,y \in X$ with $x \ne y$, there is an $f \in C(X,I)$ such that $f(x) \ne f(y)$.

We now show $E^{-1}$, the inverse of $E$, is continuous. This is where $X$ must be completely regular. Let $U$ be open in $X$ such that $x =E^{-1}(E(x)) \in U$. Since $X$ is completely regular, there is a continuous $g:X \rightarrow I$ such that $g(X-U) \subset \left\{0 \right\}$ and $g(x)=1$. Let $V=\prod \limits_{f \in C(X,I)} I_f$ where $I_f=I$ for all $f \ne g$ and $I_g=(0,1]$. Then $V_0=V \cap E(X)$ is open in $E(X)$ and $E(x) \in V_0$. We claim that $E^{-1}(V_0) \subset U$. To see this, pick $T=E(y) \in V_0$. Note that $T_g \in (0,1]$ since $T \in V_0$. If $y \in X-U$, then $T_g =0$. Thus we must have $y =E^{-1}(E(y)) \in U$.

The above discussion shows that any completely regular space $X$ is homeomorphic to a subspace of the product space $[0,1]^\mathcal{K}$ where $\mathcal{K}$ is the cardinality of $C(X,I)$. This establishes one direction of Theorem 1. The other direction is clear. Note that any cube is a compact Hausdorff space and any subspace of a compact Hausdorff space is completely regular.

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The Stone-Cech Compactification

Let $X$ be a completely regular space. Let $E$ be the evaluation as defined in $(1)$ above. According to the above discussion $E$ is a homeomorphism. Hence $E(X)$ is a topological copy of $X$ as a subspace of the product space $\prod \limits_{f \in C(X,I)} I_f$. Consider $\overline{E(X)}$ where the closure is taken in the product space. The Stone-Cech compactification of $X$ is denoted by $\beta X$ and is defined to be this closure $\beta X=\overline{E(X)}$.

See [2] and [6] for more information.

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Reference

1. Cech, E., On bicompact spaces, Ann. Math. (2) 38, 823-844, 1937.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
4. Stone, M. H., Applications of the Theorey of Boolean Rings to General Topology, Trans. Amer. Math. Soc., 41, 375-481, 1937.
5. Tychonoff, A., Uber die topologische von Raumen, Math., Ann., 102, 544-561, 1930.
6. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.