> [!definition] > > Let $I = [0, 1]$, given the [[Metric Topology|metric topology]]. For any non-empty set $A$, the [[Product Topology|product]] $I^A$ is a **cube**. By [[Tychonoff's Theorem]], $I^A$ is a [[Compactness|compact]] [[Hausdorff Space|Hausdorff space]]. > [!definition] > > Let $X$ be a [[Topological Space|topological space]] and $\cf \subset C(X, I)$, then $\cf$ **separates points and closed sets** if for any $E \subset X$ closed and $x \not \in E$, there exists $f \in \cf$ such that $f(x) \not\in \ol{f(E)}$. > 1. If $\cf$ separates points and closed sets, then there is a family $\mathcal G \subset C(X, I)$ where for any $E \subset X$ closed and $x \not\in E$, there exists $g \in \mathcal G$ such that $g(x) = 1$ and $g(E) = 0$. > 2. A [[Separation Axioms|T1]] space is [[Completely Regular Space|completely regular]] if and only if there exists $\cf \subset C(X, I)$ that separates points and closed sets. > > *Proof*. Let $E \subset X$ be closed and $x \not\in E$. By [[Urysohn's Lemma]], there exists $\phi \in C(I, I)$ such that $\phi(f(x)) = 1$ and $\phi(\ol{f(E)}) = 0$, and $g = \phi \circ f$ is the desired family. > [!theorem] > > Let $\cf \subset C(X, I)$, then there exists a canonical map > $ > e: X \to I^\cf \quad \pi_f(e(x)) = (e(x))(f) = f(x) > $ > known as the map from $X$ to $I^\cf$ **associated to** $\cf$. > 1. $e$ is [[Continuity|continuous]]. > 2. If $\cf$ separates points, then $e$ is injective. > 3. If $X$ is $T_1$ and $\cf$ separates points and closed sets, then $e$ is an embedding. > > *Proof*. Since $\pi_f \circ e = f$, $e$ is continuous in each coordinate. Therefore $e$ is continuous. > > Now suppose that $X$ is $T_1$ and $\cf$ separates points and closed sets. Since points are closed sets, $\cf$ also separates points, and $e$ is injective. Let $U \subset X$ be [[Open Set|open]], then there exists $f \in \cf$ separating $x$ from $U^c$. Let > $ > V = \pi_f^{-1}\braks{\ol{f(U^c)}}^c = \bracs{p \in I^\cf: \pi_f(p) \not\in \ol{f(U^c)}} > $ > be the image of the set of all points separated from $U^c$, then $V$ is open. Since $e(x) \in V \cap e(X) \subset e(U)$, $e(U)$ is a [[Neighbourhood|neighbourhood]] of $e(x)$ in $e(X)$. Therefore $e^{-1}$ is continuous as well. > [!theorem] > > Let $X$ be a topological space. > 1. $X$ is [[Completely Regular Space|completely regular]] if and only if it is [[Homeomorphism|homeomorphic]] to a subset of a compact Hausdorff space. > 2. If $X$ is [[Compactness|compact]] and [[Hausdorff Space|Hausdorff]], then $X$ is homeomorphic to a closed subset of a cube. > > *Proof*. Suppose that $X$ is completely regular, then $C(X, I)$ separates points and closed sets, which yields the desired embedding. Suppose that $X$ is homeomorphic to a subset of a compact Hausdorff space, then since the compact Hausdorff space is completely regular, so is $X$. > > If $X$ is compact and Hausdorff, then the image under the embedding is [[Closed Set|closed]].