[proofplan] We construct $\beta X$ by embedding $X$ into the product cube $[0,1]^{C(X,[0,1])}$ using all continuous real-valued functions and taking the closure of the image. Complete regularity makes this evaluation map a topological embedding, while [Tychonoff's theorem](/page/Tychonoff's%20Theorem) makes the ambient cube compact. For a continuous map $f: X \to K$, we embed $K$ into its own function cube and use the coordinate functions $b \circ f$ for $b \in C(K,[0,1])$ to define a continuous map from $\beta X$ into that cube; closedness of the embedded copy of $K$ forces the image to land in $K$. Density of $\eta_X(X)$ gives uniqueness. [/proofplan] [step:Embed $X$ into the cube of its continuous $[0,1]$-valued functions] Let $I := [0,1]$ with its usual compact Hausdorff topology, and define the indexing set \begin{align*} A := C(X,I). \end{align*} Let \begin{align*} P_X := \prod_{a \in A} I \end{align*} with the [product topology](/page/Product%20Topology). By [Tychonoff's theorem](/theorems/953) (citing a result not yet in the wiki: Tychonoff theorem), $P_X$ is compact. Since each factor $I$ is Hausdorff, the product $P_X$ is Hausdorff. Define the evaluation map \begin{align*} e_X: X &\to P_X \\ x &\mapsto (a(x))_{a \in A}. \end{align*} For every $a \in A$, the coordinate projection \begin{align*} \pi_a: P_X &\to I \\ (p_b)_{b \in A} &\mapsto p_a \end{align*} satisfies $\pi_a \circ e_X = a$, hence $e_X$ is continuous by the defining property of the product topology. We claim that $e_X$ is a topological embedding. If $x,y \in X$ with $x \neq y$, then $\{y\}$ is closed because $X$ is Hausdorff. Complete regularity gives a continuous map $a \in C(X,I)$ such that $a(x)=1$ and $a(y)=0$. Thus $e_X(x) \neq e_X(y)$, so $e_X$ is injective. It remains to check that the topology on $X$ is the [subspace topology](/page/Subspace%20Topology) induced by $e_X$. Let $U \subset X$ be open and let $x \in U$. Put $F := X \setminus U$, which is closed in $X$. Complete regularity gives $a \in C(X,I)$ such that $a(x)=1$ and $a(F)\subset \{0\}$. Then \begin{align*} x \in a^{-1}((1/2,1]) \subset U, \end{align*} and $(1/2,1]$ is open in the subspace topology of $I$. Since \begin{align*} a^{-1}((1/2,1]) = e_X^{-1}\left(\pi_a^{-1}((1/2,1])\right), \end{align*} the original topology on $X$ is generated by inverse images of open subsets of $e_X(X)$. Therefore $e_X: X \to e_X(X)$ is a homeomorphism. [/step] [step:Define $\beta X$ as the compact closure of the evaluation image] Define \begin{align*} \beta X := \overline{e_X(X)} \subset P_X, \end{align*} where the closure is taken in $P_X$, and define \begin{align*} \eta_X: X &\to \beta X \\ x &\mapsto e_X(x). \end{align*} Since $P_X$ is compact Hausdorff and $\beta X$ is closed in $P_X$, the space $\beta X$ is compact Hausdorff. Since $e_X$ is a topological embedding, $\eta_X$ is a continuous embedding. By definition of $\beta X$ as the closure of $e_X(X)$, the image $\eta_X(X)$ is dense in $\beta X$. [/step] [step:Embed the compact Hausdorff target $K$ into its own function cube] Let $K$ be a compact [Hausdorff space](/page/Hausdorff%20Space). Define \begin{align*} B := C(K,I), \qquad P_K := \prod_{b \in B} I. \end{align*} Define the evaluation map \begin{align*} e_K: K &\to P_K \\ y &\mapsto (b(y))_{b \in B}. \end{align*} The same coordinate-projection argument shows that $e_K$ is continuous. Because every compact Hausdorff space is completely regular by [Urysohn's lemma](/theorems/887) (citing a result not yet in the wiki: Urysohn lemma for compact Hausdorff spaces), the functions in $C(K,I)$ separate points and generate the topology of $K$. Hence $e_K$ is a topological embedding. Since $K$ is compact and $P_K$ is Hausdorff, $e_K(K)$ is compact and therefore closed in $P_K$, and \begin{align*} e_K: K \to e_K(K) \end{align*} is a homeomorphism. [/step] [step:Construct the extension of $f$ by pulling back coordinates] Let \begin{align*} f: X \to K \end{align*} be continuous. For each $b \in B$, define \begin{align*} a_b: X &\to I \\ x &\mapsto b(f(x)). \end{align*} Since $b$ and $f$ are continuous, $a_b \in A$. Define \begin{align*} R_f: P_X &\to P_K \\ (p_a)_{a \in A} &\mapsto (p_{a_b})_{b \in B}. \end{align*} For each $b \in B$, the $b$-coordinate of $R_f$ is the coordinate projection $\pi_{a_b}$ on $P_X$, so $R_f$ is continuous by the product topology. For every $x \in X$, \begin{align*} R_f(\eta_X(x)) &= R_f((a(x))_{a \in A}) \\ &= (a_b(x))_{b \in B} \\ &= (b(f(x)))_{b \in B} \\ &= e_K(f(x)). \end{align*} Thus \begin{align*} R_f(\eta_X(X)) \subset e_K(K). \end{align*} Since $\eta_X(X)$ is dense in $\beta X$, $R_f$ is continuous, and $e_K(K)$ is closed in $P_K$, it follows that \begin{align*} R_f(\beta X) \subset e_K(K). \end{align*} Define \begin{align*} \bar f: \beta X &\to K \\ z &\mapsto e_K^{-1}(R_f(z)). \end{align*} This map is continuous because $R_f|_{\beta X}$ is continuous and $e_K^{-1}: e_K(K) \to K$ is continuous. For $x \in X$, \begin{align*} \bar f(\eta_X(x)) &= e_K^{-1}(R_f(\eta_X(x))) \\ &= e_K^{-1}(e_K(f(x))) \\ &= f(x). \end{align*} Therefore $\bar f \circ \eta_X = f$. [/step] [step:Use density to prove uniqueness of the extension] Let \begin{align*} g: \beta X \to K \end{align*} be a continuous map satisfying $g \circ \eta_X = f$. Then the two continuous maps \begin{align*} e_K \circ g,\ e_K \circ \bar f: \beta X \to P_K \end{align*} agree on $\eta_X(X)$, because for every $x \in X$, \begin{align*} (e_K \circ g)(\eta_X(x)) &= e_K(f(x)) \\ &= (e_K \circ \bar f)(\eta_X(x)). \end{align*} The set $\eta_X(X)$ is dense in $\beta X$, and $P_K$ is Hausdorff. Hence two continuous maps from $\beta X$ to $P_K$ that agree on $\eta_X(X)$ agree on all of $\beta X$. Therefore \begin{align*} e_K \circ g = e_K \circ \bar f. \end{align*} Since $e_K$ is injective, $g=\bar f$. This proves uniqueness and completes the proof of the universal property. [/step]