[proofplan] The proof first handles the empty-domain case, where the uniform bound is vacuous. When $K$ is nonempty, compactness and continuity imply that the continuous envelope $g$ is bounded above on $K$. The pointwise domination $|f_n(x)| \le g(x)$ then transfers this single bound from $g$ to every function $f_n$, which is precisely uniform boundedness. [/proofplan] [step:Dispose of the empty compact set] If $K = \varnothing$, take $M := 0$. There are no pairs $(n,x) \in \mathbb N \times K$, so \begin{align*} |f_n(x)| \le M \end{align*} holds for every $n \in \mathbb N$ and every $x \in K$ by vacuity. Hence $(f_n)_{n=1}^{\infty}$ is uniformly bounded on $K$ in this case. [/step] [step:Use compactness to bound the continuous envelope] Assume now that $K \neq \varnothing$. We prove directly that $g$ is bounded above on $K$. For each $y \in K$, continuity of $g: K \to [0,\infty)$ at $y$ gives a relatively [open set](/page/Open%20Set) $U_y \subset K$ such that $y \in U_y$ and \begin{align*} g(x) < g(y) + 1 \end{align*} for every $x \in U_y$. The family $(U_y)_{y \in K}$ is an [open cover](/page/Open%20Cover) of $K$ in the [subspace topology](/page/Subspace%20Topology). Since $K$ is compact, there exist points $y_1,\dots,y_m \in K$ such that \begin{align*} K \subset \bigcup_{j=1}^{m} U_{y_j}. \end{align*} Define \begin{align*} B := 1 + \max\{g(y_1),\dots,g(y_m)\}. \end{align*} The number $B$ is finite because each $g(y_j)$ is a real number. If $x \in K$, choose $j \in \{1,\dots,m\}$ with $x \in U_{y_j}$. Then \begin{align*} g(x) < g(y_j) + 1 \le B. \end{align*} Thus $g(x) \le B$ for every $x \in K$. [guided] We need one constant that bounds $g(x)$ for all $x \in K$. Rather than assuming that $g$ attains a maximum, we extract a bound directly from compactness. Fix $y \in K$. Since $g: K \to [0,\infty)$ is continuous at $y$, the inverse image under $g$ of the open interval $(-\infty,g(y)+1)$ is open in $K$ and contains $y$. Therefore there is a relatively open set $U_y \subset K$ such that $y \in U_y$ and \begin{align*} g(x) < g(y) + 1 \end{align*} for every $x \in U_y$. The sets $U_y$ cover $K$, because each point $y \in K$ belongs to its own neighbourhood $U_y$. Compactness of $K$ means that this open cover has a [finite subcover](/page/Finite%20Subcover), so there are points $y_1,\dots,y_m \in K$ satisfying \begin{align*} K \subset \bigcup_{j=1}^{m} U_{y_j}. \end{align*} Now define the finite constant \begin{align*} B := 1 + \max\{g(y_1),\dots,g(y_m)\}. \end{align*} This number is finite because it is built from finitely many real values of the function $g$. Let $x \in K$ be arbitrary. Since the finitely many sets $U_{y_j}$ cover $K$, there exists $j \in \{1,\dots,m\}$ such that $x \in U_{y_j}$. By the defining property of $U_{y_j}$, \begin{align*} g(x) < g(y_j) + 1. \end{align*} By the definition of $B$, \begin{align*} g(y_j) + 1 \le B. \end{align*} Combining the two inequalities gives \begin{align*} g(x) < B. \end{align*} In particular, $g(x) \le B$ for every $x \in K$. Thus the continuous envelope is bounded above on the compact set $K$. [/guided] [/step] [step:Transfer the envelope bound to the whole sequence] Define \begin{align*} M := \max\{B,0\}. \end{align*} Then $M \ge 0$ and, since $g: K \to [0,\infty)$, the estimate from the previous step gives \begin{align*} 0 \le g(x) \le M \end{align*} for every $x \in K$. For every $n \in \mathbb N$ and every $x \in K$, the assumed pointwise domination gives \begin{align*} |f_n(x)| \le g(x) \le M. \end{align*} Hence there exists a constant $M \ge 0$ such that $|f_n(x)| \le M$ for all $n \in \mathbb N$ and all $x \in K$. Therefore $(f_n)_{n=1}^{\infty}$ is uniformly bounded on $K$. [/step]