[proofplan] The strategy combines a Zorn's-lemma argument with a Hahn-Banach separation. We define a *face* (sometimes called extremal subset) of $K$: a non-empty closed subset $F \subseteq K$ such that any line segment in $K$ meeting $F$ in its interior is contained in $F$. The set of faces is partially ordered by reverse inclusion. We use Zorn's lemma to produce *minimal faces* — non-empty closed extremal subsets that admit no proper extremal sub-face. We then show every minimal face is a singleton, identifying its unique element as an extreme point of $K$. To prove $K = \overline{\operatorname{conv}}(\operatorname{Ext}(K))$, we argue by contradiction: if some $x_0 \in K$ lies outside $C := \overline{\operatorname{conv}}(\operatorname{Ext}(K))$, the geometric Hahn-Banach theorem in the locally convex space $X$ separates $\{x_0\}$ from $C$ by a continuous linear functional. The set $F_0 := \{x \in K : f(x) = \max_K f\}$ is a non-empty face of $K$ (compactness gives the maximum, and the affine structure of $f$ makes it a face) disjoint from $C$. By Zorn's lemma applied inside $F_0$, there is an extreme point inside $F_0$ — but this extreme point is in $K$ but not in $C$, contradicting the definition of $C \supseteq \operatorname{Ext}(K)$. [/proofplan] [step:Define faces and verify their basic properties] A non-empty subset $F \subseteq K$ is called a **face** (or extremal subset) of $K$ if: 1. $F$ is closed in $K$. 2. For all $x, y \in K$ and $t \in (0,1)$, if $tx + (1-t)y \in F$, then $x \in F$ and $y \in F$. Equivalently, $F$ is closed and any open line segment in $K$ that meets $F$ is entirely contained in $F$. A point $z \in K$ is an extreme point of $K$ — i.e.\ $z \in \operatorname{Ext}(K)$ — iff the singleton $\{z\}$ is a face of $K$ (which by Property 2 means: $z = tx + (1-t)y$ with $x, y \in K$, $t \in (0,1)$ forces $x = y = z$). *Faces inherit closure:* by Property 1, every face is closed in the compact $K$, hence compact. *Intersection of faces is a face (when non-empty):* let $\{F_\alpha\}$ be a family of faces with $\bigcap_\alpha F_\alpha \neq \varnothing$. The intersection is closed (intersection of closed sets). Property 2: if $tx + (1-t)y \in \bigcap_\alpha F_\alpha$ with $x, y \in K$ and $t \in (0,1)$, then $tx + (1-t)y \in F_\alpha$ for each $\alpha$, so $x, y \in F_\alpha$ for each $\alpha$, giving $x, y \in \bigcap_\alpha F_\alpha$. *Face of a face is a face:* if $F$ is a face of $K$ and $G \subseteq F$ is a face of $F$, then $G$ is a face of $K$. Property 1: $G$ closed in $F$ closed in $K$ gives $G$ closed in $K$. Property 2: if $tx + (1-t)y \in G \subseteq F$ with $x, y \in K$ and $t \in (0,1)$, then by $F$ being a face of $K$, $x, y \in F$; then by $G$ being a face of $F$, $x, y \in G$. *$K$ itself is a face of $K$:* $K$ is closed in itself and Property 2 holds with $x, y \in K$ giving $x, y \in K$. (We use this only as a base for Zorn's lemma; the argument-content of the property is unused here.) [/step] [step:Apply Zorn's lemma to produce a minimal non-empty face] Let $\mathcal{F}$ be the collection of all faces of $K$. Order $\mathcal{F}$ by reverse inclusion: $F \leq F'$ iff $F \supseteq F'$. We show every chain in $(\mathcal{F}, \leq)$ has an upper bound. Let $\{F_\alpha\}_{\alpha \in A}$ be a totally ordered chain (i.e.\ for any $\alpha, \beta \in A$, either $F_\alpha \subseteq F_\beta$ or $F_\beta \subseteq F_\alpha$). Set $F_\infty := \bigcap_\alpha F_\alpha$. [claim:$F_\infty$ is non-empty] $F_\infty \neq \varnothing$. [/claim] [proof] Each $F_\alpha$ is closed in $K$ (by Property 1) and non-empty (by definition of face). The chain has the finite intersection property: any finite subcollection $F_{\alpha_1}, \ldots, F_{\alpha_n}$ has, by total order, a minimum $F_{\alpha_{i^*}} \subseteq F_{\alpha_j}$ for all $j$, so $\bigcap_{j=1}^n F_{\alpha_j} = F_{\alpha_{i^*}} \neq \varnothing$. Since $K$ is compact, every collection of closed subsets of $K$ with the finite intersection property has non-empty total intersection. Hence $F_\infty = \bigcap_\alpha F_\alpha \neq \varnothing$. [/proof] By Step 1 (intersection of faces with non-empty intersection is a face), $F_\infty$ is a face of $K$. By definition $F_\infty \supseteq F_\alpha$ is false in general — actually we have $F_\infty \subseteq F_\alpha$ for all $\alpha$, which in our reverse order means $F_\infty \geq F_\alpha$ for all $\alpha$. So $F_\infty$ is an upper bound for the chain. By **Zorn's Lemma**, $(\mathcal{F}, \leq)$ contains a maximal element $F^*$ — equivalently, a minimal (under inclusion) non-empty face of $K$. [/step] [step:Show every minimal face is a singleton] Let $F^*$ be a minimal non-empty face of $K$, as in Step 2. We show $|F^*| = 1$. Suppose for contradiction that $F^*$ contains two distinct points $a, b \in F^*$. Since $X$ is a locally convex Hausdorff space, the points $a, b$ are distinguished by a continuous linear functional: by [Hahn-Banach for Locally Convex Spaces](/theorems/2638) applied to the disjoint compact convex set $\{a\}$ and the closed convex set $\{b\}$, there is a continuous linear $\Lambda : X \to \mathbb{R}$ (or $\mathbb{C}$) and $\alpha \in \mathbb{R}$ with $\operatorname{Re}\Lambda(a) > \alpha > \operatorname{Re}\Lambda(b)$. In particular $\operatorname{Re}\Lambda(a) \neq \operatorname{Re}\Lambda(b)$. Set $f := \operatorname{Re}\Lambda : X \to \mathbb{R}$, a continuous $\mathbb{R}$-linear functional. The face $F^*$ is compact (closed in compact $K$, by Step 1), and $f$ is continuous, so $f$ attains its maximum on $F^*$. Define \begin{align*} M := \max_{x \in F^*} f(x), \quad G := \{x \in F^* : f(x) = M\}. \end{align*} *$G$ is non-empty:* the maximum is attained. *$G$ is closed in $F^*$ (hence in $K$):* $G = F^* \cap f^{-1}(\{M\})$, intersection of the closed $F^*$ with the closed $f^{-1}(\{M\})$ (preimage of a closed singleton under a continuous map). *$G$ is a face of $F^*$:* take $x, y \in F^*$ and $t \in (0,1)$ with $tx + (1-t)y \in G$. Then $f(tx + (1-t)y) = t f(x) + (1-t) f(y) = M$. Since $f(x), f(y) \leq M$ (because $x, y \in F^*$), the convex combination achieves $M$ only when $f(x) = f(y) = M$. Hence $x, y \in G$. *$G$ is a face of $K$:* by Step 1, since $G$ is a face of $F^*$ which is a face of $K$. By Step 1, $G$ is a non-empty face of $K$ with $G \subseteq F^*$. By minimality of $F^*$ (in the reverse inclusion order: $F^*$ is maximal, so no face is *strictly* larger in the order $\geq$, i.e.\ no non-empty face is *strictly contained* in $F^*$), we must have $G = F^*$. But this contradicts $f(a) \neq f(b)$ for $a, b \in F^*$: if $G = F^*$, then $f$ takes the constant value $M$ on $F^*$, contradicting $f(a) \neq f(b)$. Hence $F^*$ has at most one element. Combined with non-emptiness, $|F^*| = 1$. Write $F^* = \{z\}$. By the equivalence in Step 1, $z$ is an extreme point of $K$. So every minimal non-empty face of $K$ is a singleton consisting of an extreme point. In particular, $\operatorname{Ext}(K) \neq \varnothing$ if $K \neq \varnothing$. [guided] The argument is the core of Krein-Milman: every minimal face is a single extreme point. *Why faces have extreme points.* If a minimal face $F^*$ contained two distinct points $a, b$, we could "carve out" a smaller face inside $F^*$ by maximizing a continuous linear functional that distinguishes $a$ from $b$. The set where the maximum is attained is itself a face (a standard fact: the maximum-set of a linear functional on a face is a face — affineness of $f$ converts equality on a convex combination into equality on each summand). This carved-out face is strictly smaller (it omits whichever of $a, b$ has smaller $f$-value), contradicting minimality. *Why a separating functional exists.* The locally convex space $X$ has a rich dual: by Hahn-Banach for locally convex spaces, the topological dual $X^*$ separates points (in fact separates compact convex sets from disjoint closed convex sets). Concretely: for $a \neq b$, the singletons $\{a\}, \{b\}$ are compact disjoint, separated by some continuous linear $\Lambda$. *Why the maximum is attained.* $F^*$ is closed in compact $K$, hence compact; the continuous function $f$ attains its supremum on the compact $F^*$. *Why $G$ is a face of $F^*$.* The check is: \begin{align*} M = f(tx + (1-t)y) = t f(x) + (1-t) f(y). \end{align*} Since both $f(x) \leq M$ and $f(y) \leq M$, the convex combination $t f(x) + (1-t) f(y)$ achieves $M$ only when $f(x) = f(y) = M$. (A convex combination of numbers, each at most $M$, equals $M$ iff each summand equals $M$.) Hence $x, y \in G$. *Why $G$ is a face of $K$.* By the transitive principle: face of face is face. We checked $G$ is a face of $F^*$, and $F^*$ is a face of $K$ (it is the face we constructed in Step 2). So $G$ is a face of $K$. *Why minimality forces $G = F^*$.* The claim was that $F^*$ is minimal among faces — i.e.\ any non-empty face contained in $F^*$ equals $F^*$. Since $G$ is a non-empty face contained in $F^*$, $G = F^*$. But $G = F^*$ means every point of $F^*$ achieves the maximum $M$ of $f$, i.e.\ $f$ is constant on $F^*$. This contradicts the choice of $f$ with $f(a) \neq f(b)$. [/guided] [/step] [step:Define $C := \overline{\operatorname{conv}}(\operatorname{Ext}(K))$ and verify $C \subseteq K$] Set $C := \overline{\operatorname{conv}}(\operatorname{Ext}(K))$, the closed convex hull (in $X$) of the set of extreme points of $K$. By Step 3, $\operatorname{Ext}(K) \subseteq K$ is non-empty (assuming $K$ non-empty). Since $K$ is convex and closed (closed because it is compact in the Hausdorff space $X$ — the locally convex topology is Hausdorff by definition), $K$ contains the convex hull $\operatorname{conv}(\operatorname{Ext}(K))$ and its closure: $C \subseteq K$. (If $K = \varnothing$, then $\operatorname{Ext}(K) = \varnothing$ and $C = \overline{\operatorname{conv}}(\varnothing) = \varnothing$, so $K = C = \varnothing$.) [/step] [step:Prove $K \subseteq C$ by Hahn-Banach separation and Step 3 applied to a face] Suppose for contradiction that $K \setminus C \neq \varnothing$, and pick $x_0 \in K \setminus C$. The set $C$ is convex (closed convex hull is convex) and closed (closed convex hull is closed). The singleton $\{x_0\}$ is compact and convex. The two are disjoint. By [Hahn-Banach for Locally Convex Spaces](/theorems/2638) (geometric form, for compact-vs-closed in a locally convex space), there exist a continuous $\mathbb{R}$-linear $f : X \to \mathbb{R}$ and $\alpha \in \mathbb{R}$ with \begin{align*} f(x_0) > \alpha > \sup_{c \in C} f(c). \tag{$*$} \end{align*} (In the complex case, $f$ is the real part of a continuous $\mathbb{C}$-linear functional, but we treat $f$ purely as a real-linear continuous functional below.) The function $f|_K$ is continuous on the compact $K$, so it attains its maximum: \begin{align*} M := \max_{x \in K} f(x), \quad F_0 := \{x \in K : f(x) = M\}. \end{align*} Note $M \geq f(x_0) > \alpha$. [claim:$F_0$ is a non-empty face of $K$] The set $F_0$ is a non-empty face of $K$. [/claim] [proof] $F_0$ is non-empty by the maximum-attainment argument above. $F_0$ is closed in $K$: $F_0 = K \cap f^{-1}(\{M\})$, intersection of closed sets. Property 2 (extremality): take $x, y \in K$ and $t \in (0, 1)$ with $tx + (1-t)y \in F_0$. Then \begin{align*} M = f(tx + (1-t)y) = t f(x) + (1-t) f(y). \end{align*} Since $f(x) \leq M$ and $f(y) \leq M$, the convex combination achieves $M$ only when $f(x) = f(y) = M$, i.e.\ $x, y \in F_0$. [/proof] By Step 1 (faces are themselves convex compact sets in $X$, since closed in $K$), $F_0$ is convex and compact. The set $F_0$ is itself a compact convex subset of the locally convex space $X$. We apply the entire argument of Steps 2-3 *internally* to $F_0$: by Zorn's lemma applied to the faces of $F_0$, there exists a minimal non-empty face $F_0^* \subseteq F_0$, which by the singleton argument of Step 3 (with $F_0$ in place of $K$) is a singleton $F_0^* = \{z_0\}$ with $z_0 \in \operatorname{Ext}(F_0)$. [claim:$z_0 \in \operatorname{Ext}(K)$] Every extreme point of the face $F_0$ is an extreme point of $K$. [/claim] [proof] Suppose $z_0 = tx + (1-t)y$ with $x, y \in K$ and $t \in (0,1)$. Since $F_0$ is a face of $K$ (Claim above), $x, y \in F_0$. Now $z_0 \in \operatorname{Ext}(F_0)$, so $x = y = z_0$. [/proof] So $z_0 \in \operatorname{Ext}(K) \subseteq C$. But $z_0 \in F_0$ implies $f(z_0) = M > \alpha$, while $z_0 \in C$ implies $f(z_0) \leq \sup_C f < \alpha$ by $(*)$. Hence $M > \alpha > M$, a contradiction. Therefore $K \setminus C = \varnothing$, i.e.\ $K \subseteq C$. [guided] We have established that every compact convex subset of $X$ has at least one extreme point (Step 3). To go from "exists extreme point" to "$K$ is the closed convex hull of its extreme points," we use a Hahn-Banach separation argument applied to a hypothetical bad point. *The contradiction strategy.* If $C := \overline{\operatorname{conv}}(\operatorname{Ext}(K))$ is *strictly smaller* than $K$, pick a witness $x_0 \in K \setminus C$. The two sets $\{x_0\}$ and $C$ are disjoint, with $\{x_0\}$ compact and $C$ closed convex. The geometric Hahn-Banach theorem in locally convex spaces produces a continuous linear $f$ strictly separating them: $f(x_0) > \alpha > \sup_C f$. *The face $F_0$.* The continuous function $f$ attains its maximum $M$ on the compact $K$. Let $F_0 := \{f = M\} \cap K$. We have $f(x_0) \leq M$ (since $x_0 \in K$) and $f(x_0) > \alpha$, so $M > \alpha$. The set $F_0$ is closed (preimage of a closed singleton intersected with closed $K$), and a face of $K$ by the convex-combination argument: convex combinations of points in $K$ that lie at the maximum of $f$ force each component to lie at the maximum. *Why $F_0$ contains an extreme point of $K$.* Apply Steps 2-3 internally to the compact convex $F_0$: it has at least one extreme point $z_0$. The crucial transfer principle: $\operatorname{Ext}(F_0) \subseteq \operatorname{Ext}(K)$. Reason: if $z_0 = tx + (1-t)y$ with $x, y \in K$, then since $F_0$ is a face of $K$, $x, y \in F_0$; since $z_0 \in \operatorname{Ext}(F_0)$, $x = y = z_0$. So $z_0 \in \operatorname{Ext}(K)$. *The contradiction.* We have $z_0 \in F_0$, so $f(z_0) = M > \alpha$. But $z_0 \in \operatorname{Ext}(K) \subseteq C$, so $f(z_0) \leq \sup_C f < \alpha$. The inequalities $f(z_0) > \alpha$ and $f(z_0) < \alpha$ contradict each other. So $K \setminus C = \varnothing$. [/guided] [/step] [step:Conclude] Steps 4 and 5 together yield $C = K$, i.e.\ $K = \overline{\operatorname{conv}}(\operatorname{Ext}(K))$. The "in particular" statement — $\operatorname{Ext}(K) \neq \varnothing$ when $K \neq \varnothing$ — was established in Step 3. [/step]