[proofplan] We argue by contradiction. Suppose some extreme point $x_0 \in \operatorname{Ext}(K)$ does not lie in $S$. Replacing $S$ by its closure does not change $\overline{\operatorname{conv}}(S)$, so we may assume $S$ is closed and $x_0 \notin S$. The complement $X \setminus S$ is then an open neighbourhood of $x_0$, and the [Single Hyperplane Suffices at Extreme Points](/theorems/2664) theorem produces a slice $\{f < \alpha\} \cap K$ containing $x_0$ but disjoint from $S$. This forces $f(x) \geq \alpha$ on $S$, hence on $\overline{\operatorname{conv}}(S) = K$ by closedness and convexity. But $f(x_0) < \alpha$ for $x_0 \in K$ contradicts this. [/proofplan] [step:Reduce to the case $S$ closed] Replace $S$ with its closure $\overline{S}$. The closed convex hull is invariant under this replacement: $\overline{\operatorname{conv}}(\overline{S}) = \overline{\operatorname{conv}}(S)$. *Proof of invariance:* the inclusion $S \subseteq \overline{S}$ gives $\overline{\operatorname{conv}}(S) \subseteq \overline{\operatorname{conv}}(\overline{S})$. Conversely, $\overline{S} \subseteq \overline{\operatorname{conv}}(S)$ because $\overline{\operatorname{conv}}(S)$ is closed and contains $S$. Therefore $\operatorname{conv}(\overline{S}) \subseteq \overline{\operatorname{conv}}(S)$ (since $\overline{\operatorname{conv}}(S)$ is convex and contains $\overline{S}$), and taking closures: $\overline{\operatorname{conv}}(\overline{S}) \subseteq \overline{\operatorname{conv}}(S)$. Hence $\overline{\operatorname{conv}}(\overline{S}) = \overline{\operatorname{conv}}(S) = K$. Without loss of generality, replace $S$ by $\overline{S}$ — both inclusions $S \supset \operatorname{Ext}(K)$ and $\overline{S} \supset \operatorname{Ext}(K)$ are equivalent statements at this stage of the argument (we will derive a contradiction from $\operatorname{Ext}(K) \not\subseteq \overline{S}$, which is what we need for the original $S$). Henceforth assume $S$ is closed. [/step] [step:Suppose for contradiction that some extreme point lies outside $S$] Suppose $\operatorname{Ext}(K) \not\subseteq S$. Choose $x_0 \in \operatorname{Ext}(K) \setminus S$. The set $X \setminus S$ is open in $X$ (since $S$ is closed by Step 1), and contains $x_0$. So $V := X \setminus S$ is an open neighbourhood of $x_0$. [/step] [step:Apply the single-hyperplane theorem to obtain a slice avoiding $S$] The set $K$ is compact and convex by hypothesis, and $x_0 \in \operatorname{Ext}(K) \subseteq K$. We apply [Single Hyperplane Suffices at Extreme Points](/theorems/2664) to the locally convex space $X$, the compact convex set $K$, the extreme point $x_0$, and the open neighbourhood $V$ of $x_0$. The theorem's hypotheses are: $X$ is locally convex (given), $K$ compact and convex (given), $x_0 \in \operatorname{Ext}(K)$ (chosen in Step 2), and $V$ a neighbourhood of $x_0$ (verified in Step 2). All hypotheses are satisfied. The conclusion gives $f \in X^*$ and $\alpha \in \mathbb{R}$ with \begin{align*} x_0 \in \{x \in X : f(x) < \alpha\} \cap K \subset V = X \setminus S. \tag{$*$} \end{align*} In particular, $f(x_0) < \alpha$. [/step] [step:Derive $f \geq \alpha$ on $S$] Let $x \in S$. We show $f(x) \geq \alpha$. Since $S \subseteq K$ and $S \cap V = S \cap (X \setminus S) = \varnothing$, we have $x \in K$ and $x \notin V$. By $(*)$, every $y \in K$ with $f(y) < \alpha$ lies in $V$. Contrapositively, every $y \in K$ with $y \notin V$ satisfies $f(y) \geq \alpha$. Applying this to $x \in K \setminus V$: \begin{align*} f(x) \geq \alpha. \end{align*} Therefore $S \subseteq \{x \in X : f(x) \geq \alpha\}$. [/step] [step:Extend the inequality to $\overline{\operatorname{conv}}(S) = K$] The set $H := \{x \in X : f(x) \geq \alpha\} = f^{-1}([\alpha, \infty))$ is closed (preimage of a closed half-line under the continuous $f$) and convex (preimage of a convex set under the linear $f$). By Step 4, $S \subseteq H$. Hence $\operatorname{conv}(S) \subseteq H$ (since $H$ is convex and contains $S$), and $\overline{\operatorname{conv}}(S) \subseteq \overline{H} = H$ (since $H$ is closed). By hypothesis $\overline{\operatorname{conv}}(S) = K$, so \begin{align*} K \subseteq \{x \in X : f(x) \geq \alpha\}. \end{align*} In particular, since $x_0 \in K$, we have $f(x_0) \geq \alpha$. But Step 3 gave $f(x_0) < \alpha$. This contradicts $f(x_0) \geq \alpha$. [guided] The contradiction is the geometric heart of the partial converse: every closed convex set whose closed convex hull equals $K$ must contain all extreme points, because any "missed" extreme point is exposed by a hyperplane that cuts $K$ off from the missing point — but such a hyperplane is incompatible with the closed convex hull being all of $K$. *Why the reduction to closed $S$ is harmless.* The closed convex hull operation $\overline{\operatorname{conv}}$ is invariant under taking closures of its argument. Concretely, if $S' \subseteq \overline{S} \subseteq \overline{\operatorname{conv}}(S)$ (closure is monotone and $\overline{\operatorname{conv}}(S)$ is closed), then $\operatorname{conv}(\overline{S}) \subseteq \overline{\operatorname{conv}}(S)$ by convexity, and $\overline{\operatorname{conv}}(\overline{S}) \subseteq \overline{\operatorname{conv}}(S)$ by closedness. The reverse inclusion is monotonicity. Thus replacing $S$ by $\overline{S}$ preserves the hypothesis $\overline{\operatorname{conv}}(S) = K$. *Why a missing extreme point produces a slice.* The single-hyperplane theorem ([Single Hyperplane Suffices at Extreme Points](/theorems/2664)) is the engine: at an extreme point $x_0 \in K$, every neighbourhood of $x_0$ contains a slice $\{f < \alpha\} \cap K$. If $x_0 \notin S$ (closed), the complement $X \setminus S$ is a neighbourhood, and the theorem gives a slice fitting inside this neighbourhood — i.e.\ disjoint from $S$. *Why the slice forces a half-space inequality on $S$.* The slice contains $x_0$ but is disjoint from $S$. Equivalently: every point of $K$ where $f < \alpha$ lies outside $S$, so every point of $S \subseteq K$ satisfies $f \geq \alpha$. *Why this contradicts the closed convex hull condition.* The half-space $\{f \geq \alpha\}$ is closed and convex. It contains $S$, so it contains the closed convex hull of $S$, which by hypothesis is all of $K$. But $x_0 \in K$ with $f(x_0) < \alpha$ is excluded from this half-space — contradiction. [/guided] [/step] [step:Conclude] The contradiction in Step 5 rules out the assumption $\operatorname{Ext}(K) \not\subseteq S$ (after reduction to closed $S$ in Step 1). Hence $S \supseteq \operatorname{Ext}(K)$. For the "in other words" claim: by [Krein-Milman](/theorems/2661), $\overline{\operatorname{conv}}(\operatorname{Ext}(K)) = K$, so $\operatorname{Ext}(K)$ itself is a subset whose closed convex hull is $K$. The statement just proved shows that any other subset $S$ with $\overline{\operatorname{conv}}(S) = K$ contains $\operatorname{Ext}(K)$; therefore $\operatorname{Ext}(K)$ is the (unique) smallest such subset. [/step]