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An extreme point $x$ of a convex set $K$ is one for which $x=(1-t)y+tz$, $y,z\in K$ and $0<t<1$ imply $y=z=x$. It is enough to test midpoint decompositions: a nontrivial interior point of a segment is itself the midpoint of a shorter nontrivial segment contained in that segment.

We prove the Krein-Milman theorem. Assume first that $K$ is nonempty. A face of a convex set $K$ means a convex subset $F$ with the property that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. The family of nonempty compact faces of $K$, including $K$ itself, has a minimal member by the Zorn lemma: every inclusion-decreasing chain has a nonempty intersection by compactness and the finite intersection property, and that intersection is again a compact face.

Let $M$ be such a minimal face. If it contains two distinct points, the continuous dual of a Hausdorff locally convex space separates them, by the permitted Hahn-Banach theorem. Choose a continuous real linear functional $\ell$ that is not constant on $M$. It attains a maximum on the compact set $M$, andis a nonempty proper compact face of $M$, and hence a face of $K$. Indeed, equality at the maximum in a convex combination forces both endpoint values to be maximal. This contradicts minimality. Thus $M$ is a singleton, and its point is extreme in $K$. The same argument applied within any nonempty compact face proves that every such face contains an extreme point of $K$.

Let $C=\overline{\mathrm{conv}}(\mathrm{ext}K)$. It is contained in $K$, since the latter is closed and convex, and extreme points exist when $K$ is nonempty. If $x_0\in K\setminus C$, the permitted separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_C\ell$. Let $F$ be its maximizer face on $K$. This nonempty compact face contains an extreme point $v$ of $K$, butcontradicting $v\in \mathrm{ext}K\subseteq C$. ThereforeFor empty $K$, the equality holds with the usual empty convex-hull convention. This proves both the extreme-point existence step and the closed-convex-hull step; no compactness of the set of extreme points itself is assumed.

The real absolutely summable sequence space is ${\ell }^1=\{x=(x_j):{\sum }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_1={\sum }_j|x_j|$. The real l-infinity sequence space is ${\ell }^{\infty }=\{x=(x_j):{\sup }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_{\infty }={\sup }_j|x_j|$. Their completeness is allowed as given; their extreme points still require proof.

For ${\ell }^1$, an interior point of the unit ball is not extreme because small opposite perturbations in one coordinate stay in the ball. At a boundary point with two nonzero coordinates $i,j$, choose $0<\epsilon <\min (|x_i|,|x_j|)$ and putThe two distinct vectors $x+z,x-z$ both have norm one and average to $x$, so $x$ is not extreme. A remaining boundary point must be $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, the $j$th coordinate forces $y_j=z_j=1$, exhausting both norm budgets and forcing all other coordinates to zero. The negative case is identical. ThusFor ${\ell }^{\infty }$, if $|x_j|<1$ in even one coordinate, small opposite changes to that coordinate give a nontrivial midpoint decomposition within the ball. If every $x_j=\pm 1$, a midpoint decomposition forces both summands to have that same value in every coordinate, so they equal $x$. HenceThese are precisely the extreme points of real sequence-space unit balls for the two spaces under consideration.

A point $x$ of a convex set $K$ is an extreme point if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.

Here is a proof of the Krein-Milman theorem. The empty-set case is immediate, so take $K\ne \varnothing$. A face of a convex set is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the finite intersection property. The intersection remains a compact face. The Zorn lemma, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.

If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real linear functional $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same minimal compact face argument inside any nonempty compact face shows that each such face contains an extreme point of $K$.

Let $H=\overline{\mathrm{conv}}(\mathrm{ext}K)$. Since $K$ is compact, hence closed in the Hausdorff space, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous linear functional withThe maximizer face of $\ell$ on $K$ contains an extreme point $e$ by the preceding argument. Then $e\in H$ but $\ell (e)={\max }_K\ell \ge \ell (p)>{\sup }_H\ell$, a contradiction. ThereforeAll closures refer to the given locally convex topology, and compactness is what justified both maxima and chain intersections.

For the real l-infinity sequence space, if $|x_j|<1$ for any coordinate, choose $0<ϵ\le 1-|x_j|$. The two distinct sequences $x\pm ϵe_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. ThusFor the space of sequences converging to zero, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, givingThese are extreme points of real sequence-space unit balls. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not compact in the norm topology, as the coordinate vectors have mutual distance one.

A point $x$ of a set $S$ is an extreme point if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a convex set it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set ${\mathbb{R}}^2$ has no extreme points, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.

For the real l-infinity sequence space, a vector in the closed unit ball is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta \le 1-|x_j|$. The distinct vectors $x\pm \delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\Vert y{\Vert }_{\infty },\Vert z{\Vert }_{\infty }\le 1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore

For the real absolutely summable sequence space, any point with $\Vert x{\Vert }_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\Vert x{\Vert }_1$. If $\Vert x{\Vert }_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta <\min (|x_j|,|x_k|)$ and putBoth perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\Vert x+h{\Vert }_1=\Vert x-h{\Vert }_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus

For the real l2 sequence space, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\Vert x{\Vert }_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The parallelogram law givesThe left side is at most one, so $h=0$ and $y=z=x$. ConsequentlyThese are the extreme points of real sequence-space unit balls: signs in ${\ell }^{\infty }$, coordinate spikes in ${\ell }^1$, and the whole sphere in ${\ell }^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.