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A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

If an open set meets $\overline{A}$, it meets $A$. Thus if two open sets meet the closure of an irreducible set, they both meet $A$, their intersection meets $A$ by irreducibility, and therefore it also meets $\overline{A}$. This proves irreducibility of the closure. A singleton is irreducible, so its closure is an irreducible closed subset.

In a Hausdorff space, two distinct points have disjoint open neighborhoods. An irreducible set cannot contain both, so every nonempty irreducible closed subset is a singleton, which is closed and has a unique generating point. Hence every Hausdorff space is a sober topological space. An infinite set with the cofinite topology is itself irreducible, because any two nonempty open sets meet, but every singleton is closed. Its whole space is therefore not the closure of a singleton, proving that it is not sober.

For the proposed sobrification, let $\hat{X}$ consist of nonempty irreducible closed subsets. The basic operations areThe last identity uses irreducibility of each member of $\hat{X}$. Hence these sets already form a topology. The map $U\mapsto \hat{U}$ is injective: $\overline{\{x\}}$ meets an open $U$ exactly when $x\in U$, so equality of the hats forces equality of the original opens. It is surjective onto the topology just constructed by definition.

Now let $C$ be a nonempty irreducible closed subset of $\hat{X}$. Its complement is $\hat{W}$ for a unique open $W\subseteq X$. Put $F=X\setminus W$. ThenIf opens $U,V$ meet $F$, choose points in those intersections. Their singleton closures lie in $C$, so $\hat{U},\hat{V}$ meet $C$. Irreducibility of $C$ forces $\widehat{U\cap V}$ to meet $C$, and hence $U\cap V$ meets $F$. Thus $F$ is nonempty, closed and irreducible, so $F$ is itself a point of $\hat{X}$.

The closure of this point in $\hat{X}$ is exactly $\{A:A\subseteq F\}$: inclusion makes every open meeting $A$ meet $F$, whereas a point of $A\setminus F$ is separated by the open $X\setminus F$. Hence $C=\overline{\{F\}}$. Any other point $G$ with the same closure satisfies $G\subseteq F$ and $F\subseteq G$, forcing $F=G$. This proves the constructed space is sober.

A completely regular Hausdorff space is a Hausdorff space in which each point $x$ outside a closed set $F$ is separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. Put $E=C_b(X)$ and define the evaluation character by ${\delta }_x(g)=g(x)$. The weak-star topology on $E^{\prime }$ is the topology of pointwise convergence on $E$: its coordinate maps $p\mapsto p(g)$ are continuous, and generate the topology. It is Hausdorff, and the dual unit ball is compact in it by the Banach-Alaoglu theorem.

For nonempty $X$, $\Vert {\delta }_x\Vert =1$ by the supremum norm bound and the constant function $1$. All the coordinates of $x\mapsto {\delta }_x$ are continuous, so $\delta$ is continuous. Complete regularity separates distinct points, proving injectivity. To prove continuity of the inverse onto its image, let $x\in O$ with $O$ open in $X$. Separating $x$ from $X\setminus O$ gives a bounded continuous $u$ withSuch neighborhoods prove that $\delta$ is a homeomorphism onto $\delta (X)$. This qualification is essential: it is not onto the whole dual, since the zero functional is not an evaluation on a nonempty space.

Define the Stone-Čech compactification to beThe Banach-Alaoglu theorem makes this a compact Hausdorff space, with $X$ identified with its dense evaluation copy. For $g\in C_b(X)$, the coordinate function $\tilde{g}(p)=p(g)$ is a continuous extension. Density gives $\Vert \tilde{g}{\Vert }_{\infty }=\Vert g{\Vert }_{\infty }$. Conversely any $h\in C(\beta X)$ restricts to an element $g$ of $C_b(X)$, and $h=\tilde{g}$ because they agree on the dense subset $X$. Restriction and coordinate extension are inverse linear isometries, henceThey also preserve products: the corresponding continuous functions agree on $X$ and hence everywhere.

For the universal extension property, let $f:X\to K$ be continuous and $K$ a compact Hausdorff space. Pullback defines a bounded linear operator $T:C(K)\to C_b(X)$ by $Tg=g\circ f$. Its dual map $T^{\prime }:E^{\prime }\to C(K{)}^{\prime }$ is weak-star continuous, since $(T^{\prime }p)(g)=p(Tg)$. Evaluation ${\delta }_K:K\to C(K{)}^{\prime }$ is a homeomorphism onto a compact, thus closed, subset: a compact Hausdorff space is completely regular, by the Urysohn lemma. Now $T^{\prime }{\delta }_x={\delta }_K(f(x))$, soTherefore $\overline{f}={\delta }_K^{-1}\circ T^{\prime }{|}_{\beta X}$ is the required continuous extension. Two continuous maps into the Hausdorff space $K$ agreeing on the dense subset $X$ agree everywhere, proving uniqueness.

If $X$ is open in $\beta X$, then it is a locally compact space: around any $x$, regularity of the compact Hausdorff space $\beta X$ gives an open neighborhood whose closure is compact and lies in $X$. Conversely, suppose $X$ is locally compact. Choose an open neighborhood $U$ of $x$ in $X$ with compact closure $C\subseteq X$. There is an open $O\subseteq \beta X$ with $O\cap X=U$. Density of $X$ implies $O\subseteq {\overline{U}}^{\beta X}$, while compactness of $C$ makes it closed in $\beta X$, so ${\overline{U}}^{\beta X}\subseteq C$. Thus $x\in O\subseteq X$. This provesThe empty-space cases are immediate.

The final printed assertion, for an arbitrary subset of $\beta X$, is false. Take the infinite discrete space $X=\mathbb{N}$. It is not compact, so its compactification has a point $p\in \beta \mathbb{N}\setminus \mathbb{N}$. The singleton $\{p\}$ is closed, hence equals its own closure. It is not open, since every nonempty open subset of $\beta \mathbb{N}$ meets the dense copy of $\mathbb{N}$. This is a counterexample.

The valid result holds for subsets $D\subseteq X$, and more generally for open subsets of $\beta X$. For $D\subseteq X$, its indicator is bounded continuous because $X$ is discrete. Its extension $\widetilde{1_D}$ is still idempotent by density, so it takes only the values zero and one. The set $H=\{\widetilde{1_D}=1\}$ is clopen and has $H\cap X=D$. Density then gives $H={\overline{D}}^{\beta X}$. If $O$ is open in $\beta X$, the set $O\cap X$ is dense in $O$, so $\overline{O}=\overline{O\cap X}$ is clopen by the result just proved. Consequently $\beta X$ is extremally disconnected, which means the closure of every open set is open. This is the discrete Stone-Čech compactifications are extremally disconnected theorem, with the essential open-set qualification.

A topological group is a group equipped with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous maps. The product uses the product topology. Translation and inversion are therefore homeomorphisms. Hausdorffness is sometimes included in the definition; the two open-subgroup arguments below do not require it.

For the inverse system, take a directed set $I$, topological spaces $X_i$, and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=\mathrm{id}$ and $f_{ik}=f_{ij}{f}_{jk}$. An inverse limit is a space $X$ with compatible continuous maps ${\pi }_i:X\to X_i$ such that every compatible family $q_i:Y\to X_i$ factors uniquely as $q_i={\pi }_{i}q$ through a continuous map $q:Y\to X$. Existence follows from the concrete constructionwith the subspace topology of the product topology. The only possible map $q$ is $q(y)=(q_i(y){)}_i$. Compatibility puts its image in $X$, and coordinatewise continuity makes it continuous. This proves the universal property of an inverse limit, even when $X$ is empty. Given two inverse limits, their universal properties supply maps each way; the composites have the same coordinates as the identity, so uniqueness makes them identities. The inverse limit exists and is unique up to the unique homeomorphism respecting its projections.

A profinite group can be defined as an inverse limit of finite discrete groups, or as a compact Hausdorff space that is a totally disconnected space, equipped with continuous group operations. For the first implication, use the Tychonoff theorem. The compatibility equalities define a closed subgroup of the compact Hausdorff product. Two distinct points are distinguished by a finite discrete coordinate, so no connected subset contains both. Consequently the inverse limit is a compact Hausdorff totally disconnected topological group.

For the converse, use the topological result that a compact Hausdorff totally disconnected space has a basis of clopen sets. Choose a clopen set $U$ containing $1$ inside any specified identity neighbourhood. For each $u\in U$, continuity gives an identity neighbourhood $V_u$ and an open neighbourhood $W_u$ of $u$ with $V_u{W}_u\subseteq U$. Finitely many $W_u$ cover the compact set $U$. Intersect their $V_u$ and its inverse to obtain a symmetric identity neighbourhood $V$ with $VU\subseteq U$. Symmetry also gives $V^{-1}U\subseteq U$, and hence $vU=U$ for every $v\in V$. The setis a subgroup containing $V$, so it is open. It is contained in $U$, since $1\in U$ and $h=h\cdot 1\in hU=U$. By part (ii), $H$ has finite index. Its normal core of a subgroup is a finite intersection of conjugates of $H$, hence an open normal subgroup $N$ contained in $U$. This proves the open normal subgroup basis of a profinite group.

The natural group homomorphismis injective because the intersection of these $N$ is $\{1\}$. For a compatible family of cosets, any finitely many have nonempty intersection: their intersection subgroup is another indexing subgroup, and its prescribed coset lies in each. The finite intersection property and compactness give a point belonging to all the cosets, proving surjectivity. Finally a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The two definitions of profinite group are equivalent.

The general adjoint functor theorem says that, for a functor $U:\mathcal{A}\to \mathcal{B}$ between locally small categories, with $\mathcal{A}$ a complete category, $U$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The solution-set condition means that for each $B\in \mathcal{B}$ there is a set of arrows $b_i:B\to U{A}_i$ such that every $b:B\to UA$ factors as $U(a)b_i$ for some $i$ and some $a:A_i\to A$.

Here is the smallness argument behind the theorem. A complete category that is locally small and has a weakly initial set $(W_i)$ has an initial object. Form the product in a category $W={\prod }_i{W}_i$. It is weakly initial, since a projection $W\to W_i$ followed by a chosen arrow $W_i\to X$ gives an arrow $W\to X$. Take the simultaneous equalizer $e:E\to W$ of all endomorphisms of $W$ with $1_W$; the family is a set by local smallness. The object $E$ is also weakly initial. For parallel arrows $a,b:E\rightrightarrows X$, take their equalizer $j:Y\to E$ and a map $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ gives $ejte=e$. Cancellation of the monomorphism $e$ gives $jte=1_E$. Thus the monomorphism $j$ has a right inverse and is an isomorphism, so $a=b$. Existence of arrows out of $E$ was already established, proving the initial-object lemma for complete categories with a weakly initial set.

If $U$ preserves small categorical limits, the comma category $(B\downarrow U)$ is complete: take the limiting object in $\mathcal{A}$, and use preservation to assemble the arrows from $B$. It is locally small, and the solution-set condition provides a weakly initial set. Its initial object is an arrow ${\eta }_B:B\to ULB$ with a natural bijectionFor a morphism $v:B\to B^{\prime }$, define $Lv$ by $U(Lv){\eta }_B={\eta }_{B^{\prime }}v$. Uniqueness proves functoriality and naturality, giving a left adjoint $L⊣U$. Conversely, a right adjoint preserves categorical limits, because applying the adjunction converts a limiting cone over a diagram into the corresponding limiting cone of hom-sets. Its adjunction unit at $B$ is a singleton solution set. This proves both directions of the general adjoint functor theorem.

Apply it to the inclusion $J:\mathbf{KHaus}\hookrightarrow \mathbf{Top}$, where $\mathbf{KHaus}$ is the full subcategory of compact Hausdorff spaces. A small product in a category of compact Hausdorff spaces is again compact Hausdorff by the Tychonoff theorem. An equalizer of two continuous functions into a Hausdorff space is closed, hence compact Hausdorff. Products and equalizers construct all small categorical limits, and the inclusion preserves these constructions. Both categories are locally small.

Fix a topological space $X$ and a continuous function $f:X\to K$ into a compact Hausdorff space. The closure $K_f=\overline{f(X)}$ is compact Hausdorff and has a dense subset of cardinality at most $|X|$. The permitted cardinal estimate givesFor $X=\varnothing$, use $K_f=\varnothing$. Choose a set of representatives of all compact Hausdorff topologies on underlying sets of cardinality at most this bound, and all continuous functions from $X$ to those representatives. Every $f$ factors through one of them, using the inclusion $K_f\hookrightarrow K$. This is the solution-set condition. Consequently the inclusion of compact Hausdorff spaces has a left adjoint, the compact Hausdorff reflection:The universal arrow $X\to RX$ has dense image: factoring it through its closed image and applying uniqueness supplies an inverse to that image inclusion. It need not be an embedding when $X$ is not sufficiently separated; the assertion is a compact Hausdorff reflection, rather than a claim that every topological space embeds in a compact Hausdorff space.

For completeness, the alternative Special adjoint functor theorem can be proved from the same initial-object argument. Its limit form assumes that $\mathcal{A}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$; then $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{B}$ locally small, has a left adjoint precisely when it preserves small limits. To prove sufficiency, start with $b:B\to UA$ and intersect all subobjects $M\hookrightarrow A$ through which $b$ factors after applying $U$. Well-poweredness makes this a small intersection of subobjects, and preservation of limits gives a smallest supporting object $(M,b_M)$. For maps $s,t:M\rightrightarrows Q_i$, equality $U(s)b_M=U(t)b_M$ makes their equalizer another supporting subobject. Minimality forces that equalizer to be invertible, so $s=t$. Thereforeis an injective function. The evaluation embedding into cogenerator products embeds $M$ in ${\prod }_i{Q}_i^{S_i}$, where $S_i$ is the realized subset of the fixed set $\mathcal{B}(B,U{Q}_i)$. There are only a set of choices of these subsets, only a set of subobjects of each resulting product, and only a set of arrows from $B$ to each $U$-image. They give a solution set, so the already proved general adjoint functor theorem applies. Necessity is again preservation of limits by a right adjoint. For $\mathbf{KHaus}$, monomorphisms are embeddings onto closed subspaces, and $[0,1]$ is a coseparator because continuous functions to it separate points. Thus the Special adjoint functor theorem gives the same reflection directly.

Take $\mathbb{N}=\{1,2,\dots \}$. We first prove the Ellis–Numakura lemma in the one-sided continuity convention needed here. Let $S$ be a nonempty compact Hausdorff space with an associative multiplication for which each map $x\mapsto xa$ is continuous. Among its nonempty closed subsemigroups, there is an inclusion-minimal one, $M$: a decreasing chain has nonempty intersection by compactness, and that intersection is a closed subsemigroup, so Zorn's lemma applies.

Fix $a\in M$. The set $Ma$ is nonempty and compact, hence closed in the Hausdorff space, and is a subsemigroup becauseMinimality gives $Ma=M$. Thus the set $T=\{x\in M:xa=a\}$ is nonempty. It is closed by the stated continuity, and it is a subsemigroup, since $(xy)a=x(ya)=xa=a$ for $x,y\in T$. Minimality again gives $T=M$, so $a\in T$ and $\boxed{a^2=a}$, giving a semigroup idempotent. If one instead uses the opposite one-sided continuity convention, the same proof uses $aM$ and $\{x:ax=a\}$; no joint continuity is required.

Apply this lemma to the Stone-Čech compactification of the natural numbers with its given addition. In the addition on the Stone-Čech compactification of the natural numbers convention,The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply there exists an idempotent ultrafilter $p$ with $p+p=p$.

On the positive integers such a $p$ is nonprincipal: a principal ultrafilter at $n$ adds to itself to give the one at $2n$, which is different. Hence every cofinite set belongs to $p$. If zero is included in one's convention for $\mathbb{N}$, use the same compact semigroup argument on the closed space of nonprincipal ultrafilters: it is nonempty by compactness of the infinite discrete set's compactification, and the displayed addition sends two free ultrafilters to a free ultrafilter. This avoids the trivial principal idempotent at zero.

To deduce Hindman's theorem, let $A$ be the unique colour class belonging to $p$. DefineIdempotence gives $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma also gives $A^{\ast }-n\in p$ whenever $n\in A^{\ast }$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ givesIntersecting this set with $A-n$ produces exactly $A^{\ast }-n$.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_r$ have been chosen with every nonempty finite sum in $A^{\ast }$. The finite intersectionbelongs to $p$, and is therefore nonempty. Choose $x_{r+1}$ from it. The new sums are $x_{r+1}$ and $s+x_{r+1}$, and they all lie in $A^{\ast }$. Induction yieldsThis proves every finite colouring of the positive integers admits a monochromatic finite-sums set generated by a strictly increasing infinite sequence, the required Hindman theorem.

False without a separation assumption. Give $X=\{0,1\}$ the indiscrete topology $\{\varnothing ,X\}$, and take $Y=\{0\}$. Both spaces are finite and hence compact, but $Y$ is not a closed set because $X\setminus Y=\{1\}$ is not open. In a Hausdorff space, a compact subset would be closed; that hypothesis is absent here.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

A compact space is one for which every open cover has a finite subcover. A Hausdorff space has disjoint open neighbourhoods of any two distinct points.

Let $K$ be a compact subspace of a Hausdorff space $X$, and let $x\notin K$. For each $y\in K$, choose disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$. Finitely many $U_{y_1},\dots ,U_{y_m}$ cover $K$. Their corresponding intersection $V_{y_1}\cap \cdots \cap V_{y_m}$ is an open neighbourhood of $x$ disjoint from $K$. Thus $X\setminus K$ is open, proving compact subspaces of a Hausdorff space are closed.

In particular, $C_1\cap C_2$ is closed as a subspace of compact $C_1$. A closed subset of a compact space is compact: add its open complement to an open cover, apply compactness to the whole space, and discard that extra open set. Hence $C_1\cap C_2$ is compact.

In the cocountable topology on $\mathbb{R}$, two nonempty open sets always intersect. Their two countable complements have countable union, which cannot exhaust the uncountable real line. Therefore this space is not Hausdorff.

Every finite subset is compact in any topology. Conversely, from any infinite subset $C$ select distinct points $c_1,c_2,\dots$ and defineEach $U_n$ is cocountable and open; the sets increase and their union is all of $\mathbb{R}$. They therefore cover $C$, but a finite selection is contained in some $U_N$, which misses the infinitely many $c_j$ with $j\ge N$. No finite subcover exists. Thus the compact subsets of a cocountable space are

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

Let $K$ be a compact set of a Hausdorff space $X$. For any $x\notin K$ and each $y\in K$, the Hausdorff property gives disjoint open sets $U_y\ni y$ and $V_y\ni x$. The sets $U_y$ cover $K$; compactness supplies a finite subcover $U_{y_1},\dots ,U_{y_r}$. The open set $V={\bigcap }_i{V}_{y_i}$ contains $x$ and misses every $U_{y_i}$, hence misses $K$. Thus $X\setminus K$ is open, and every compact subset of a Hausdorff space is closed. The empty-set case is immediate.

Since $\mathbb{R}$ with its usual topology is Hausdorff, its compact sets are closed. The open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has a finite subcover of any compact set $K$. Taking the largest index in that subcover gives $K\subset (-N,N)$, so $K$ is a bounded set. This proves the required direction of the Heine-Borel theorem without assuming its converse.

A quotient topological space of a topological space $X$ is obtained by an equivalence relation $\sim$: its points are the classes $X/\sim$, and the canonical surjection is $q(x)=[x]$. The quotient topology declares a subset $U$ of the quotient open exactly when $q^{-1}(U)$ is open in $X$. More generally a surjection onto a set specifies the same topology by this condition.

For a Hausdorff source with a non-Hausdorff quotient, use $X=\mathbb{R}$ and $x\sim y$ when $x-y\in \mathbb{Q}$. Any nonempty open inverse image is a nonempty open subset of $\mathbb{R}$ invariant under rational translations. It contains an interval $I$, and for every real $t$ there is a rational $r$ with $t-r\in I$; hence it contains every $t$. Thus $\mathbb{R}/\mathbb{Q}$ has the indiscrete topology.

It has distinct points, for instance the classes of $0$ and $\sqrt{2}$, but neither has a proper nonempty open neighbourhood. The quotient is not Hausdorff, even though $\mathbb{R}$ is a Hausdorff space.

The definitions in a unital C-star algebra areThus a Hermitian element of a C-star algebra is self-adjoint, a Unitary element of a C-star algebra has inverse $x^{\ast }$, and a Normal element of a C-star algebra commutes with its adjoint. We derive the needed C-star algebra facts directly from $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, as required.

First $\Vert 1\Vert =\Vert 1{\Vert }^2$ gives $\Vert 1\Vert =1$, andgives $\Vert a\Vert \le \Vert a^{\ast }\Vert$; applying the same argument to $a^{\ast }$ gives equality. Thus the involution is an isometry. For a Unitary element of a C-star algebra $u$,The general Banach algebra spectrum bound gives $|\lambda |\le 1$ for $\lambda \in \sigma (u)$. Such a $\lambda$ is nonzero, and the inverse spectral mapping theorem gives ${\lambda }^{-1}\in \sigma (u^{-1})$, so also $|\lambda {|}^{-1}\le 1$. Hence

If $h=h^{\ast }$, the convergent exponential series and the isometric involution give $(e^{ith}{)}^{\ast }=e^{-ith}$. These commuting exponentials multiply to $1$, so $e^{ith}$ is unitary for real $t$. By the exponential spectral mapping theorem, if $\lambda \in \sigma (h)$ then $e^{it\lambda }\in \sigma (e^{ith})\subseteq \mathbb{T}$. Taking $t=1$ gives $e^{-\mathrm{Im}\lambda }=1$, soOnly general Banach algebra spectral mapping theorem and the defining C-star identity were used.

Now prove spectral permanence for C-star algebras. Let $B\subseteq A$ be the norm-closed C-star subalgebra with the same identity. The algebraic inclusion always gives ${\sigma }_A(b)\subseteq {\sigma }_B(b)$. For $h=h^{\ast }\in B$, both spectra lie in $\mathbb{R}$. If $\lambda \notin {\sigma }_A(h)$, choose nonreal ${\lambda }_n\to \lambda$. Since ${\sigma }_B(h)\subseteq \mathbb{R}$, each $({\lambda }_{n}1-h{)}^{-1}$ belongs to $B$. By continuity of inversion in a Banach algebra, these inverses converge in $A$ to $(\lambda 1-h{)}^{-1}$, which belongs to $B$ because $B$ is closed. Thus $\lambda \notin {\sigma }_B(h)$, proving equality for hermitian elements.

For the requested normal $x\in B$, suppose $b=x-\lambda 1$ is invertible in $A$. The element $b^{\ast }b$ is hermitian and invertible in $A$, so the equality just proved places $(b^{\ast }b{)}^{-1}$ in $B$. ConsequentlySince $b$ already has a two-sided inverse in $A$, multiplying $yb=1$ by that inverse gives $y=b^{-1}$. Hence its inverse belongs to $B$, and

We need one more elementary consequence of the C-star identity: the spectral radius norm equality for normal elements. If $h=h^{\ast }$, then $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $a$ is normal, commutativity of $a,a^{\ast }$ givesEvery power of $a$ is a Normal element of a C-star algebra. Induction gives $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The general spectral radius formula therefore implies

The continuous functional calculus is the unique unital $\ast$-homomorphism $C(K)\to \mathcal{B}(H)$ taking the coordinate function to $T$. For bounded linear operators on a Hilbert space, the adjoint identity gives $\Vert S\Vert ={\sup }_{\Vert v\Vert =\Vert w\Vert =1}|⟨Sv,w⟩|=\Vert S^{\ast }\Vert$. ThereforeThis proves the C-star identity for $\mathcal{B}(H)$ directly; completeness and submultiplicativity come from the operator norm. LetBecause $T$ is a normal operator, this is a commutative unital C-star algebra. Every element of $C$ is a Normal element of a C-star algebra, so the preceding norm identity says its Gelfand transform is isometric:Here $\Delta (C)$ is the compact character space from the general Gelfand representation theorem for commutative Banach algebras.

Each algebra character preserves the involution. Indeed, write $a=h+ik$ with $h,k$ hermitian; $\chi (h)\in {\sigma }_C(h)\subseteq \mathbb{R}$ and similarly for $k$, so $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the continuous mapis injective: its value determines $\chi (T^{\ast })$ and hence its value on all the dense polynomials. It is surjective because the general character description of the spectrum gives $\{\chi (T):\chi \in \Delta (C)\}={\sigma }_C(T)$, and spectral permanence for C-star algebras identifies this with ${\sigma }_{\mathcal{B}(H)}(T)=K$. A continuous function that is a bijection from a compact space to a Hausdorff space is a homeomorphism, so we identify $\Delta (C)$ with $K$.

Under this identification, the Gelfand transform sends $T$ to $u(z)=z$ and $T^{\ast }$ to $\overline{u}$. Its image is an isometric, and therefore closed, unital self-adjoint subalgebra of $C(K)$. It separates points because it contains $u$. The complex Stone-Weierstrass theorem makes the image dense, hence equal to all of $C(K)$. Inverting the Gelfand transform and including $C$ into $\mathcal{B}(H)$ givesan isometric unital C-star homomorphism with $u(T)=T$.

To prove uniqueness without assuming automatic continuity of C-star homomorphisms, let $\Psi$ be any other such map. Since every $f\in C(K)$ is normal, $\Psi (f)$ is normal. A unital algebra homomorphism preserves inverses, soUsing the normal-element norm identity yields $\Vert \Psi (f)\Vert =r(\Psi (f))\le \Vert f{\Vert }_{\infty }$. Thus $\Psi$ is contractive. It agrees with $\Phi$ on polynomials in $u,\overline{u}$, since both send them to the same polynomials in $T,T^{\ast }$. These polynomials are dense by the Stone-Weierstrass theorem, so continuity proves $\Psi =\Phi$. This establishes the continuous functional calculus with no unproved theorem specific to C-star algebras.

A disconnected spectrum gives a nontrivial closed invariant subspace. Write $K=K_1\bigsqcup K_2$ with $K_1,K_2$ nonempty and both open and closed in $K$. The indicator function $f=1_{K_1}$ is continuous on $K$, even though no such continuity is needed across the gap outside $K$. Let $P=f(T)$. The continuous functional calculus givesIts isometry gives $\Vert P\Vert =1$ and $\Vert I-P\Vert =1$, so $P\ne 0,I$. The linear projection has closed range $Y=PH=\ker (I-P)$. Its range is nonzero and proper, and $T(Px)=P(Tx)\in Y$ proves invariance. Since it also commutes with $T^{\ast }$, the same subspace is reducing. This proves disconnected spectrum gives a reducing subspace.

The figure can be realized without any eigenvalues: take $H=L^2(K,dA)$, where $dA$ is planar area on the two closed disks, and let $T$ multiply by $z$. Its adjoint multiplies by $\overline{z}$, so it is a normal operator. An eigenvector for $\lambda$ would be supported on the area-zero singleton $\{\lambda \}$, hence would be zero in $L^2$. Outside $K$, multiplication by $1/(z-\lambda )$ is a bounded inverse to $T-\lambda I$. For $\lambda \in K$, normalized indicator functions of $K\cap \{|z-\lambda |<\epsilon \}$ have $\Vert (T-\lambda I)h\Vert \le \epsilon$, excluding a bounded inverse. Thus its spectrum is exactly $K$. The linear projection in the figure multiplies by $1_{K_1}$, and its range consists of functions supported on the left disk.