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Theorems matter at least as much as definitions: they show when a categorical description captures actual mathematical structure. Definitions in category theory are exceptionally economical, but economy alone does not explain their applications.

An adjunction compresses a universal mapping property into natural hom-set bijections. The free-group adjunction, for example, records that assigning images to generators extends uniquely to a homomorphism. That is a useful definition, but proving that a free group exists and satisfies this property still requires mathematics. The definition organizes that proof rather than replaces it.

The Yoneda lemma offers a sharper example. A representable functor is easy to define; the theorem identifying natural transformations out of it with elements of another functor gives the definition its computational force. The representation of a set-valued functor as a colimit of representables is then a substantive consequence, and the theorem that filtered colimits commute with finite limits turns it into a criterion for finite-limit preservation. None of these conclusions follows merely from choosing attractive terminology.

Likewise a monad packages operations and their compatibility, while the Eilenberg-Moore category defines the corresponding algebras. The Beck monadicity theorem answers a different question: whether a given category is really recovered by that algebraic description. Applying it to compact Hausdorff spaces needs a topological argument about closed quotient relations. The definitions would not by themselves predict or justify the result.

Finally, abelian categories put kernels in a category, cokernels in a category and exactness into a setting where homological arguments can be reused. Some consequences, such as image factorization, become short because strong structural content has deliberately been included in the definition. Yet left derived functors, long exact sequences and the detection of projective objects supply information about obstructions to exactness which the terminology alone cannot provide.

Definitions and theorems therefore play complementary roles. A good definition states what should be invariant across examples; a theorem proves which constructions preserve it, which examples satisfy it, and what can be deduced from it. Category theory makes that partnership unusually visible, rather than making proofs less important.

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.

False. We construct a compact Hausdorff space with a countable dense discrete subspace but no compatible metric. Letfor each subset $S\subseteq \mathbb{N}$. Give $P$ its product topology, and let $X=\overline{j(\mathbb{N})}$ in $P$. The Tychonoff theorem makes $P$ compact, and the product is Hausdorff; hence $X$ is a compact Hausdorff space. By construction, $Y=j(\mathbb{N})$ is countable and dense in $X$, so $X$ is a separable topological space. The coordinate $S=\{n\}$ isolates $j(n)$ within $Y$. Thus the subspace topology on $Y$ is discrete and is induced by the discrete metric.

Suppose $X$ were metrizable. An infinite sequence of distinct points of $Y$ would have a convergent subsequence, by sequential compactness of a compact metric space. Write this subsequence as $j(n_k)$ with distinct $n_k$. For $S=\{n_2,n_4,n_6,\dots \}$, its $S$ coordinate alternates between zero and one and therefore does not converge. But coordinate projections are continuous in the product topology, so every coordinate of a convergent sequence must converge. This contradiction proves that $X$ is not metrizable. This is a separable compactification need not be metrizable counterexample.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal space admit a continuous $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. Using the usual convention, normal spaces are also $T_1$.

The Tietze extension theorem states that a continuous real-valued function on a closed subset of a normal space extends continuously to the whole space. If its range lies in a closed bounded interval, the extension may be chosen in that same interval.

Prove first the interval version for $f:S\to [-1,1]$. If a continuous residual $r$ on $S$ obeys $|r|\le M$, the sets $\{r\le -M/3\}$ and $\{r\ge M/3\}$ are disjoint closed subsets of $X$. Urysohn's lemma gives a continuous $h:X\to [-M/3,M/3]$ taking the endpoint values on those two sets. On each of the three ranges of $r$, it follows that $|r-h|\le 2M/3$ on $S$. Repeating with $M_j=(2/3{)}^j$ produces $h_j$ with $\Vert h_j{\Vert }_{\infty }\le M_j/3$ and a residual bounded by $M_{j+1}$. The series $\tilde{f}={\sum }_{j\ge 0}{h}_j$ converges uniformly, has absolute value at most $\sum M_j/3=1$, is continuous and restricts to $f$ as the residual tends to zero. Rescaling proves every closed bounded interval version.

For unbounded $f$, first extend $f/(1+|f|)$ to $g:X\to [-1,1]$. The closed set $B=g^{-1}(\{-1,1\})$ is disjoint from $S$. Urysohn's lemma gives $\psi :X\to [0,1]$ with $\psi {|}_S=1$ and $\psi {|}_B=0$. Then $h=\psi g$ takes values strictly between minus one and one everywhere and equals $f/(1+|f|)$ on $S$. Thus $h/(1-|h|)$ is a continuous real-valued extension of $f$, proving the full theorem.

Statement 1 is true: metric spaces are normal, and the bounded interval form applies. Statement 2 is true: compact Hausdorff spaces are normal, and the unrestricted real-valued form applies. Statement 3 is false. For a counterexample, let $X=\{a,b,c\}$ have open sets $\varnothing ,\{a\},\{a,b\},\{a,c\},X$. The subset $S=\{b,c\}$ is closed and discrete, so $f(b)=-1$, $f(c)=1$ is continuous on it. Every open neighborhood of $b$ or $c$ in $X$ contains $a$. A continuous extension would pull back disjoint neighborhoods of minus one and one to disjoint open sets containing $b$ and $c$, yet both would contain $a$. This is impossible.

For positive linear operators on continuous functions, a general Korovkin theorem, in its compact Korovkin test space form, can be stated using a test space $H\subset C(K,\mathbb{R})$, where $K$ is a compact Hausdorff space. Assume $H$ contains a strictly positive function $h_0$. Require that, for every $x\in K$, the only positive linear functional $L$ on $C(K)$ with $L(h)=h(x)$ for all $h\in H$ is evaluation at $x$. Then convergence on the test space implies convergence on every continuous function:By the Riesz-Markov-Kakutani representation theorem, the functional condition equivalently says that a positive measure with these test moments must be ${\delta }_x$. This permits arbitrary test families, not just the interval tests $1,x,x^2$.

For completeness, let $c={\min }_K{h}_0>0$. Positivity gives $U_{n}1\le c^{-1}{U}_n{h}_0$, so the operator norms are uniformly bounded. If the conclusion failed for some $f$, choose $x_n$ where its error is bounded away from zero. The evaluation functionals $L_n(g)=U_{n}g(x_n)$ have uniformly bounded norm. Compactness of $K$ and the Banach-Alaoglu theorem give a subnet on which $x_n\to x$ and $L_n$ converges in the weak-star topology to a positive $L$. For every $h\in H$, uniform test convergence gives $L(h)=h(x)$. The hypothesis forces $L(f)=f(x)$, contradicting the chosen errors. Complex-valued functions follow by applying the real result to their real and imaginary parts.

A useful concrete sufficient condition is that $1\in H$ and for each $x$ there is $g_x\in H$ with $g_x\ge 0$ and zero set exactly $\{x\}$. Its zero moment forces the representing measure to be supported at $x$, while the constant moment fixes its mass.

For the circle, take $H=\mathrm{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos (t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus the periodic Korovkin test set isUniform convergence on these three functions is sufficient for uniform convergence on $C(\mathbb{T})$.

The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subset C(K,\mathbb{R})$ which contains the constants and separates points is dense in $C(K,\mathbb{R})$ for the uniform norm, when $K$ is compact Hausdorff space. If $A$ is also closed, $A=C(K,\mathbb{R})$. The complex version needs closure under complex conjugation; that extra condition is unnecessary here.

Use the usual unital-subalgebra convention $I_H\in A$. The spectral theorem for a commutative operator algebra states that the character space ${\Phi }_A$, equipped with the Gelfand topology, is a compact Hausdorff space, and there is a unique regular projection-valued measure $E$ on its Borel sigma-algebra satisfyingwhere $\hat{a}(\chi )=\chi (a)$ is the Gelfand transform. The map $a\mapsto \hat{a}$ is an isometric unital star-isomorphism from $A$ onto $C({\Phi }_A)$ by the Commutative Gelfand--Naimark theorem.

More explicitly, every $E(B)$ is an orthogonal projection, $E(\varnothing )=0$, $E(B\cap C)=E(B)E(C)$, and for disjoint Borel sets $B_n$,with convergence in the norm of the Hilbert space. Regularity means that each scalar spectral measure $B\mapsto ⟨E(B)v,w⟩$ is a regular Borel measure. The integral identity then implies $\Vert a\Vert ={\sup }_{\chi \in {\Phi }_A}|\hat{a}(\chi )|$. No separability assumption on $H$ is needed.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

A family $\mathcal{S}\subseteq C(K)$ is equicontinuous if for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such thatfor every $y\in U$ and every $f\in \mathcal{S}$. The Arzelà-Ascoli theorem says that, for a compact Hausdorff space $K$, a subset of $C(K)$ has compact closure in the uniform norm if and only if it is equicontinuous and pointwise bounded. On compact $K$, equicontinuity and pointwise boundedness together imply uniform boundedness.

For necessity, a compact closure is totally bounded. Given $\epsilon >0$, choose a finite $\epsilon /3$-net $f_1,\dots ,f_m$. The finitely many continuous functions $f_j$ are simultaneously continuous near each point, and comparison with a nearby $f_j$ proves equicontinuity of the whole family. Evaluation at any fixed point is a continuous map $C(K)\to \mathbb{R}$ or $\mathbb{C}$, so compactness also gives pointwise boundedness.

Conversely, equicontinuity gives, around each $x\in K$, a neighborhood on which every $f\in \mathcal{S}$ oscillates by less than $\epsilon /3$. Compactness supplies a finite such cover with selected points $x_1,\dots ,x_m$. Pointwise boundedness makes the set of vectorsa bounded subset of a finite-dimensional normed vector space, so it has a finite $\epsilon /3$-net. Choose one function of $\mathcal{S}$ for every nonempty cell of this net. If two functions have evaluation vectors in the same cell, then comparison at a selected $x_j$ and the two oscillation bounds show that their uniform distance is less than $\epsilon$. Thus $\mathcal{S}$ is totally bounded. Since $C(K)$ is a Banach space, its closure is complete; a complete totally bounded metric space is compact, proving the theorem.

A finite positive Borel measure on a compact Hausdorff space is regular when its values on Borel sets can be approximated from inside by compact sets and from outside by open sets. For signed or complex measures require this property of the variation measure. These are the measures representing the continuous dual space of $C(K)$ in the Riesz-Markov-Kakutani representation theorem.

Embed $\mathbb{N}$ into the compact product $\{0,1{\}}^{\mathcal{P}(\mathbb{N})}$ by $j(n{)}_S=1_S(n)$ and take its closure $X$. The Tychonoff theorem makes $X$ a compact Hausdorff space, while $j(\mathbb{N})$ is countable, dense and discrete in its subspace topology. Yet $X$ is not metrizable. Any convergent subsequence $j(n_k)$ of distinct points would have a convergent coordinate for $S=\{n_2,n_4,\dots \}$, whereas that coordinate alternates between zero and one. This contradicts sequential compactness of a compact metric space. Thus a dense metrizable subspace does not force a separable compact Hausdorff ambient space to be metrizable.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

For a completely regular Hausdorff space, the Stone-Čech compactification is the weak-star closure of the evaluation functionals in the dual unit ball of $C_b(X)$. The Banach-Alaoglu theorem gives compactness; complete regularity makes evaluation a topological embedding. Every bounded real continuous function extends by $\tilde{f}(p)=p(f)$, yielding an isometric isomorphism $C_b(X)\cong C(\beta X)$. For a continuous map into a compact Hausdorff space $K$, the adjoint of pullback sends this closure into the closed evaluation copy of $K$, yielding the unique continuous extension. Uniqueness follows from density. It is a universal compactification, not a homeomorphism between $X$ and the whole dual space.

A subsemigroup is a nonempty subset of a semigroup closed under its associative operation. In a compact Hausdorff space carrying a left-topological semigroup, descending chains of nonempty compact subsemigroups have a nonempty compact subsemigroup as their intersection. This supplies the minimal object used in the Ellis–Numakura lemma.