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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.

Work first over an affine neighbourhood $\mathrm{Spec}A$ of $y=\mathfrak{p}$, with an affine chart $\mathrm{Spec}B$ in its inverse image. The scheme-theoretic fibre on this chart has ringwhere $\kappa (\mathfrak{p})$ is the residue field. The prime ideal correspondence for localization and the correspondence for a quotient identify its points with primes $\mathfrak{q}\subset B$ containing $\mathfrak{p}B$ and disjoint from $S$. These two conditions are exactly $\mathfrak{q}\cap A=\mathfrak{p}$.

Under this bijection, each principal open $D(b\otimes 1)$ corresponds to $D(b)\cap f^{-1}\{y\}$. These are bases for the two topologies: every element of the fibre ring is represented by $b/s$, with the image of $s$ invertible. Thus the bijection is a homeomorphism. The identifications agree on overlapping charts and glue, provingThe scheme-theoretic fibre may nevertheless carry nilpotents, so this topological statement does not identify its structure sheaf with a naive restriction of ${\mathcal{O}}_X$.

Let $a,b\in {\mathbb{R}}^2\setminus T$. If $a=b$, use the constant path. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a countable set of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.

Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear pathlies in the complement. This proves that the plane minus a countable set is path connected.

If a homeomorphism $h:{\mathbb{R}}^2\to \mathbb{R}$ existed, its restriction after removing a point $a$ would give a homeomorphism between ${\mathbb{R}}^2\setminus \{a\}$ and $\mathbb{R}\setminus \{h(a)\}$. The first is a path-connected space and therefore connected; the second consists of two separated intervals. Since homeomorphisms preserve connectedness, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every bounded linear functional continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<ϵ,1\le j\le m\}$ for finite families in $X^{\ast }$.

Mazur theorem says that for any convex subset $C$ of a real or complex normed vector space,The norm closure is convex. If $x$ is outside it, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $a$ with $\mathrm{Re}f(x)>a\ge {\sup }_{y\in C}\mathrm{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the norm closure. The other inclusion follows because the weak topology is weaker than the norm topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.

The sequential formulation, Mazur lemma, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the norm closure of its convex hull. Select a finite convex combination of the $n$th tail at norm distance less than $1/n$ from $x$.

A weakly bounded set $D$ satisfies ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^{\ast }$. The completeness of the dual space holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the unit ball. The Uniform boundedness principle therefore gives ${\sup }_{x\in D}\Vert Jx\Vert <\infty$, and $\Vert Jx\Vert =\Vert x\Vert$ proves norm boundedness.

One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^{\ast }:{\sup }_{x\in D}|f(x)|\le n\}$ cover the Banach space $X^{\ast }$. The Baire category theorem makes some $F_n$ contain a ball $f_0+r{B}_{X^{\ast }}$, after shrinking the radius. For $\Vert g\Vert \le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so ${\sup }_D|g(x)|\le 2n$. Scaling and the dual norm formula yield ${\sup }_D\Vert x\Vert \le 2n/r$. A weakly compact set is weakly bounded because each functional has compact, hence bounded, image. Consequently it is norm bounded.

For a Banach space, let $J:X\to X^{\ast \ast }$ be its canonical isometry. We use two explicitly stated weak-star facts: Banach-Alaoglu theorem makes a dual unit ball weak-star compact, and Goldstine theorem makes $J(B_X)$ weak-star dense in $B_{X^{\ast \ast }}$. The weak topology on $X$ is carried by $J$ to $\sigma (X^{\ast \ast },X^{\ast })$ on its image, since their coordinates are the same evaluations $f(x)$.

If $X$ is a reflexive Banach space, $J(B_X)=B_{X^{\ast \ast }}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the Hausdorff space $X^{\ast \ast }$. Goldstine density then forces $J(B_X)=B_{X^{\ast \ast }}$, which implies $JX=X^{\ast \ast }$. This proves the weak compactness characterization of reflexivity:Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^{\ast }}$ separate points. If $K$ is empty there is nothing to prove; otherwise defineThe summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Thus $d$ induces precisely the weak topology on $K$, the countable separating family metrizes a weakly compact set result. No norm-density of the separating family in $X^{\ast }$ is claimed or required.

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 $T^{\ast }$.

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.

First construct the spectral measure of a normal operator and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^{\ast }(I,T)$ generated by $T$ and $T^{\ast }$ is commutative because $T$ is a normal operator. The Gelfand transform identifies $A$ with $C({\Phi }_A)$. By spectral permanence for C-star algebras,is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal{B}(H)$. It is injective because a character is determined by its values on $T,T^{\ast }$ and $I$, and $\chi (T^{\ast })=\overline{\chi (T)}$. Thus $h$ is a homeomorphism. Apply part (a) and setThis is a projection-valued measure with $P(K)=I$ andThe associated continuous functional calculus is isometric: $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }$ for $u\in C(K)$.

If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq \mathbb{C}$. If $P(U)=0$, then $u(T)={\int }_{K}udP=0$, contradicting $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }>0$. ThereforeThis proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.

Now take an isolated point $\lambda \in K$. Its singleton is relatively open, so $P(\{\lambda \})\ne 0$. Multiplication of spectral integrals givesThus its range is contained in the eigenspace $\ker (T-\lambda I)$. Conversely, for $v$ in that kernel, the positive scalar spectral measure ${\mu }_v(B)=⟨P(B)v,v⟩$ satisfiesIt follows that ${\mu }_v(K\setminus \{\lambda \})=0$, and hence $\Vert (I-P(\{\lambda \}))v{\Vert }^2=0$. ConsequentlyThe identification of this projection with the eigenspace holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.

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):

The elliptization theorem says that a closed oriented three-manifold with finite fundamental group has spherical geometry. When the fundamental group is a cyclic group, it is a lens space. Consequently the candidates are $L(5,q)$ with $q\in \{1,2,3,4\}$, using the same surgery orientation convention throughout.

The oriented classification of lens spaces says that $L(p,q)$ and $L(p,q^{\prime })$ are related by an orientation-preserving homeomorphism exactly when $q^{\prime }\equiv q^{\pm 1}(\mathrm{mod}p)$. Since $1^{-1}=1$, $4^{-1}=4$, and $2^{-1}=3$ modulo five, the complete list isThese are three distinct oriented homeomorphism classes. Reversing orientation interchanges $L(5,1)$ and $L(5,4)$; the class containing $L(5,2)$ and $L(5,3)$ admits an orientation-reversing homeomorphism. The elliptization theorem and the stated oriented classification of lens spaces are the general classification results used here.

The fiber monodromy is periodic when isotopy is allowed to move the boundary, in the sense of a periodic surface homeomorphism. Here is a justification from the Seifert fibered space structure, rather than a special formula about torus knots. Its base is a disk with cone-point orders $p,q$. The fiber Seifert surface is an incompressible surface because its inclusion in the mapping torus is injective on fundamental groups. It is also a boundary-incompressible surface: lift a proposed boundary-compressing disk to the infinite cyclic cover $\Sigma \times \mathbb{R}$ and project to $\Sigma$. This would homotope an essential arc of $\Sigma$ into $\partial \Sigma$, contradicting its essentiality. The classification of incompressible surfaces in Seifert fibered spaces therefore applies. A vertical surface in a Seifert fibered space has zero intersection with a regular Seifert fiber, whereas the fibration class evaluates on $f=pqm$ as $pq$. Thus the Seifert surface is isotopic to a horizontal surface in a Seifert fibered space, with $pq$ intersections per regular orbit.

Following the oriented Seifert fibers from one intersection to the next gives a first-return homeomorphism of the horizontal topological surface. It is a representative of the fiber monodromy, and its $pq$-th power is the identity. It has order $pq$: a regular orbit meets the connected horizontal surface in a Seifert fibered space in $pq$ cyclically ordered points. This realizes the periodic case of the Nielsen–Thurston classification theorem. It is neither an Anosov homeomorphism nor a pseudo-Anosov homeomorphism.

There is a boundary convention here. A representative fixed pointwise on the boundary is not periodic relative to the boundary: its $pq$-th power is a boundary Dehn twist (with sign determined by the return-map convention). Capping the boundary, or allowing it to rotate during isotopy, gives the finite-order representative intended by “periodic”. This distinguishes periodic surface type from literal finite order in the mapping class group relative to the boundary.

For any fibered knot, its Alexander module is $H_1(\Sigma ;\mathbb{Z})$ with the deck transformation given by homological monodromy, so its Alexander polynomial of a knot is $\det (tI-{\varphi }_{\ast })$ up to a unit. The general Turaev torsion relation $\Delta (t)\doteq (1-t)\tau (X)$ and part (b) therefore giveThis rational expression is a polynomial: in its factorization into cyclotomic polynomials, $\gcd (p,q)=1$ cancels all denominator factors. It is monic, has constant coefficient one, and has polynomial degree $pq+1-p-q=(p-1)(q-1)$. The characteristic polynomial has polynomial degree $2g$, is monic, and has constant coefficient $\det {\varphi }_{\ast }=1$, because homological monodromy preserves the intersection form. These facts remove the unit ambiguity and giveThe sign change between $\det ({\varphi }_{\ast }-tI)$ and $\det (tI-{\varphi }_{\ast })$ is trivial because $\mathrm{rank}{H}_1(\Sigma )=2g$ is even. As a separate geometric check, the orbifold Euler characteristic of the base is $1/p+1/q-1$, and its $pq$-sheeted horizontal surface in a Seifert fibered space has Euler characteristic $\chi (\Sigma )=p+q-pq=1-2g$.