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For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

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

For the reality of the eigenvalues, one may initially allow a nonzero complex eigenvector $\mathbf{v}$ with $M\mathbf{v}=\lambda \mathbf{v}$. Since $M$ is real symmetric, it is a Hermitian matrix, andis real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf{v}$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.

For two such eigenvectors, symmetry givesDistinct eigenvalues therefore give zero dot product. Together with the prescribed unit lengths,These $n$ vectors form an orthonormal basis of ${\mathbb{R}}^n$.

Expand a real unit vector as $\mathbf{x}={\sum }_a{c}_a{\mathbf{e}}_a$. Then ${\sum }_a{c}_a^2=1$, and its Rayleigh quotient isMoreover,Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge 2$. Equality occurs precisely along the top eigendirection:

To obtain a unit vector in $S$ using just ${\mathbf{e}}_1,{\mathbf{e}}_2$, let $u=({\mathbf{e}}_1{)}_1$ and $v=({\mathbf{e}}_2{)}_1$ denote their first coordinates. If $u^2+v^2>0$, takeThen ${\alpha }_{1}u+{\alpha }_{2}v=0$ and ${\alpha }_1^2+{\alpha }_2^2=1$. Hence ${\alpha }_1{\mathbf{e}}_1+{\alpha }_2{\mathbf{e}}_2$ has first coordinate zero and unit length. If $u=v=0$, simply take ${\alpha }_1=1,{\alpha }_2=0$. In either case the Rayleigh quotient of the constructed vector isThe continuous quadratic form attains its maximum on the compact space $S$, so

For $\mathbf{x}=(0,\mathbf{y}{)}^T$, deletion of the first row and column gives ${\mathbf{x}}^{T}M\mathbf{x}={\mathbf{y}}^{T}A\mathbf{y}$, while $|\mathbf{x}|=|\mathbf{y}|$. The spectral theorem for real symmetric matrices applied to $A$ identifies its largest eigenvalue with this maximum:Combining the upper bound for every unit vector with the constructed lower bound provesThis is largest-eigenvalue interlacing for a principal submatrix: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

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.

Use the canonical embedding into the bidual $J_{X}x(f)=f(x)$. It is an isometry by the Hahn-Banach theorem. The weak-star topology $\sigma (X^{\ast \ast },X^{\ast })$, restricted to $J_{X}X$, is exactly the weak topology $\sigma (X,X^{\ast })$: both are determined by the same scalar evaluations $f(x)$. Therefore if $K\subseteq X$ is a weakly compact set, $J_{X}K$ is compact in the Hausdorff weak-star topology of $X^{\ast \ast }$, and hence is closed for the weak-star topology. This explains the first assertion without treating the whole embedded space $J_{X}X$ as weak-star closed.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint isFor each $x\in X$, the scalar evaluation $y^{\ast }\mapsto (T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$ is weak-star continuous on $Y^{\ast }$. These are exactly the defining evaluations of the weak-star topology on $X^{\ast }$, so $T^{\ast }$ is weak-star-to-weak-star continuous. Equivalently, if $y_{\alpha }^{\ast }\stackrel{w^{\ast }}{\to }{y}^{\ast }$, then $(T^{\ast }{y}_{\alpha }^{\ast })(x)\to (T^{\ast }{y}^{\ast })(x)$ for every $x$.

The Goldstine theorem states that $J_X(B_X)$ is dense in $B_{X^{\ast \ast }}$ for $\sigma (X^{\ast \ast },X^{\ast })$, where the balls are closed unit balls. The Banach-Alaoglu theorem states that for any normed vector space $E$, the closed dual unit ball $B_{E^{\ast }}$ is compact for $\sigma (E^{\ast },E)$. Neither assertion claims norm compactness or, in general, sequential compactness.

For the weakly compact operator equivalences below, identify $X$ and $Y$ with their canonical images only when stated. In particular, the range condition means $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. We shall prove $(i)⟺(ii)$, $(ii)⟹(iii)⟹(iv)⟹(ii)$. The key identities, obtained by evaluating on $y^{\ast }\in Y^{\ast }$, are

Let $\{O_i\}$ be an open cover of $Z$ in its subspace topology. Write $O_i=Z\cap U_i$ with each $U_i$ open in $X$. Since $Z$ is a closed set, $X\setminus Z$ is open, and $\{U_i\}\cup \{X\setminus Z\}$ is an open cover of the compact space $X$. A finite subcover restricts to a finite subcover of $Z$. Hence the closed subset $Z$ is compact.

If $\{V_j\}$ is an open cover of $f(Z)$ in its subspace topology, the inverse images under the restricted continuous map $f{|}_Z$ form an open cover of $Z$. Take a finite subcover; the corresponding $V_j$ cover $f(Z)$. Thus the continuous image $f(Z)$ is compact. Neither argument assumes a Hausdorff space or asserts that a compact set must be closed set in an arbitrary topological space.

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $[0,1]$, and letThe set is nonempty. If $s=\sup A<1$, choose $U\in \mathcal{U}$ containing $s$. An interval about $s$ lies in $U$; a point of $A$ just to the left of $s$ then extends the finite cover beyond $s$, contradicting the definition of the supremum. Hence $s=1$, and the same neighbourhood argument covers $1$, proving $[0,1]$ compact.

The continuous map $t\mapsto (\cos 2\pi t,\sin 2\pi t)$ maps $[0,1]$ onto $S^1$. A continuous image of a compact space is compact: pull an open cover back, choose a finite subcover, and push the corresponding sets forward. Thus $S^1$ is compact.

In the unusual topology on $X={\mathbb{R}}^2\setminus \{0\}$, a proper closed set is a finite union of punctured lines through the origin. Take any nonempty member $U$ of an open cover. Its complement is such a finite union. Each punctured line has the indiscrete subspace topology, because every other vector subspace meets it only at the removed origin, and is therefore compact. Finitely many finite subcovers, together with $U$, cover $X$. Hence $X$ is compact.

A compact space is one for which every open cover has a finite subcover. A metric space is sequentially compact when every sequence in it has a convergent subsequence whose limit belongs to the space.

No. Give an infinite set $X$ the discrete metricThis space is bounded. Every Cauchy sequence is eventually constant, so it is complete. However, the open cover by the singleton balls $B_{1/2}(x)=\{x\}$ has no finite subcover. Hence $X$ is not compact.

A filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes the empty set, is upward closed, and is closed under finite intersections. An ultrafilter is a proper filter that contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To prove the ultrafilter lemma, order the proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is again a proper filter, so Zorn lemma gives a maximal extension $\mathcal{U}$. If neither $A$ nor $X\setminus A$ belonged to $\mathcal{U}$, adjoining either one would generate an improper filter. There would then be $U,V\in \mathcal{U}$ with $U\cap A=\varnothing$ and $V\cap (X\setminus A)=\varnothing$. But $U\cap V=\varnothing$, contradicting propriety. Hence $\mathcal{U}$ is an ultrafilter.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters on $\mathbb{N}$, with basic setsThe identitiesshow that these sets form a basis of clopen sets. Distinct ultrafilters disagree on some $A$; one lies in $\overline{A}$ and the other in the disjoint set $\overline{\mathbb{N}\setminus A}$. Thus $\beta \mathbb{N}$ is a Hausdorff space.

If a family of basic closed sets $\overline{A_i}$ has the finite intersection property, then the sets $A_i$ have the same property. They generate a proper filter, which extends to an ultrafilter lying in every $\overline{A_i}$. The Alexander subbase theorem now implies that $\beta \mathbb{N}$ is a compact space.

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzelà-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.

Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the open set $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a compact space is compact.

Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint open sets $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. PutThen $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Thereforeare disjoint open neighbourhoods of $A$ and $B$. This proves the normality of a compact Hausdorff space.

Let ${\rho }_{n,j}^{\prime }$ be the $j$th one-system marginal of ${\rho }_n^{\prime }$, and put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1$. Contractivity of trace distance under partial trace givesuniformly in $j$. Because the one-system state space is a compact space, continuity of $f(\cdot \Vert \sigma )$ implies uniform continuity. Hence there is a function $\delta (\epsilon )\to 0$ such thatfor every $j$.

Multipartite superadditivity and additivity now implyThereforewhich proves the required lower asymptotic semicontinuity of quantum relative entropy argument.

Choosewhere $r_p$ is a normal radius at $p$. Take piecewise smooth curves $c_j$ from $p$ to $q$ such that $L_g(c_j)\to d_g(p,q)$. Each $c_j$ first leaves the normal ball $B_g(p,\delta )$ at a point $p_j$. The Gauss lemma gives $d_g(p,p_j)=\delta$. The geodesic sphereis compact, because the tangent-space sphere is compact and ${\exp }_p$ is defined on it. After taking a convergent subsequence, let $p_j\to p_0$.