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A metric dissimilarity must satisfy, for all $x,y,z$, the four propertiesThey are nonnegativity, identity of indiscernibles, symmetry and the triangle inequality. The following two transformations of the simple matching coefficient behave differently at the identity.

Assume $X$ is nonempty, as is necessary for a probability measure to exist. Give its isometry group $I_X$ the uniform metricEvery distance-preserving self-map of a compact metric space is onto. Indeed, if $x_0\notin gX$, then $\delta =d(x_0,gX)>0$, and for $n>m$ the iterates satisfy $d(g^m{x}_0,g^n{x}_0)=d(x_0,g^{n-m}{x}_0)\ge \delta$. This infinite separated set contradicts compactness. Consequently $I_X$ is a group under composition, with the identity and inverses again isometries.

The metric $D$ is invariant under both left and right multiplication: left invariance uses preservation of distances, and right invariance uses surjectivity. Inversion is an isometry for $D$, and $D(gh,g^{\prime }{h}^{\prime })\le D(g,g^{\prime })+D(h,h^{\prime })$ proves continuity of multiplication. Moreover, $I_X$ is compact. A sequence of isometries has a subsequence converging at every point of a countable dense subset of $X$, by a diagonal argument. Their common Lipschitz constant $1$ and finite epsilon-nets of $X$ make this subsequence uniformly Cauchy. Its uniform limit preserves distances and is onto by the preceding argument. Thus $I_X$ is a compact metrizable topological group. A transitive group action means that for every $x,y\in X$ some $g\in I_X$ has $gx=y$.

We construct the invariant Borel probability measure directly. The Hall marriage theorem says that a finite bipartite graph with equal-sized vertex classes $A,B$ has a perfect matching precisely when every subset $S\subseteq A$ has at least $|S|$ neighbours in $B$. Its following consequence is useful. If $A,B$ are $\epsilon$-separated subsets of maximum cardinality in a compact metric space, join points whose distance is less than $\epsilon$. If some $S\subseteq A$ had fewer than $|S|$ neighbours, replacing those neighbours in $B$ by $S$ would produce a larger $\epsilon$-separated set. This contradicts maximal cardinality. Hence there is a bijection between $A$ and $B$ moving each point by less than $\epsilon$.

Choose ${\epsilon }_n\downarrow 0$ and maximum-cardinality ${\epsilon }_n$-separated sets $A_n\subset I_X$. Such sets exist: a finite covering by balls of radius less than ${\epsilon }_n/2$ bounds every separated set's cardinality. Fix $x_0\in X$ and define the empirical probability measuresFor $f\in C(X)$ let ${\omega }_f(r)={\sup }_{d(x,y)\le r}|f(x)-f(y)|$, which tends to zero with $r$ by uniform continuity. For every $h\in I_X$, left invariance of $D$ makes $h{A}_n$ another maximum-cardinality separated set. Match it to $A_n$ as above. ThenThe allowed compactness of probability measures on a compact metric space supplies a subsequence converging to $\mu$ in the weak-star topology. Passing to that limit for each fixed $h,f$ gives $\int fd\mu =\int f\circ hd\mu$. The same subsequence works for every $h,f$, since the displayed estimate held for all of them before taking the limit.

For uniqueness, put $F_n(x)=|A_n{|}^{-1}{\sum }_{a\in A_n}f(ax)$. For $x=h{x}_0$, right invariance of $D$ lets us match $A_{n}h$ with $A_n$. Transitivity therefore gives the uniform estimate $|F_n(x)-F_n(x_0)|\le {\omega }_f({\epsilon }_n)$ for every $x\in X$. If $\nu$ is any invariant Borel probability measure, thenTwo invariant probability measures consequently differ on the integral of $f$ by at most $2{\omega }_f({\epsilon }_n)$, hence agree on every continuous function. Continuous functions determine Borel probability measures on a compact metric space, so the invariant probability measure exists and is unique.

For a $d$-dimensional normed vector space, with unit ball $K$ and John ellipsoid $\mathcal{E}$, its volume ratio isIt is at least $1$ and invariant under invertible linear coordinate changes.

For ${\ell }_1^n$, the unit ball has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its John ellipsoid is $n^{-1/2}{B}_2^n$. To verify maximum volume, let $A{B}_2^n$ be any centered inscribed ellipsoid. For each sign vector $\epsilon \in \{-1,1{\}}^n$, containment gives $|A^T\epsilon |\le 1$. Averaging over independent Rademacher random variables yields $\mathrm{tr}(A{A}^T)\le 1$. The arithmetic-geometric mean inequality for its squared singular values gives $|\det A|\le n^{-n/2}$. The scalar matrix $n^{-1/2}I$ attains this bound because $\Vert x{\Vert }_1\le \sqrt{n}|x|$. Centered ellipsoids suffice: if $a+A{B}_2^n$ is contained in the symmetric convex unit ball, averaging it with $-a+A{B}_2^n$ gives the centered copy with the same volume.

Using the stated Euclidean ball volume and $n=2k$,since $(2k)!/k!={\prod }_{j=k+1}^{2k}j\ge k^k$. Therefore $\mathrm{vr}({\ell }_1^{2k})\le \sqrt{8/\pi }$ uniformly in $k$.

The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed. Here is a counterexample independent of any asymptotic section theorem. Take $E={\ell }_{\infty }^{2k}$. Its John ellipsoid is $B_2^{2k}$: an inscribed $A{B}_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the Hadamard inequality gives $|\det A|\le 1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\dots ,f_k$ an orthonormal basis of it. The random unit vectorhas $|x|=1$. Since ${\sum }_j{f}_j(i{)}^2\le 1$ for each coordinate $i$, the moment-generating function bound for independent Rademacher random variables givesChoose $r=\sqrt{2\log (8k)/k}$; the latter probability is at most $1/2$, so some unit vector in $F$ has $\Vert x{\Vert }_{\infty }\le r$. The proposed section inequality would therefore forceThus even one half-dimensional section cannot have the printed universal constant.

The valid orthogonal splitting with bounded volume ratio is that $L$ may depend on $V=\mathrm{vr}(E)$, and $L=32V^2$ suffices. This supplies a dimension-independent constant for the preceding ${\ell }_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its epsilon-net ingredient.

In John position, $\Vert x{\Vert }_E\le |x|$. Polar integration of the radial function $\rho (u)=1/\Vert u{\Vert }_E$ giveswhere $\sigma$ is normalized spherical measure.

A maximal $\epsilon$-separated subset of $S^{k-1}$ is an epsilon-net. Balls of radius $\epsilon /2$ about its points have disjoint interiors and lie in the ball of radius $1+\epsilon /2$. Comparing volumes bounds its cardinality by $(1+2/\epsilon {)}^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.

Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^{\perp }$ and one such net on each unit sphere. Rotate both by a common uniformly distributed orthogonal matrix $U$. This distribution can be obtained from Question 1 applied to the compact orthogonal group with its Frobenius metric; left and right multiplication are transitive isometries. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The union bound shows that the probability that some rotated net point has norm less than $2\epsilon$ is at mostSet $\epsilon =(32V^2{)}^{-1}$. Because $V\ge 1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\ge 1$. Some rotation therefore makes every point of both nets have norm at least $2\epsilon$.

The triangle inequality and $\Vert x{\Vert }_E\le |x|$ imply that $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric. Every unit vector of $U{F}_0$ or $U{F}_0^{\perp }$ is within $\epsilon$ of a good net point and thus has norm at least $\epsilon$. By homogeneity, on both orthogonal subspaces,For ${\ell }_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-volume ratio restriction is essential; the literal unrestricted assertion cannot be proved.

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.

True. Choose a norm-dense sequence $(x_n)$ in the unit ball of the separable normed vector space $E$. On the dual unit ball $B^{\prime }$, defineEach summand is at most $2^{1-n}$, so the series converges uniformly. Symmetry and the triangle inequality follow term by term. If $d(\varphi ,\psi )=0$, the two continuous functionals agree on the dense sequence, hence on the unit ball and then on all of $E$. Thus $d$ is a metric.

A relative weak-star neighbourhood controls finitely many evaluations. Conversely, to make $d$ small, first choose a tail whose sum is small and then control the finitely many evaluations at $x_1,\dots ,x_N$. This proves that each metric ball is weak-star open. For the other direction, let $x\in E$ and $ϵ>0$. Approximate $x/\Vert x\Vert$ by $x_n$ when $x\ne 0$. Since $\Vert \varphi -\psi \Vert \le 2$ on $B^{\prime }$,The second term can be made smaller than $ϵ/2$ by choosing $n$; the first is smaller than $ϵ/2$ whenever $d(\varphi ,\psi )$ is sufficiently small, because $|\varphi (x_n)-\psi (x_n)|\le 2^{n}d(\varphi ,\psi )$. Every evaluation is therefore metric-continuous. These two inclusions show that $d$ induces exactly the weak-star topology, proving weak-star metrizability of the dual ball without a completeness assumption on $E$.

For each integer $n\ge 0$, put $Z_n=\{z:f^{(n)}(z)=0\}$. Every derivative $f^{(n)}$ is an entire function and cannot be identically zero: otherwise repeated integration would make $f$ a polynomial. By the identity theorem for holomorphic functions, its zero set has empty interior, and continuity makes $Z_n$ a closed set. Thus $Z_n$ is nowhere dense.

The complex plane with its usual metric is a complete metric space, so the Baire category theorem gives $z_0\notin {\bigcup }_{n\ge 0}{Z}_n$. The Taylor series there has coefficients

Fix any estimator $\hat{\eta }$ and put $A(x)=d(\hat{\eta }(x),f)$, $B(x)=d(\hat{\eta }(x),g)$ and $D=d(f,g)$. The triangle inequality for the metric gives $A+B\ge D$. HenceThe larger of the two risk functions dominates their average. Using nonnegativity to replace both probability density functions by their minimum,The right side is independent of the estimator, so taking the infimum provesThis metric squared-loss two-point bound uses overlap of the observation laws to quantify how hard it is to distinguish the two separated parameters.

There is a genuine qualification: the first printed inequality is valid as a useful general assertion for $0<\eta <1$, but is false for arbitrary $\eta >0$. Also the displayed likelihood ratio presupposes $P_{f_1}\ll P_{f_0}$. We first prove the intended bound and then give a counterexample to the unrestricted quantifier.

For any estimator $T$, set $A=\{d(T,f_0)<r_n\}$. By the triangle inequality, $d(T,f_1)\ge r_n$ on $A$, while $d(T,f_0)\ge r_n$ on $A^c$. HenceLet $C=\{|Z-1|\le \eta \}$ with $0<\eta <1$. On $C$, $Z\ge 1-\eta$. ThereforeThe last step is Markov inequality. Taking the supremum over the model and then the infimum over estimators proves the intended metric two-point risk bound. At $\eta =1$ the right side is zero. For $\eta >1$ a negative right side is also harmless, but a positive right side need not be a lower bound.

Here is an explicit counterexample with positive probability density functions and $n=3$. Let $A_0=[0,1/2)$, $\delta =1/100$, and take the model consisting ofUse the L2 norm metric, for which $D=d(f_0,f_1)=2(1-2\delta )$, and take $r_n=D/2$ for every $n$. The majority-half decision selects $f_0$ if at least two observations lie in $A_0$. Each error probability is $p=3{\delta }^2-2{\delta }^3=0.000298$, so its maximum normalized risk function is $2p=0.000596$. This is also the likelihood-ratio test, giving ${\mathbb{E}}_{f_0}|Z-1|=2(1-2p)$. At $\eta =3/2$ the proposed right side is $1/12-2p/3>0.083$, contradicting the risk of this explicit estimator. Thus the intended range restriction cannot be omitted.

For the second request use the intended L2 norm metric $d(f,g)=({\int }_0^1(f-g{)}^2{)}^{1/2}$. The PDF defines its square, not the square itself as a metric. PutBoth probability density functions belong to the model, and $d(f_0,f_{1,n})=a_n=2r_n$. The two-point choice may depend on $n$, because the lower bound is applied separately at each sample size. Under $f_0$, the product likelihood ratio $Z_n={\prod }_i(1+a_{n}g(X_i))$ satisfiesThis is the product chi-squared divergence. By Cauchy-Schwarz inequality, ${\mathbb{E}}_0|Z_n-1|<1/4$. With $\eta =1/2$, the proved metric two-point risk bound has normalized right side greater than $1/8$. ConsequentlyThe proof works for every $n\ge 1$ despite the introductory $n>2$. Measurability of the functions defining the probability density functions is understood. This establishes the requested lower bound; it does not assert a matching upper bound over this unrestricted class.

A map $\Phi :S_D\to \mathbb{R}$ is Hadamard differentiable at $s_0$ if there is a continuous linear functional ${\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\to 0$ with $t_j\ne 0$, $a_j\to a$ in $S$, and $s_0+t_j{a}_j\in S_D$,For tangential Hadamard differentiability, the limit directions $a$ are restricted to a specified tangent subspace and the derivative is continuous linear on that subspace. The domain condition is indispensable.

Assume $X_n\in S_D$ almost surely and use the usual measurable, separably supported setting for norm-valued convergence in distribution, with the limit in the tangent space. The functional delta method givesTo justify this by Skorokhod representation theorem, a sufficient precise version is: for weakly convergent random elements in a separable complete metric space, one can construct copies of the whole sequence and its limit on one probability space which converge almost surely in the metric. More generally the separable-limit-support version of the representation theorem can be used. Apply it to $Z_n=r_n(X_n-s_0)$, obtaining ${\tilde{Z}}_n\to \tilde{X}$ almost surely. The variables $s_0+r_n^{-1}{\tilde{Z}}_n$ have the same laws as $X_n$ and lie in $S_D$ on a common probability-one event. The definition of Hadamard differentiability, with $t_n=1/r_n$, gives convergence almost surely of the transformed quotients to ${\dot{\Phi }}_{s_0}(\tilde{X})$, and hence the claimed convergence in distribution. If $X$ is a Gaussian measure, the limit is Gaussian because the derivative is linear. Without membership of $X_n$ in $S_D$, the expression $\Phi (X_n)$ in the question is not defined.

The final function space is $C_b(\mathbb{R})$ with supremum norm; it is not separable as a whole. This does not affect the deterministic differentiability proof below. Any statistical application of the preceding representation argument must use its separability or tightness hypotheses, rather than assuming that every normed space automatically satisfies them.

Fix admissible $F,G$ and set $T(F,G)={\int }_0^{u}G(x)F^{\prime }(x)dx$. The candidate derivative, defined even for continuous directions $a,b$ without derivatives, isThis follows formally from the product rule and integration by parts. We now establish the uniform-direction limit required by Hadamard differentiability.

Let $F_j=F+t_j{a}_j$ and $G_j=G+t_j{b}_j$ belong to the stated domain, with $a_j\to a$, $b_j\to b$ in supremum norm. The domain gives $\int |F_j^{\prime }|\le 1$ and $\int |F^{\prime }|\le 1$, while $F_j\to F$ uniformly. We first claim that, for every continuous $b$ on $[0,u]$,For a continuously differentiable test function $v$, integration by parts gives convergence directly from uniform convergence of $F_j$: both boundary terms and $\int v^{\prime }{F}_j$ converge. By the Weierstrass approximation theorem, approximate $b$ uniformly on $[0,u]$ by a polynomial $v$. The difference between the two integrals introduced by this approximation is at most $2\Vert b-v{\Vert }_{\infty }$, by the uniform bounded variation bounds. Letting the approximation error decrease proves the claim. This is weak convergence of bounded-variation integrators.

Now decompose the quotient in the order that retains the bounded-variation bound:The first term differs from $\int b{F}_j^{\prime }$ by at most $\Vert b_j-b{\Vert }_{\infty }$, so it tends to $\int b{F}^{\prime }$. For the second term, integration by parts yieldswhich tends to the corresponding expression with $a$. Hence the quotient tends to ${\dot{T}}_{F,G}(a,b)$ for every admissible perturbation sequence. Finally,This proves that the derivative is a continuous linear functional on the product space. Therefore $T$ is Hadamard differentiable at every point of the stated domain. The bounded-variation restriction is what controls the apparently dangerous interaction between uniformly small perturbations and possibly large perturbation derivatives; no convergence of $a_j^{\prime }$ or $b_j^{\prime }$ has been assumed.

Work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. A Fréchet space is a Hausdorff, complete, metrizable locally convex space. Its topology is generated by a countable separating family of seminorms $(p_n{)}_{n\ge 1}$. A neighbourhood base at zero consists of finite intersectionsReplacing $p_n$ by ${\max }_{j\le n}{p}_j$ makes the family increasing. One compatible translation-invariant metric isCompleteness means that a sequence Cauchy for all these seminorms converges in the space. The separating condition ensures that $d(x,y)=0$ only when $x=y$.

Let $E^{\prime }$ denote the continuous dual space for the original topology $\tau$. The weak topology $\sigma (E,E^{\prime })$ is the coarsest topology making all $f\in E^{\prime }$ continuous. Its basic zero-neighbourhoods areEquivalently, it is generated by the seminorms $x\mapsto |f(x)|$.

To prove the Hausdorff property, fix $z\ne 0$. Some continuous seminorm $p$ has $p(z)>0$. On the one-dimensional span of $z$, define $f_0(az)=ap(z)$. Then $|f_0(az)|=p(az)$. The real or complex Hahn-Banach theorem extends this to $f:E\to \mathbb{F}$ with $|f(x)|\le p(x)$. Thus $f$ is originally continuous and $f(z)\ne 0$. This proves continuous-dual point separation. For distinct $x,y$, inverse images under such an $f$ of disjoint small scalar neighbourhoods of $f(x)$ and $f(y)$ are disjoint weak neighbourhoods. Hence the weak topology is Hausdorff.

If a linear functional is originally continuous, it belongs to $E^{\prime }$ and is weakly continuous by definition. Conversely, $\sigma (E,E^{\prime })$ is coarser than $\tau$, so every weakly continuous scalar function is originally continuous. In particular,This special case of the continuous dual of a weak topology needs no completeness argument.

For the linear map $T:E_1\to E_2$, original continuity implies that $g\circ T\in E_1^{\prime }$ whenever $g\in E_2^{\prime }$. Each such composition is weakly continuous on $E_1$, so the defining property of the target weak topology makes $T$ weak-to-weak continuous.

Conversely, suppose $T$ is weak-to-weak continuous. Every $g\circ T$, $g\in E_2^{\prime }$, is then weakly continuous, and hence continuous for ${\tau }_1$. We show that the graph of a linear operator $\Gamma (T)$ is closed in the original product topology. If $x_n\to x$ in $E_1$ and $T{x}_n\to y$ in $E_2$, thenBy continuous-dual point separation on $E_2$, this forces $y=Tx$. The product is metrizable, so sequential closedness is closedness. The closed graph theorem for Fréchet spaces now gives original continuity of $T$.

The relevant version of that theorem does use completeness of both spaces: a closed graph is a Fréchet space, its first-coordinate projection is a continuous linear bijection onto $E_1$, and the open mapping theorem for Fréchet spaces gives a continuous inverse. Composing with the second-coordinate projection gives $T$. ThusEquality of continuous scalar functionals does not imply equality of the two topologies; the closed graph theorem for Fréchet spaces is the extra step that makes this conclusion valid.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

True. Choose a countable dense sequence $(x_j)$ in the separable Banach space $E$. For $\varphi ,\psi$ in its closed dual unit ball, putThis is a metric: positivity follows because continuous functionals agreeing on a dense subset agree everywhere; symmetry is immediate; and the triangle inequality follows from subadditivity of $t/(1+t)$ for $t\ge 0$. The tail of the series is uniformly small, so convergence on these coordinates implies convergence for this metric, and conversely metric convergence controls each individual coordinate.

The uniform bound $\Vert \varphi -\psi \Vert \le 2$ upgrades coordinate convergence to pointwise convergence on all of $E$. For any $x\in E$, choose $x_j$ close to $x$ and useFor neighbourhoods involving finitely many $x$, choose finitely many such approximations. This proves equality of the induced topologies, including for arbitrary nets. Thus the weak-star metrizability of the dual ball gives

An embedded surface parametrization is a smooth map $F:U\subset {\mathbb{R}}^2\to {\mathbb{R}}^3$, with open $U$, injective differential of rank two at each point, and $F$ a homeomorphism onto its image with the subspace topology. Its induced metric isIn coordinates, an arc-length-parametrized geodesic satisfieswhere the Christoffel symbols are ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$. The normalization fixes unit speed; it is compatible with the differential equations because geodesic speed is constant.

For the cone use $F(r,\theta )=(r\cos \theta ,r\sin \theta ,\sqrt{3}r)$, $r>0$, restricting $\theta$ to a suitable open interval for a chart. Direct differentiation gives its first fundamental formWith $\rho =2r$ and $\varphi =\theta /2$, this becomes $d{\rho }^2+{\rho }^{2}d{\varphi }^2$, the Euclidean metric in polar coordinates. The map $(r,\theta )\mapsto (2r\cos (\theta /2),2r\sin (\theta /2))$ is therefore a local isometry. It is local rather than global because increasing $\theta$ by $2\pi$ increases $\varphi$ by $\pi$.

Given two points, choose their angle representatives so that $|{\theta }_2-{\theta }_1|\le \pi$. Their developed plane angles then differ by at most $\pi /2$. The straight segment between their developed positions stays away from the origin: it lies in an angular sector of width at most $\pi /2$ containing both positive-radius endpoints. Develop that segment back onto the cone. Since local isometries preserve geodesics, it gives a geodesic joining the two points; it can be parametrized by arc length. This constructs geodesics on a punctured circular cone. For a repeated point the constant curve is the degenerate geodesic. In particular, this argument does not appeal to completeness of the punctured cone.

The joining geodesic need not be unique. For endpoints with equal positive radius and opposite azimuth, one may use the angular changes $+\pi$ and $-\pi$. In the plane these give two straight segments from $(\rho ,0)$ to $(0,\rho )$ and $(0,-\rho )$, respectively. Both avoid the origin. Their lifts pass on opposite sides of the cone and have distinct images, so a reparametrization cannot identify them.

A null geodesic congruence is a smooth local family of null geodesics, with one generator through each point of the region being described. Choose an affine parameter $\lambda$ on each generator and write $U^a=d{x}^a/d\lambda$. Before a caustic, $U$ is a smooth, nonzero null vector field satisfying $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$.

With $B^a{}_b={\nabla }_b{U}^a$, compatibility of the Levi-Civita connection with the metric givesThe geodesic equation with affine parameter gives the other contraction:Thus both contractions vanish. Nullness supplies the first identity, and affine parametrization supplies the second; a general nonaffine tangent would instead have $B^a{}_b{U}^b=\kappa U^a$.

A subspace is topologically complete if its subspace topology is induced by some complete metric. The compatible complete metric need not be the restriction of the displayed ambient metric. For example, an open interval is topologically complete although its usual metric is incomplete.

Suppose $\rho$ is a compatible complete metric on $Y$. For each $n\ge 1$ and $y\in Y$, choose an ambient open set $V_{n,y}$ containing $y$, contained in $B_d(y,1/n)$, and with $\rho$-diameter of $V_{n,y}\cap Y$ at most $1/n$. Such sets exist because the two topologies on $Y$ agree. Put $U_n={\bigcup }_{y\in Y}{V}_{n,y}$. Clearly $Y\subseteq {\bigcap }_n{U}_n$.

If $x\in {\bigcap }_n{U}_n$, choose $y_n\in Y$ with $x\in V_{n,y_n}$. Then $d(x,y_n)<1/n$, so $y_n\to x$ in the ambient metric. For fixed $m$, the point $x$ belongs to some open $V_{m,z}$, and eventually all $y_n$ belong to this set. Their pairwise $\rho$-distances are then at most $1/m$. Hence $(y_n)$ is $\rho$-Cauchy, and completeness gives a limit $y\in Y$. Compatibility gives $d(y_n,y)\to 0$, forcing $x=y$. ThusThis proves the G-delta criterion for topological completeness in the required direction.

The converse is true in a complete metric ambient space. If $Y={\bigcap }_n{U}_n$ with each $U_n$ open in complete $X$, define $a_n(y)=1/d(y,X\setminus U_n)$, omitting any term whose complement is empty. A compatible complete metric on $Y$ isContinuity of each $a_n$ and the uniformly small series tail show that $D$ has the original topology. A $D$-Cauchy sequence is $d$-Cauchy, hence converges to some $x\in X$. For each $n$, its real coordinates $a_n$ form a Cauchy sequence and remain bounded. Since distance to a closed set is continuous, $d(x,X\setminus U_n)>0$, so $x\in U_n$ for every $n$. The same coordinate convergence and series-tail argument give convergence in $D$. Thus $D$ is complete.