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The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

The linear independence of the first $n$ elements of each orthonormal system shows that $T_n$ and $S_n$ have the same finite dimension $n$ and are closed subspaces of a Hilbert space. Apply part (b) with $U=T_n$ and $V=S_n^{\perp }$. Since $(S_n^{\perp }{)}^{\perp }=S_n$, the positive directed subspace angle cosine givesLet $Q$ be the oblique projection onto $T_n$ along $S_n^{\perp }$. Define ${\tilde{f}}_n=Qf$. Its residual $f-Qf$ is orthogonal to every ${\psi }_j$ with $1\le j\le n$, so the required measurements agree. Conversely, if $g\in T_n$ has the same measurements, then $f-g\in S_n^{\perp }$; uniqueness of the direct sum decomposition gives $g=Qf$. Thus this is finite-dimensional Hilbert sampling reconstruction.

There is also an explicit coefficient description. Use the inner product convention linear in its first entry and putThen $Cc=b$. The orthonormal systems show $\Vert P_{S_n}{\sum }_k{c}_k{\varphi }_k\Vert =\Vert Cc{\Vert }_2$ and $\Vert {\sum }_k{c}_k{\varphi }_k\Vert =\Vert c{\Vert }_2$, so the smallest singular value of $C$ is $\cos {\theta }_{T_n,S_n}>0$. Hence $c=C^{-1}b$, another direct proof of existence and uniqueness.

For $n\ge 1$, both summands of this direct sum are nonzero: the infinite orthonormal system $({\psi }_j)$ contains ${\psi }_{n+1}\in S_n^{\perp }$. The permitted oblique projection norm formula therefore applies without its degenerate exception, givingThe operator norm immediately yields the stability estimate $\Vert {\tilde{f}}_n\Vert \le \sec {\theta }_{T_n,S_n}\Vert f\Vert$. For the approximation bounds, put $e=f-P_{T_n}f$. Because $Q$ fixes $T_n$, we haveso the operator norm estimate gives $\Vert f-Qf\Vert \le \sec {\theta }_{T_n,S_n}\Vert e\Vert$. For the lower bound, $e\perp T_n$ while $P_{T_n}f-Qf\in T_n$. The Pythagorean identity givesThe unique measurement-matching reconstruction is stable and within the secant function factor of the best orthogonal projection approximation:If a zero-dimensional reconstruction is admitted, it is simply ${\tilde{f}}_0=0$ and its error equals the norm of $f$; that case is best stated directly instead of using the angle of a zero space.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

For a finite collection $\mathcal{L}$ of distinct affine lines in a vector space in ${\mathbb{R}}^n$, $n\ge 2$, a joint is a point incident to $n$ lines whose direction vectors are linearly independent. The joints theorem assertsIn the customary three-dimensional formulation this is $\#J\lesssim L^{3/2}$, with three noncoplanar incident lines at every joint. We prove the general form, which includes that formulation.

Let $L=\#\mathcal{L}$ and $M=\#J$. The conclusion is immediate when $M=0$. Otherwise set $D=\lceil n{M}^{1/n}\rceil$. Suppose for contradiction that $M>LD$. Repeatedly delete any line incident to at most $D$ of the currently retained joints, deleting those joints at the same time. Each deleted line loses at most $D$ current joints, so even deleting all $L$ lines could lose at most $LD<M$ joints. Therefore the process must stop with a nonempty set $J^{\prime }$ and a line collection ${\mathcal{L}}^{\prime }$ such that each retained line contains more than $D$ retained joints. Every retained joint still has its original $n$ independent incident lines: if any line through it had been deleted, the joint would have been deleted too.

There is a nonzero multivariate polynomial of total degree at most $D$ vanishing on $J^{\prime }$, becauseChoose such a polynomial $P$ of smallest possible total degree $d\le D$. This is an application of the polynomial method in combinatorics. Every line of ${\mathcal{L}}^{\prime }$ contains more than $D\ge d$ roots of a polynomial of its polynomial restriction to a line, so $P$ vanishes identically on every such line.

At a retained joint $x$, differentiating along each of its independent line directions $v_1,\dots ,v_n$ gives $v_i\cdot \nabla P(x)=0$. Their linear independence therefore forces $\nabla P(x)=0$. Each partial derivative of $P$ vanishes on all of $J^{\prime }$ and has smaller total degree. Minimality of $d$ forces every partial derivative to be the zero polynomial. Over the real numbers, a polynomial with all partial derivatives zero is constant; a nonzero constant cannot vanish on the nonempty $J^{\prime }$. This is the required contradiction.

It follows that $M\le LD$. Since $D\le (n+1)M^{1/n}$ for $M\ge 1$,This proves the joints theorem by the pruning and minimal-degree polynomial argument.

The exponent is sharp. Take all axis-parallel lines passing through the grid $\{1,\dots ,m{\}}^n$. There are $n{m}^{n-1}$ distinct lines and $m^n$ joints; the coordinate directions span ${\mathbb{R}}^n$ at every grid point. Thus no smaller power of the number of lines can bound all joint configurations.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

The Steinitz exchange lemma says that if $u_1,\dots ,u_r$ are linearly independent and $v_1,\dots ,v_s$ span the same vector space, then $r\le s$ and, after reordering the $v_j$, the list $u_1,\dots ,u_r,v_{r+1},\dots ,v_s$ still spans it.

To prove this, suppose $u_1,\dots ,u_k$ have already replaced $k$ of the spanning vectors. Express $u_{k+1}$ in that spanning list. At least one coefficient of an unreplaced $v_j$ is nonzero: otherwise $u_{k+1}$ lies in the span of $u_1,\dots ,u_k$, contradicting linear independence. Solve this expression for that $v_j$ and replace it by $u_{k+1}$. The new list still spans. If all $s$ vectors were replaced before the $u_i$ were exhausted, the next $u_i$ would be dependent on its predecessors. Hence $r\le s$, and the replacement can continue through $u_r$.

For two bases of sizes $r,s$, apply the Steinitz exchange lemma in both directions to get $r\le s\le r$. Thus every basis has the same size. Given a linearly independent set and any finite basis, the replacement procedure produces a finite spanning list containing that set. Successively delete any remaining vector dependent on the others; none of the prescribed independent vectors need be deleted, since one can instead delete a dependent supplementary vector. Equivalently, greedily retain only supplementary vectors outside the span already built. The resulting list is a basis containing the prescribed set.

For the three-dimensional example, the columns have matrixTherefore the three vectors are a basis. For the four-dimensional example,This is a nontrivial linear dependence, so the four vectors are not a basis. The distinction is an instance of the adjacent sums of a cyclic basis criterion: alternating coefficients close consistently around an even cycle, but not an odd one.

Let $Dx=\lambda x$ with $x\ne 0$. Commutation givesThus $Cx$ belongs to the eigenspace $E_{\lambda }(D)$, with the zero vector allowed. Since $D$ has $n$ distinct eigenvalues, each eigenspace is one-dimensional. Hence $Cx=\mu x$ for some scalar $\mu$. This is the fact that commuting maps preserve eigenspaces.

The standard linear independence theorem for eigenvectors belonging to distinct eigenvalues supplies a basis $x_1,\dots ,x_n$ of eigenvectors of $D$. The preceding argument makes every one an eigenvector of $C$ as well. Their column matrix therefore simultaneously diagonalizes $C$ and $D$. This is simple spectrum gives simultaneous diagonalization.

For real matrices, this conclusion always holds over $\mathbb{C}$; a real common basis follows when the distinct eigenvalues of $D$ are real. Real matrix entries alone do not ensure a real eigenbasis: a quarter-turn of ${\mathbb{R}}^2$, with $C=D$, has eigenvalues $\pm i$. The matrices in this question have real spectra, so their common basis is real.

Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.

The independence of eigenvectors for distinct eigenvalues ensures that such eigenvectors form a linearly independent set. For completeness, apply ${\prod }_{j\ne k}(M-{\lambda }_{j}I)$ to a proposed relation ${\sum }_i{c}_i{\mathbf{e}}_i=0$. It leaves $c_k{\prod }_{j\ne k}({\lambda }_k-{\lambda }_j){\mathbf{e}}_k=0$, so every $c_k=0$. Thus the three eigenvectors form a basis over the relevant field and $\mathbf{x}={\sum }_i{a}_i(t){\mathbf{e}}_i$. Substitution into the linear system of ordinary differential equations gives ${\sum }_i({\dot{a}}_i-{\lambda }_i{a}_i){\mathbf{e}}_i=0$; linear independence implies ${\dot{a}}_i={\lambda }_i{a}_i$. HenceA real matrix may have a complex conjugate pair of eigenvalues. In that case the calculation is over $\mathbb{C}$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three eigenvalues are real is needed.

The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

We prove the stronger finite branching bound in a complete market: modulo null sets, ${\mathcal{F}}_t$ is generated by at most $n^t$ atoms. Here an atom is a positive-probability event which cannot be split into two positive-probability measurable pieces.

At time zero there is one atom because ${\mathcal{F}}_0$ is trivial. Suppose ${\mathcal{F}}_{t-1}$ has at most $n^{t-1}$ atoms, and fix one such atom $A$. Any predictable process holdings over $(t-1,t]$ are constant on $A$, so the restriction to $A$ of every time-$t$ replicable payoff lies ina vector space of dimension of a vector space at most $n$. If $A$ contained $n+1$ disjoint positive-probability ${\mathcal{F}}_t$ events, their indicators would have linear independence on $A$. Market completeness would replicate each indicator, contradicting that dimension bound.

For clarity, the absence of $n+1$ disjoint positive pieces implies that $A$ is a union of at most $n$ atoms: start with $A$ and repeatedly split any non-atom into two positive pieces. Each split increases the count by one. The process must stop before the count exceeds $n$, and at termination every piece is an atom. Thus each time-$(t-1)$ atom has at most $n$ successors. Induction gives at most $n^t$ atoms at time $t$.

Each disjoint positive-probability ${\mathcal{F}}_t$ event contains at least one distinct atom. ThereforeThe argument includes $t=0$ and uses the ability to replicate the indicator claims at every intermediate maturity.