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The equal-variance form of Slepian's lemma is the following. If $U,V$ are centered vectors with multivariate normal distributions, $\mathbb{E}{U}_i^2=\mathbb{E}{V}_i^2$ for every $i$, and $\mathbb{E}{U}_i{U}_j\ge \mathbb{E}{V}_i{V}_j$ whenever $i\ne j$, thenfor every threshold vector $(u_i)$. In particular $\mathbb{P}({\max }_i{U}_i\le u)\ge \mathbb{P}({\max }_i{V}_i\le u)$. No nonsingularity of the covariance matrices is required.

Here the pointwise variances $2t^{2H}$ and $2t^{2H^{\prime }}$ differ. To make the comparison rigorous, use the increment form, namely the Sudakov-Fernique inequality: if centered multivariate normal distributions satisfy $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then $\mathbb{E}{\max }_i{U}_i\le \mathbb{E}{\max }_i{V}_i$. Unlike the preceding form of Slepian's lemma, this statement has no equal-pointwise-variance hypothesis.

For completeness, the increment form follows by Gaussian integration by parts. Make $U,V$ independent, define $Z_r=\sqrt{1-r}U+\sqrt{r}V$ for $0<r<1$, and letThe $p_i$ are softmax function weights, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. Direct differentiation gives ${\partial }_i{F}_{\tau }=p_i$ and ${\partial }_{ij}{F}_{\tau }=\tau (p_i{1}_{i=j}-p_i{p}_j)$. The Gaussian integration by parts identity therefore yieldsThe last summands are nonnegative because $D_{ii}+D_{jj}-2D_{ij}=\mathbb{E}(V_i-V_j{)}^2-\mathbb{E}(U_i-U_j{)}^2$. Sinceintegration in $r$ and the limit $\tau \to \infty$ prove the Sudakov-Fernique inequality. All derivatives needed here are bounded for a fixed $\tau$. Singular covariance matrices can be handled by adding small independent centered normal noise with the same variance to both vectors, proving the identity for the nonsingular vectors, and letting the noise variance decrease to zero.

Use the printed normalization of fractional Brownian motion, which omits the conventional factor $1/2$. Its increment variance isFor $s,t\in [0,1]$ and $H>H^{\prime }$, $|t-s{|}^{2H}\le |t-s{|}^{2H^{\prime }}$. Enumerate $T$ by increasing finite subsets $T_n$ whose union is $T$, with $0\in T_n$. The Sudakov-Fernique inequality givesBoth Gaussian processes vanish at zero almost surely. Thus these maxima are nonnegative and increase with $n$; monotone convergence theorem givesTheir continuous modifications agree with the original Gaussian processes simultaneously on the countable set $T$, after discarding a single null event. Continuity and density of the rational parameters make each rational supremum equal to the supremum over $[0,1]$ for these modifications.

The expected suprema are finite. Indeed, the canonical distance of $Y_{H^{\prime }}$ is $d(s,t)=\sqrt{2}|t-s{|}^{H^{\prime }}$. A Euclidean grid supplies a closed-ball metric covering number bound $N(\epsilon )\le 1+\lceil (\sqrt{2}/\epsilon {)}^{1/H^{\prime }}\rceil$ for $0<\epsilon \le \sqrt{2}$. The Dudley entropy integral bound for an anchored centered separable Gaussian process isThe integral is finite since its integrand grows at most like a constant times $\sqrt{1+\log (1/\epsilon )}$ near zero. The same argument applies to $X_H$.

For any real path vanishing at zero,A centered Gaussian process has the same probability law as its negative. Combining this symmetry with the increment comparison gives the stronger boundIn particular the requested comparison isThe common normalization factor $\sqrt{2}$ affects both sides equally. The comparison is valid for every $0<H^{\prime }<H<1$, including the equality cases $s=t$ or $|s-t|=1$ in the increment bounds.

For $X$ in the dual cone $C$, copositivity of $Q-\lambda I$ gives $⟨X,Q-\lambda I{⟩}_F\ge 0$. An upper-bounding Lagrangian for the maximization is thereforeIts supremum over unrestricted $\lambda$ is finite precisely when $\mathrm{tr}X=1$. Minimizing that upper bound gives the conic programThis is completely positive optimization, not simply semidefinite programming: $C$ is the completely positive cone.

There is also a direct equality certificate. Every feasible $X={\sum }_j{u}_j{u}_j^T$, $u_j\ge 0$, has ${\sum }_j\Vert u_j{\Vert }_2^2=1$. Its objective is a weighted average of the nonnegative-sphere Rayleigh quotients, hence is at least $\alpha$. For a minimizing unit vector $x_{\ast }$ from part (g), $X_{\ast }=x_{\ast }{x}_{\ast }^T$ is feasible and has objective $\alpha$. Thus $\boxed{X_{\ast }=x_{\ast }{x}_{\ast }^T}$ attains the dual and both values agree, without relying on unverified regularity assumptions.

Let $(u_n)$ be increasing and bounded above, and let $L=\sup \{u_n:n\in \mathbb{N}\}$. For every $\epsilon >0$, the definition of the supremum supplies $N$ with $L-\epsilon <u_N\le L$. Monotonicity gives $L-\epsilon <u_n\le L$ for every $n\ge N$, proving $u_n\to L$. This is the monotone bounded sequence theorem.

For the recurrence, $f(x_n)>0$ makes $(x_n)$ increasing. If it were bounded above, it would converge to some $L$. Since $x_n\le L$ and $f$ is decreasing, $f(x_n)\ge f(L)>0$, whencea contradiction. Therefore $x_n\to \infty$.

Use the standard convention that the finite-variation process $X$ has càdlàg paths and that the stochastic integral starts from zero. Write $V(X)$ for its total-variation process. The necessary admissibility condition is ${\int }_{(0,t]}|H_s|d{V}_s(X)<\infty$ almost surely for each finite $t$; it holds, for example, for locally bounded $H$. Define the pathwise Lebesgue-Stieltjes integralAn atom at $s>0$ contributes $H_s\Delta X_s$, so this convention also fixes the treatment of jumps.

Fix $t$. For a generator $H_s=1_A{1}_{(a,b]}(s)$ of the predictable sigma-algebra, with $A\in {\mathcal{F}}_a$, the Lebesgue-Stieltjes integral isIt vanishes when $t\le a$ and is ${\mathcal{F}}_t$-measurable otherwise. The total-variation process is adapted: for a càdlàg path its value at $t$ is the supremum over a countable collection of partitions of an interval using rational interior points and endpoint $t$. Thus the positive and negative Lebesgue-Stieltjes measures $d{X}^{\pm }=(dV(X)\pm dX)/2$ have ${\mathcal{F}}_t$-measurable masses on the generator intervals.

The Monotone class theorem, applied separately to these positive random measures on $[0,t]$, extends this measurability from the generators of the predictable sigma-algebra to all nonnegative predictable processes. Truncation and positive/negative decomposition then cover every admissible $H$. Consequently $H\cdot X$ is adapted.

The integrability condition is essential. Taken literally, an arbitrary previsible process cannot always be integrated against a finite-variation process: with $X_t=t$ and deterministic $H_s=s^{-1}$ for $s>0$, $H_0=0$, the proposed value at every $t>0$ diverges. The usual càdlàg convention is also what identifies $dX$ by $dX((a,b])=X_b-X_a$.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.

No. For each $x>0$, $e^{-nx}\to 0$, whereas $e^{-n0}=1$ for every $n$. The pointwise limit on $[0,1]$ is thereforeThis limit is not continuous at $0$. Since every $f_n(x)=e^{-nx}$ is continuous, the uniform limit theorem rules out uniform convergence. Directly, $|f_n(x)-f(x)|=e^{-nx}$ for $x>0$, whose supremum over $(0,1]$ is $1$ for every $n$.

Fix $a\in L$. Inside the ambient transitive model $M$, the Axiom of power set makes the collection of constructible subsets of $a$ a set. For each such subset $x$, choose the least stage of the constructible hierarchy at which $x$ appears. The Axiom schema of replacement and the supremum of a set of ordinals give an ordinal $\beta$ bounding all these stages; enlarge $\beta$ so that $a\in L_{\beta }$.

Nowis definable over $L_{\beta }$ with parameter $a$. It therefore belongs to the definable power set $\mathcal{D}(L_{\beta })=L_{\beta +1}\subseteq L$. This set contains exactly the subsets of $a$ that belong to the constructible universe, so it witnesses the Axiom of power set in $L$. Therefore $L\models$ Power Set.

True. For each $y\in x$, the singleton set $\{y\}$ is a finite subset of $x$, so the hypothesis givesTaking the supremum over all $y\in x$ yields

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

Part i gives $\Vert D\Vert \le B$. Conversely, choose indices $j_k$ with $|d_{j_k}|\to B$. Since $\Vert e_{j_k}{\Vert }_2=1$,Taking the limit proves $\boxed{\Vert D\Vert =B}$, whether or not the supremum is attained.

For a nonzero ordinal $\alpha$, letThe set is nonempty because $0\in S$, and it is an initial segment because ordinal exponentiation by $\omega$ is strictly increasing. Put $\beta =\sup S$.

If $\beta$ is a successor, the definition of the supremum forces $\beta \in S$. If $\beta$ is a limit ordinal, then continuity of ordinal exponentiation givesso again $\beta \in S$. Thus ${\omega }^{\beta }\le \alpha$, while ${\omega }^{\beta +1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.

Apply division by an additively indecomposable ordinal with $\lambda ={\omega }^{\beta }$:Since $\alpha \ge {\omega }^{\beta }$, the quotient $q$ is nonzero. It must be finite: if $q\ge \omega$, thencontradicting the maximality of $\beta$. Writing $q=n$ givesThis is the leading-term decomposition.

A càdlàg process $A$ is a finite-variation process when, almost surely, for every $T<\infty$,where the supremum is over every finite partition of an interval $0=t_0<\cdots <t_n=T$.

Uniform convergence on compacts in probability of $X^n$ to $X$ means that, for every $T<\infty$ and $\epsilon >0$,

The Bonnet-Myers theorem states that if a complete connected $n$-dimensional Riemannian manifold satisfiesfor some $k>0$, thenIn particular, $M$ is compact and has finite fundamental group.

By the Hopf-Rinow theorem, points $p,q\in M$ are joined by a unit-speed length-minimizing geodesic $\gamma :[0,\ell ]\to M$. Choose a parallel orthonormal frame $E_1,\dots ,E_{n-1}$ normal to $T=\dot{\gamma }$ and setThe endpoint-vanishing fields $V_i$ arise from fixed-endpoint variations. Since $\gamma$ minimizes length, its Riemannian index form is nonnegative on each $V_i$. Summing the second variation of Riemannian arc length givesUsing the Ricci curvature bound and integrating ${\sin }^2$ and ${\cos }^2$ yieldsso $\ell \le \pi /\sqrt{k}$. Taking the supremum over $p,q$ proves the diameter bound. Hopf-Rinow now makes the closed bounded space $M$ compact. Finally, the same bound applies to the complete universal cover; a compact universal cover has finite fibres over $M$, so ${\pi }_1(M)$ is finite.

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

True. Choose $N$ such that $\ell /2<a_n<3\ell /2$ for every $n\ge N$. All the finitely many preceding terms are also positive. Define the positive minimum and finite maximumEvery term lies between $\eta$ and $B$, so every ratio is at most $B/\eta$. Consequently the required supremum satisfies

The radius of convergence of a power series ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ is the number $R\in [0,\infty ]$ such that the series has absolute convergence for $|z|<R$ and diverges for $|z|>R$. Equivalently,Use $r^0=1$, including at $r=0$. The supremum may be $+\infty$. There is no general conclusion at $|z|=R$: boundary convergence depends on the coefficients. If $R=0$, only the centre is guaranteed; if $R=\infty$, every finite $z$ lies inside the disk of convergence.

Put $s=\sup (-A)$. For every $x\in A$, $-x\le s$, hence $-s\le x$, so $-s$ is a lower bound of $A$. If $l$ is any lower bound of $A$, then $-x\le -l$ for every $x\in A$, making $-l$ an upper bound of $-A$. The definition of supremum gives $s\le -l$, or $l\le -s$. Thus $-s$ is the greatest lower bound, and

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

False. Take $x_{1,n}=0$ for every $n$, and $x_{m,n}=1$ for every $m\ge 2$ and every $n$. Each row is constant, so its infimum and supremum coincide. Nevertheless,The two outer extrema can select different rows.