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Let $Z_t={\int }_0^{t}X(s)ds$ and set $a=q+\lambda$. Starting from $-1$, the first holding time has exponential distribution of rate $a$. Its joint density with a jump to $+1$ is $q{e}^{-au}du$. Before such a jump $Z$ decreases to $-u$, so the remaining rise needed to exceed $c$ is $c+u$. If instead the jump is to the absorbing state zero, the level stays nonpositive and never exceeds $c>0$.

Conditioning on this first jump and using the Strong Markov property gives

For the natural filtration ${\mathcal{F}}_t=\sigma (X_s:0\le s\le t)$, a nonnegative random time $\tau$ is a stopping time when $\{\tau \le t\}\in {\mathcal{F}}_t$ for every $t\ge 0$: whether it has occurred can be decided from the observations already made.

For a time-homogeneous finite-state continuous-time Markov chain, the Strong Markov property says that, on $\{\tau <\infty \}$, conditional on ${\mathcal{F}}_{\tau }$, the process $(X_{\tau +s}{)}_{s\ge 0}$ has the law of a fresh chain started at $X_{\tau }$ and is independent of the earlier history given that state. In particular, with transition matrix $P_s=e^{sQ}$,The property applies also to bounded functions of several future times or of the entire future path.

Finite independent Gaussian sums are Gaussian; directly, multiplication of their characteristic functions givesBy Markov inequality, $\mathbb{P}(|Y_n-Y|>ϵ)\le {ϵ}^{-2}\Vert Y_n-Y{\Vert }_2^2\to 0$. Part (ii) makes the same sequence converge in distribution to $Y$, and the characteristic-function estimate there makes ${\phi }_{Y_n}\to {\phi }_Y$. Uniqueness of distributions determined by characteristic functions therefore yieldsIf ${\sigma }^2=0$, all coefficients vanish and $Y=0$ almost surely; $N(0,0)$ here denotes that degenerate distribution.

The independent standard Gaussian random variables have mean zero and variance one. For $m>n$, the mixed products have expectation zero, givinguniformly over $m>n$ as $n\to \infty$. Thus $(Y_n)$ is Cauchy in the L2 space. Its completeness, proved in (i), suppliesMoreover $L^2$ convergence implies convergence of the means and second moments: $\mathbb{E}Y=0$ and $\mathbb{E}{Y}^2={\lim }_n{\sum }_{j\le n}{\alpha }_j^2={\sigma }^2$.

Suppose $X_n\to X$ in probability. For a fixed real $t$ and any $\delta >0$, split according to $|X_n-X|\le \delta$. The supplied exponential inequality and the bound $|e^{it{X}_n}-e^{itX}|\le 2$ giveLet $n\to \infty$, then $\delta \downarrow 0$. Thus the characteristic functions converge pointwise to ${\phi }_X$. The Lévy continuity theorem states that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution. Every characteristic function is continuous at zero, so

The Lp space is the vector space of equivalence classes, modulo equality almost everywhere, of measurable functions $f$ with $\int |f{|}^{p}d\mu <\infty$, equipped with $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. To prove it is a Banach space, let $(f_n)$ be Cauchy in this norm. Choose a subsequence $(f_{n_j})$ such thatTake measurable representatives and put $g_j=|f_{n_{j+1}}-f_{n_j}|$, $G_M={\sum }_{j=1}^M{g}_j$. The Minkowski inequality gives $\Vert G_M{\Vert }_p\le {\sum }_j{2}^{-j}\le 1$. By the monotone convergence theorem, $G={\sum }_{j\ge 1}{g}_j$ satisfies $\int G^{p}d\mu \le 1$. Consequently $G$ is finite almost everywhere, and the telescoping seriesconverges absolutely almost everywhere. Define it arbitrarily on the null exceptional set. The bound $|f|\le |f_{n_1}|+G$ puts $f$ in $L^p$.

For the tails, monotone convergence and Minkowski again giveSince the original sequence is Cauchy, the triangle inequality then gives $f_n\to f$ in norm, not merely along the subsequence. HenceThis proof needs no finiteness or sigma-finiteness assumption on the measure space.

Each entry of ${\mathrm{char}}_A(A)$ is a polynomial in the entries of $A$, so their simultaneous zero set $Z$ is Zariski closed. A matrix in $D(\Delta )$ has distinct eigenvalues and is diagonalizable. If $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$, thenThus $D(\Delta )\subseteq Z$. Affine space is irreducible because its coordinate ring is an integral domain; by part (i), its nonempty open subset $D(\Delta )$ is dense. The closed set $Z$ containing it must be all of affine space. This proves the Cayley-Hamilton theorem:

The coefficients of $\det (TI-A)$ are polynomial functions of the matrix entries. Its polynomial discriminant $\Delta (A)$ is therefore a polynomial in those entries, and it is nonzero exactly when all roots of the characteristic polynomial are distinct. Thus the required set is the principal open subset $D(\Delta )$ of affine space. It is nonempty: an algebraically closed field is infinite, so choose a diagonal matrix with $n$ pairwise distinct entries.

It is also affine as a variety. The map $A\mapsto (A,1/\Delta (A))$ identifies it with the closed hypersurfaceProjection to $A$ is its regular inverse. This proves the requested affineness while retaining the open embedding into matrix space.

An affine variety is irreducible when it is nonempty and cannot be expressed as the union of two proper Zariski-closed sets. If a nonempty open subset $U$ were not dense, $X=\overline{U}\cup (X\setminus U)$ would be such a union: the first set is proper by assumption, and the second is proper because $U$ is nonempty. Hence every nonempty open subset of an irreducible variety is dense.

False. Take a bouquet of two circles and the two-dimensional torus. Both are connected polyhedra and have first homology group ${\mathbb{Z}}^2$. Their fundamental groups are respectively the nonabelian free group $F_2$ and the abelian group ${\mathbb{Z}}^2$, so they are not isomorphic. First homology only remembers the abelianization of the fundamental group.

True. Coordinate projections define ${\pi }_1(X\times Y,(x,y))\to {\pi }_1(X,x)\times {\pi }_1(Y,y)$. A pair of based loops gives the inverse by $t\mapsto ({\gamma }_X(t),{\gamma }_Y(t))$. Homotopies combine and project coordinatewise, and loop concatenation does too. These maps are well-defined inverse group homomorphisms, requiring no local connectedness assumptions.

False. Changing the connecting path changes the isomorphism by an inner automorphism. Take $x=y$, one path constant and the other representing $a$ in the Klein bottle group $⟨a,b:ab{a}^{-1}=b^{-1}⟩$. The first induced isomorphism is the identity, whereas the second sends $b$ to $b^{-1}\ne b$. The element $b$ has infinite order, as is also witnessed by mapping it to a translation and $a$ to a reflection in the infinite dihedral group. A fixed path to a different endpoint can be appended if distinct basepoints are desired.

True. A path $u:x\to y$ gives the basepoint-change isomorphism of fundamental groups $[\gamma ]\mapsto [\overline{u}\ast \gamma \ast u]$. Concatenating paths is compatible with homotopy and loop multiplication. Using the reversed path gives its inverse, because $u\ast \overline{u}$ and $\overline{u}\ast u$ contract relative to endpoints.

If $\rho$ is a faithful representation, $g\mapsto {\lambda }_g$ embeds $Z$ into the multiplicative group of complex numbers. Its image is a finite group of roots of unity, hence lies in the cyclic group of $n$th roots of unity and is itself cyclic. Thus $Z$ is cyclic.

Let $\chi$ be the character of a representation of $\rho$. The character orthogonality identity for an irreducible character is ${\sum }_{g\in G}|\chi (g){|}^2=|G|=n$. For $g\in Z$, the scalar formula from part (i) gives $|\chi (g)|=d$. Keeping these $m$ nonnegative summands yields

If $g$ belongs to the centre of a group, then $\rho (g)$ commutes with $\rho (h)$ for every $h\in G$. The Schur lemma says that an endomorphism of a finite-dimensional complex irreducible representation commuting with the representation is scalar. ThereforeSince $g$ has finite order, ${\lambda }_g$ is a root of unity and has modulus one.

An isomorphism fixing $L$ fixes its subfield $K$, so an $L$-isomorphism is a $K$-isomorphism. The converse is false because an isomorphism over $K$ may act nontrivially on $L$.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha =2^{1/4}>0$, $F=L(\alpha )$ and $F^{\prime }=L(i\alpha )$. Both are degree-two extensions of $L$. The polynomial $X^4-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion, and both $\alpha$ and $i\alpha$ are roots. Sending $\alpha$ to $i\alpha$ therefore gives a $\mathbb{Q}$-isomorphism $F\to F^{\prime }$; on $L$ it sends $\sqrt{2}$ to $-\sqrt{2}$.

There is no $L$-isomorphism. The two quadratic extensions adjoin square roots of $\sqrt{2}$ and $-\sqrt{2}$. Their square classes would agree only if $-1$ were a square in $L$. To check this criterion directly, a square root of $\sqrt{2}$ in $L(i\alpha )$ would have form $a+b(i\alpha )$; its square forces $ab=0$. The case $b=0$ would put $\alpha$ in $L$, while $a=0$ would require $b^2=-1$. Both are impossible, the latter since $L\subseteq \mathbb{R}$. Thus the extensions are isomorphic over $K$ but not over $L$.

A field homomorphism fixing $K$ is injective, since its kernel is an ideal of a field and cannot contain one. It is also $K$-linear. Equal finite field extension degrees then make it a bijection, hence a $K$-isomorphism.

Equal degrees alone do not ensure that any such homomorphism exists. For example $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree two, but a $\mathbb{Q}$-isomorphism would require a square root of two in $\mathbb{Q}(\sqrt{3})$. Squaring $a+b\sqrt{3}$ with rational $a,b$ would give $2ab=0$ and $a^2+3b^2=2$. Neither $a^2=2$ nor $b^2=2/3$ has a rational solution. Thus the two degree-two extensions are not isomorphic over $\mathbb{Q}$.

False. Take $\alpha ={\omega }_1+1$. Continuity of ordinal multiplication in the right argument gives $({\omega }_1+1){\omega }_1={\omega }_1^2$. For a precise bound, for each countable ordinal $\beta$ monotonicity givesBoth bounding expressions have supremum ${\omega }_1^2$ as $\beta$ ranges below ${\omega }_1$, because $2\beta$ is still countable and these indices are cofinal. But the successor rule gives

False. Take $\alpha ={\omega }^2+1$, which satisfies the proposed lower bound. Associativity of ordinal addition and $1+{\omega }^2={\omega }^2$ giveThe final successor survives only in the second expression.