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For each $s\in S\setminus T$, maximality implies a nontrivial relationHere $a_s\ne 0$, since otherwise this would contradict the linear independence in a module of $T$. Thus $a_{s}s\in N$. Take $r={\prod }_{s\in S\setminus T}{a}_s$, with empty product equal to one. Because $R$ is an integral domain, $r\ne 0$. Every $s\in S\setminus T$ satisfies $rs\in N$, and this is also true for $s\in T$. Expressing any element of the finitely generated module as an $R$-linear combination of $S$ now gives $rM\subseteq N$. This is clearing denominators relative to an independent module subset; it does not require $R$ to be a field.

False without a separation assumption. Give $X=\{0,1\}$ the indiscrete topology $\{\varnothing ,X\}$, and take $Y=\{0\}$. Both spaces are finite and hence compact, but $Y$ is not a closed set because $X\setminus Y=\{1\}$ is not open. In a Hausdorff space, a compact subset would be closed; that hypothesis is absent here.

True. In a discrete topology, the singletons form an open cover. If $X$ is compact, a finite subcover exists, so $X$ is finite. Conversely, any open cover of a finite set has a finite subcover: choose one member containing each point. This converse does not require the discrete topology.

For an active monster, independence givesBoth probabilities vanish before time $100$ and at odd times. By convolution and Cauchy-Schwarz, $p_{2m}(a,0)\le p_{2m}(0,0)<1/m$: indeed, write the $2m$-step probability as the inner product of two translates of the $m$-step probability mass function. ThereforePositivity follows by prescribing $100$ simultaneous steps taking both walkers straight to the origin; that event has probability $4^{-200}$. An active monster thus leaves a positive escape probability.

For a sleepy monster, the origin capture by a sufficiently sleepy planar random walk can be proved using the printed schedule: during $I_j=\{n_j+1,\dots ,n_{j+1}\}$ it occupies $Y_{j+1}$. For any $M$, let $J_M$ be the first $j\ge M$ with $Y_{j+1}=0$. Markov-chain recurrence of $Y$ makes $J_M$ finite almost surely, and it is independent of $X$. By conditioning on $J_M$ and using the stated block bound,As $M\to \infty$, continuity of probability for decreasing events shows that the event $C_{\infty }$ of infinitely many simultaneous origin visits has probability at least $1/2$. Regard the independent trajectories as continuing even after a possible first capture.

To turn this into probability one, use the eventual coupling of planar symmetric random walks; the block events need not be independent. Two planar simple symmetric random walks begun at sites of the same parity can be coupled to agree eventually. Run them independently until they meet. Their difference walk is irreducible on the even-parity sublattice, and its $m$-step return probability is $p_{2m}(0,0)$, by convolution and symmetry. The preceding divergent-series criterion makes this difference walk a recurrent Markov chain, so it hits zero almost surely. After that stopping time, use identical increments for both copies, preserving their marginal laws by the Strong Markov property.

Fix any finite calendar time $N$ and any two possible histories up to that time. Corresponding princess positions have the same parity, as do the corresponding monster positions at its internal walk time. Apply the eventual probabilistic coupling separately to the two princess futures and the two monster futures, keeping the princess and monster walks independent in each pair. Both the calendar clock and the monster's internal clock tend to infinity. Hence after a finite calendar time the two pairs of trajectories agree, so either both realize $C_{\infty }$ or neither does. The conditional probability of $C_{\infty }$ is therefore the same for every possible finite history.

Consequently $\mathbb{E}(1_{C_{\infty }}\mid {\mathcal{F}}_N)=\mathbb{P}(C_{\infty })$ for every $N$. By the martingale convergence theorem, these conditional expectations converge to $1_{C_{\infty }}$ as the histories exhaust the trajectories. Thus $\mathbb{P}(C_{\infty })$ is zero or one. Since it is at least $1/2$, it is one. In particular,This argument uses the supplied unconditional block probabilities; it does not silently strengthen them to conditional probabilities given past failures.

For a planar simple symmetric random walk write its coordinates as $(S_n,T_n)$ and rotate to $(S_n+T_n,S_n-T_n)$. Each step now has four equally likely values $(1,1),(1,-1),(-1,1),(-1,-1)$. Thus the two rotated coordinates are independent one-dimensional simple symmetric random walks.

A return is impossible at odd times, and at time $2n$ its probability isFor $n\ge 2$, the supplied estimate gives $1/(4n)<p_{2n}(0,0)<1/n$. At $n=1$ the lower bound is actually equality, $p_2(0,0)=1/4$; this harmless strict-endpoint typo does not affect the argument. The harmonic series diverges, so the criterion proved above shows that the origin is recurrent, in the terminology of a recurrent Markov chain. Translation invariance gives the same conclusion for every state. The Markov chain is irreducible, so from any starting point it hits any fixed state almost surely and revisits it infinitely often. One can see the hitting assertion directly: repeated returns to the starting point give repeated positive-probability opportunities to follow a finite path to the target.

Let $T_j=\inf \{n\ge 1:X_n=j\}$. By the Strong Markov property, the probability of hitting $j$ and subsequently returning to $i$ is $f_{ij}{f}_{ji}$. This event implies a return to $i$, so $f_{ii}\ge f_{ij}{f}_{ji}$. For $i=j$, interpret the event as two successive returns.

Set $T_i^{(0)}=0$ and let $T_i^{(r)}$ be the $r$th successive return. Repeated use of the Strong Markov property gives ${\mathbb{P}}_i(T_i^{(r)}<\infty )=f_{ii}^r$. The total number $N_i$ of visits, including time zero, is both ${\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ and ${\sum }_{r\ge 0}{1}_{\{T_i^{(r)}<\infty \}}$. By Tonelli theorem,The identity is valid with value $+\infty$. In particular, divergence of the return-probability sum is equivalent to Markov-chain recurrence.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

The time-independent Schrödinger equation gives the even interior wavefunction $\psi =A\cos (\alpha x)$, with ${\alpha }^2=2m(U+E)/{\hslash }^2$. Outside, normalizability of a bound state requires $\psi =C{e}^{-\beta (|x|-l)}$, with ${\beta }^2=-2mE/{\hslash }^2$.

Continuity of the wavefunction and its derivative at $x=l$ gives $C=A\cos (\alpha l)$ and $-\beta C=-\alpha A\sin (\alpha l)$. A nonzero solution cannot have $\cos (\alpha l)=0$, so division givesAt the formally allowed endpoint $E=-U$, the even interior solution is constant; derivative matching would force the exterior amplitude to vanish, leaving the zero solution. Nontrivial bound states therefore have $-U<E<0$. The matching formula is the even-state condition for a finite square well.

For $z=R+iy$, $0\le y\le \pi$, the numerator modulus is $e^{-2Ry/\pi }$ and the denominator modulus is at least $1-e^{-2R}$. Hence the magnitude of the right-side contour integral is at mostOn $z=-R+iy$ the corresponding bounds are $e^{2Ry/\pi }$ and $e^{2R}-1$, giving exactly the same bound $\pi /(2R)$. Thus the total vertical contribution is $O(R^{-1})$ and vanishes. This integral estimate handles the endpoints where a pointwise uniform-decay argument would fail.

On the top side put $z=x+i\pi$, with $x$ running from $R$ to $-R$. Since $e^{i(x+i\pi {)}^2/\pi }=-e^{i{x}^2/\pi }{e}^{-2x}$ and $e^{-2(x+i\pi )}=e^{-2x}$, the reversed orientation cancels the minus sign. The two horizontal integrals therefore sum toIf the vertical contributions vanish, the previous residue theorem evaluation yields the Fresnel integralThis also is an ordinary oscillatory improper integral, not merely a symmetric principal value: integration by parts, using $(e^{i{x}^2/\pi }{)}^{\prime }=(2ix/\pi )e^{i{x}^2/\pi }$, bounds either tail beginning at $R$ by a constant times $1/R$.

The denominator vanishes when $z=(2j+1)i\pi /2$. For $R>0$, only $z_0=i\pi /2$ lies inside the rectangle. This is a simple pole, since the denominator derivative there is $2$. Its residue isThe positive orientation and the residue theorem consequently give

Factor $X^3+X=X(X-i)(X+i)$. These factors generate pairwise comaximal ideals, so the Chinese remainder theorem givesThe three copies of $\mathbb{C}$ have different module structures: $X$ acts by $0,i,-i$, respectively. Thus $q(X)$ acts on $(a,b,c)$ as $(q(0)a,q(i)b,q(-i)c)$, making $\Phi$ a module homomorphism.

The inverse, obtained by Lagrange interpolation, isEvaluating at $0,i,-i$ recovers $(a,b,c)$. Conversely, a polynomial whose three evaluations vanish is divisible by $X(X-i)(X+i)$, so these maps are mutual inverses on the quotient module. Three copies with identical trivial $X$-action would not give the required isomorphism.

The annihilating polynomial excludes any free $\mathbb{C}[X]$-summand. In a cyclic primary decomposition summand $\mathbb{C}[X]/(P^a)$, the same condition requires $P^a$ to divide $(X-2{)}^4$. Thus $P=X-2$, up to a scalar unit, and $1\le a\le 4$. The complex dimension of this summand is $a$.

Write $A_j=\mathbb{C}[X]/((X-2{)}^j)$. The exponents must form a partition of an integer $4$, giving exactlyThey are distinct module isomorphism types by uniqueness in the structure theorem for finitely generated modules over a principal ideal domain; equivalently they are the five possible Jordan block patterns for eigenvalue $2$ in dimension four.

The structure theorem for finitely generated modules over a principal ideal domain applies because a Euclidean domain is a principal ideal domain. For a finitely generated $R$-module,where the $d_i$ are nonzero nonunits. The free rank $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. Equivalently, the torsion submodule decomposes into cyclic modules $R/(P^a)$ for irreducible $P$, with uniquely determined prime-power elementary factors.

For the Legendre transform, maximize $px-f(x)$ over $x>0$. Its derivative is $p-1-\log x$, and strict concavity makes the stationary point $x=e^{p-1}$ the unique maximizer. Substitution gives the convex conjugate of x log xThe range of $f^{\prime }$ is all of $\mathbb{R}$, so all real dual arguments have this finite value. Every point of the original graph is supported by its tangent of slope $f^{\prime }(x)$; the conjugate supremum therefore recovers $f$ from these supporting lines. Equivalently, the inverse derivative relation for this strictly convex differentiable Legendre transform gives $f^{\ast \ast }=f$ on $x>0$. If an extended-real function on all of $\mathbb{R}$ is wanted, its closed convex extension has value zero at $x=0$ and infinity for $x<0$; biconjugation includes that endpoint closure.

A convex function satisfies $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$ for all $x,y$ in its domain and $0\le t\le 1$. If the second derivative exists on an interval, this is equivalent to $g^{\prime \prime }\ge 0$ there. The equivalence follows from the nondecreasing first derivative, or from the supporting-line inequalities obtained by integration. HereThus it is a strictly convex function, and in particular convex.

At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.

On the stated branch of the principal complex logarithm, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1{)}^2)$. ThereforeZero is a logarithmic branch point, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to 0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen branch cut; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.

For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently zero is a non-isolated singularity, as described by reciprocal cosine with accumulating poles. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos (1/z)$ itself has an essential singularity at zero, taking its reciprocal introduces this accumulation of poles.

The zeroes of $\cosh z$ occur at $z=i\pi (n+\frac{1}{2})$, $n\in \mathbb{Z}$. They are simple because $\sinh z\ne 0$ there. Squaring the denominator therefore givesThere are no other finite singularities; this follows from the local zero-order criterion for a pole.