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Let $\theta =T-T_0>0$ and ${\theta }_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming volume flux, equal to the outgoing volume flux, isOnly one of these equal exchange fluxes multiplies the temperature difference in the heat budget: incoming fluid brings $q{T}_0$, outgoing fluid removes $qT$. With constant mass density and specific heat capacity, those factors cancel and the well-mixed ventilation temperature balance isIntegrating from ${\theta }_1$ givesThe cooling is algebraic rather than exponential because the buoyancy-driven single-opening exchange flow weakens as its driving reduced gravity decreases. The formula assumes negligible wall heat capacity, no heat input, the specified aperture geometry, and a uniform interior temperature maintained by mixing.

With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving stable density stratification. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of buoyancy, that changes the physical outcome.

Optical cluster observations give galaxy positions and line-of-sight redshifts. Subtract the mean recession to obtain the line-of-sight velocity dispersion ${\sigma }_{\mathrm{los}}$. If the tracers are in approximate virial equilibrium with isotropic velocities, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$ and $K\simeq 3M{\sigma }_{\mathrm{los}}^2/2$. Write the gravitational energy as $U_G=-\alpha G{M}^2/R$, with $\alpha$ determined by the density profile and radius convention. The virial theorem givesFor a uniform sphere $\alpha =3/5$, this becomes $5R{\sigma }_{\mathrm{los}}^2/G$. Projected separations are commonly used to estimate a gravitational radius rather than literally assume a uniform sphere. Interlopers, substructure, anisotropy, tracer bias and a nonzero surface term affect the estimate. The inferred mass is total gravitating mass, not merely the galaxies' stellar mass.

For X-ray observations, the hot intracluster gas emits thermal bremsstrahlung and atomic lines. Spectroscopy determines its temperature, and the surface brightness, after deprojection, determines a gas-density profile. Under spherical hydrostatic equilibrium with thermal gas pressure, $d{P}_g/dr=-{\rho }_{g}GM(<r)/r^2$ and $P_g={\rho }_g{k}_{B}T/(\mu m_p)$. ConsequentlyThe order-of-magnitude virial version is $k_{B}T\sim \mu m_{p}GM/R$. X-ray gas temperature is therefore a second measure of the depth of the gravitational potential. Nonthermal pressure, mergers, clumping and temperature gradients must be considered before equating the thermal estimate to the true mass.

For dense clusters the vacuum force is small compared with their internal gravity. If retained in the hydrostatic equation, it changes $GM/r^2$ to $GM/r^2-\Lambda r/3$ and adds $\Lambda r^3/(3G)$ to the mass formula above. The optical virial relation likewise uses the separate vacuum contribution derived in part (c). Comparing optical and X-ray estimates tests the dynamical assumptions and exposes a gravitating mass much larger than the luminous stellar component.

Use the physical root of the cubic which tends to $1/2$ as $\eta \to 0$. For $0<\eta <1$, the cubic has value $1$ at $x=0$ and value $-\eta /4$ at $x=1/2$, and its derivative $6\eta x^2-(2+\eta )$ is negative throughout that interval. Thus it has exactly one root there. Implicit differentiation givesThe smallest limiting value occurs as $\eta \to 1^{-}$, since larger $\eta$ cannot have a finite turnaround. At $\eta =1$ the equation factors:The continuation of the collapsing root is thereforeThe lower endpoint is an infimum, not an attained minimum for a shell that turns around in finite time. This is the minimum virial-radius ratio with positive cosmological constant. The separate root $x=1$ at $\eta =1$ describes the limiting static situation and is not the collapse branch.

The smaller ratio does not mean vacuum repulsion makes turnaround easier. At fixed turnaround data, $U_m=-A-B$ is more negative than for $\Lambda =0$. If one tried $x=1/2$, the virialized energy would be $-A-B/2$, which is not negative enough to equal $U_m$. Extra contraction is necessary to increase the magnitude of the gravitational binding energy. Thus vacuum repulsion inhibits formation of a bound shell, while the shell that does form has a smaller idealized final-to-turnaround ratio under the stated energy and profile assumptions.

No. Assign the two free generators to $(1,0)$ and $(0,1)$ in the lexicographically ordered group ${\mathbb{Z}}^2$. The universal property gives a lattice-ordered group homomorphism onto this group: its image contains both ordinary group generators. It is not an Archimedean ordered group, sinceThus a lattice-homomorphic image can contain a nonzero positive infinitesimal even when the original free object has an Archimedean pointwise function representation.

No. The free Abelian lattice-ordered group on two generators is countable, since every element is represented by a finite term over a countable alphabet of operation symbols and the two generators. The additive group of all real-valued functions on the given half-plane is uncountable: it already contains a distinct constant function for every real number. A lattice-ordered group homomorphism that is an isomorphism would be a bijection, contradicting these cardinalities. The concrete free-group model consists of specially constrained integer-linear term functions, not all real-valued functions on a domain.

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

No: there are at most three, including zero and the whole group. A two-generator ordered Abelian group is torsion-free and has rank of an abelian group at most two. A convex subgroup of an ordered group is pure: if $ng\in C$ with $n>0$, then $g$ lies between zero and $ng$ or their negatives, so $g\in C$. Therefore a proper inclusion $C<D$ of convex subgroups increases rational rank: an element of $D\setminus C$ cannot have a nonzero integer multiple in $C$. Convex subgroups form a chain, so a rank-two group permits at most two strict inclusions. This is the bound for convex subgroups of a finite-rank ordered Abelian group. It is sharp: lexicographically ordered group ${\mathbb{Z}}^2$ has exactly $0$, $\{0\}\times \mathbb{Z}$, and ${\mathbb{Z}}^2$.

Construct the conjugacy of positive real-line bumps using $u=v=\alpha$. Since both $\alpha f$ and $\alpha g$ are greater than $\beta$, choose the fundamental-interval map to be identity on $[\alpha ,\beta ]$ and then increase from $\beta$ to the respective right endpoints. Equivariant extension along $f$ and $g$ gives a conjugator on their supports fixing $[\alpha ,\beta ]$. Extend it over the remaining portions of $J=[\gamma ,\gamma h]$, fixing the endpoints of $J$; both support intervals lie strictly inside $J$. Call the resulting interval bijection $b_0$. It satisfies $f{b}_0=b_{0}g$ on $J$.

The point $\gamma$ lies in a positive orbital of $h$, because $\gamma h>\gamma$. Its half-open cells $J{h}^n$, $n\in \mathbb{Z}$, partition that orbital. Extend byand set $b$ to identity outside the orbital. Since $b_0$ fixes the cell endpoints, this is an order automorphism and satisfies $bh=hb$. Each cell is preserved; $f$ and $g$ are identity on all cells other than the original one, so $fb=bg$ holds everywhere. Finally $b$ fixes $[\alpha ,\beta ]$ pointwise, and $a$ is supported inside that interval. Hence $ba=ab$. ThusThis is conjugating positive bumps while preserving a fundamental interval; merely choosing an arbitrary conjugator would not ensure either commutation relation.

Write the two supports as $I_f=(l_f,r_f)$ and $I_g=(l_g,r_g)$. Every point in a support moves strictly right: a positive nonidentity order automorphism has no fixed point inside its one support interval. Choose $u\in I_f$, $v\in I_g$ and an increasing bijection $\theta :[u,uf]\to [v,vg]$ matching endpoints. The half-open intervals $[u{f}^n,u{f}^{n+1})$, for $n\in \mathbb{Z}$, partition $I_f$, because the iterates tend to its endpoints. DefineThe endpoint choices make adjacent pieces fit, giving an increasing bijection $b:I_f\to I_g$ with $fb=bg$. Extend over the two complementary closed rays by increasing maps taking $l_f,r_f$ to $l_g,r_g$. Outside the supports both automorphisms are identity, so $fb=bg$ still holds. Therefore $\boxed{b^{-1}fb=g}$. This gives the conjugacy of positive real-line bumps.

For a faithful transitive ordered action, primitivity here means that the only invariant equivalence relations with convex classes are equality and the universal relation. The trichotomy is: a regular Archimedean action, an order two-transitive action, or a periodic action. In the regular case the chain is an ordered additive subgroup of $\mathbb{R}$, acting by translations. In the periodic case there is a fixed-point-free period $z$ on the Dedekind completion, commuting with the action and having cofinal and coinitial integer iterates; inside a period interval the relevant stabilizer acts order two-transitively. These are the cases of the McCleary trichotomy theorem.

Only the regular Archimedean case satisfies the inequality. It is a linearly ordered group. For positive $a,b$, if $a\le b$, then $ab\le b^2\le b^2{a}^2$; if $b\le a$, then $ab\le a^2\le b^2{a}^2$. Taking $a=|f|$, $b=|g|$ proves the displayed inequality.

To exclude an order two-transitive action, choose $x<y<z<t$. Choose an automorphism taking $(x,y)$ to $(y,z)$ and replace it by its pointwise join with the identity, giving positive $a$ with $xa=y$, $ya=z$. Similarly obtain positive $b$ with $xb=x$, $yb=t$. ThenThus $ab\nleqq b^2{a}^2$, while $|a|=a$ and $|b|=b$. The same construction inside a period interval excludes the periodic case, using the order two-transitive stabilizer supplied by the classification.

Here transitivity is order n-transitivity: increasing tuples are compared, rather than arbitrary reorderings. For increasing tuples ${\alpha }_1<\cdots <{\alpha }_n$ and ${\beta }_1<\cdots <{\beta }_n$, choose $h_{ij}$ sending both ${\alpha }_i$ to ${\beta }_i$ and ${\alpha }_j$ to ${\beta }_j$ when $i\ne j$, using order two-transitivity. For $i=j$, choose an element sending the single point to its target, using transitivity. Define

This belongs to the lattice-ordered permutation group. At ${\alpha }_i$, all terms in its $i$th inner meet equal ${\beta }_i$. At ${\alpha }_k$, every inner meet is at most ${\beta }_k$, because its $j=k$ term takes that value. The outer join is consequently exactly ${\beta }_k$ there. Thus $h$ maps the whole tuple to its target. This finite interpolation by lattice operations proves every finite order-transitivity without assuming the group permits arbitrary cut-and-paste restrictions of its elements.

For each $g\ne 1$, choose a value in a lattice-ordered group $P_g$, using Zorn's lemma. Such a value is a prime convex lattice subgroup. Otherwise the last construction in Question 1(i) would give two larger convex lattice subgroups whose intersection is $P_g$; maximality among subgroups omitting $g$ would put $g$ in both, hence in their intersection. Thus the right coset set ${\Omega }_g=P_g\backslash G$ is a totally ordered set.

Right multiplication defines an action ${\rho }_g(a):P_{g}h\mapsto P_{g}ha$. It is well-defined, preserves the total order, and has inverse ${\rho }_g(a^{-1})$. The coset formulas from Question 1(i) show pointwise preservation of both join and meet, becauseand likewise for the meet.

Take the disjoint union of these chains, arranged in any fixed order sum of the indices $g\ne 1$. Let every $a\in G$ act on each component by ${\rho }_g(a)$. This gives a lattice-ordered group homomorphism into the order automorphisms of one chain. It is injective: a nonidentity $a$ moves the identity coset in ${\Omega }_a$, since $P_{a}a\ne P_a$. Its image is closed under pointwise join and meet, giving exactly the required lattice-ordered permutation group. This is the Holland representation theorem.

Let $T_e$ be the theory of the extension axioms for an unrestricted binary relation. It has quantifier elimination. To prove this, consider a conjunction of literals $\theta (y,\mathbf{x})$. A witness $y$ equal to one of the parameters is handled by substitution. A witness distinct from them is governed entirely by a finite pattern of incoming edges, outgoing edges and its loop. For each equality pattern of the parameter tuple, incompatible literals make the pattern impossible, and every compatible pattern is supplied by the appropriate extension axiom. Thus existence of a witness is a finite disjunction of quantifier-free conditions on the parameters. Disjunctive normal form and induction on formulas eliminate all remaining quantifiers.

Given any formula $\varphi (\mathbf{x})$, choose a quantifier-free $\psi (\mathbf{x})$ withBy compactness theorem, finitely many extension axioms ${\sigma }_1,\dots ,{\sigma }_s$ already entail this particular universally quantified equivalence. Each has asymptotic probability one by the assumption. The union bound givesTherefore the universal equivalence also has asymptotic probability one. We obtain asymptotic equivalence to a quantifier-free formula, uniformly over every tuple in each structure satisfying those finitely many axioms. An infinite conjunction of probability-one events is not used; the finite consequence supplied by compactness is essential.

For an unrestricted binary relation $R$, the extension axioms permit arbitrary directed edges and loops. For each $m\ge 0$ and each choice of bits ${ϵ}_i,{\delta }_i,\eta \in \{0,1\}$, require that every tuple of distinct $x_1,\dots ,x_m$ has a fresh element $y$ withHere $P^1$ denotes $P$ and $P^0$ denotes $\neg P$; the notation is a literal abbreviation, not exponentiation of a relation. Incoming and outgoing choices are independent, and the loop bit is also specified. These are the extension axioms for an unrestricted binary relation. If one instead restricts to undirected loop-free graphs, the allowable patterns reduce to the random-graph extension axioms of Question 2; no such restriction is imposed here.