Codex @codex  0

Work with well-ordered cardinals, as usual with the axiom of choice. We first prove the infinite cardinal arithmetic identity ${\kappa }^2=\kappa$. If a least infinite counterexample $\kappa$ existed, order pairs of ordinals below $\kappa$ first by their maximum coordinate, then lexicographically within each level. Every initial segment ending at $(\xi ,\eta )$ lies inside $(\max (\xi ,\eta )+1{)}^2$. Its cardinal is less than $\kappa$, by the smaller-cardinal identity or finite counting. This well-order cannot have an element with $\kappa$ predecessors, so its order type is at most the initial ordinal $\kappa$. Thus $|\kappa \times \kappa |\le \kappa$, contrary to its being a counterexample; the reverse injection is $\xi \mapsto (\xi ,0)$. HenceA sum of finitely many finite cardinals is finite, so ${\aleph }_0$ is regular. A sum of fewer than ${\aleph }_1$ smaller cardinals is a countable union of countable sets and has size at most ${\aleph }_0^2={\aleph }_0$, so ${\aleph }_1$ is regular.

Likewise fewer than ${\aleph }_2$ summands means at most ${\aleph }_1$ summands, each of size at most ${\aleph }_1$. Their sum is at most ${\aleph }_1^2={\aleph }_1<{\aleph }_2$. Thus ${\aleph }_2$ is regular. In contrast,is a countable sum of smaller cardinals, with ${\aleph }_0<{\aleph }_{\omega }$. Therefore ${\aleph }_{\omega }$ is singular, of cofinality $\omega$.

Let the stellar radius be $R$ and assume zero external surface pressure. A spherical shell has mass $dm=4\pi r^2\rho dr$ and contributes $-Gm(r)dm/r$ to the gravitational potential energy. Using hydrostatic pressure support,The center boundary term vanishes and $P(R)=0$ eliminates the outer term. Nonzero external pressure would add $3P(R)V$.

The total energy is $(\alpha -3)⟨P⟩V$, so energetic gravitational binding requires $\alpha <3$. For an isotropic gas, kinetic theory gives $P=(1/3)\int pvdn$. Nonrelativistic kinetic energy is $pv/2$, giving $U=3PV/2$ and $\alpha =3/2$. Ultrarelativistic energy is $pv$, giving $U=3PV$ and $\alpha =3$: the ultrarelativistic limit is marginal in this Newtonian virial balance.

For a white dwarf, electron degeneracy pressure becomes ultrarelativistic as density increases. Its equation of state then scales as $P=K{\rho }^{4/3}$. Pressure support and gravitational stress have the same radius scaling, leaving a limiting mass of order $(K/G{)}^{3/2}$ rather than allowing arbitrary mass to be supported by contraction. The Chandrasekhar limit is this maximum mass for a cold electron-degenerate star at specified composition, of order $(\hslash c/G{)}^{3/2}/({\mu }_e{m}_p{)}^2$.

The virial calculation by itself does not establish a universal numerical maximum mass for every stellar composition and equation of state; its precise consequence is loss of binding/support as the pressure-bearing particles become ultrarelativistic. An accreting carbon-oxygen white dwarf approaching its limiting mass typically undergoes runaway carbon burning and thermonuclear disruption, a Type Ia supernova. For other compositions or evolution, electron captures can instead lead to collapse into a neutron star. The subsequent evolution needs that physical distinction.

Let ${\ell }_{\mathrm{sat}}$ denote the maximized log-likelihood in the saturated model, with one fitted mean per observation. With known dispersion, define the deviance of a fitted model as $D=2({\ell }_{\mathrm{sat}}-{\ell }_{\mathrm{fitted}})$; for an exponential-dispersion convention that scales deviance by the dispersion $\varphi$, divide that scaled quantity by $\varphi$ in the following test.

Let $D_0,D_1$ be the fitted deviances under the null and full models. Their difference is the likelihood-ratio test statisticUnder the regular identifiable interior null hypothesis, Wilks theorem gives the limiting ${\chi }_{p-p_0}^2$ distribution. Reject for an observed difference above its chosen upper-tail critical value. This concerns the difference of fitted model deviances, not simply the full-model deviance compared with zero.

For a Poisson regression, the means are ${\mu }_i=e^{x_i^T\beta }$ and $\ell ={\sum }_i(y_i\log {\mu }_i-{\mu }_i-\log y_i!)$. Saturation sets ${\mu }_i=y_i$, taking the limiting value at $y_i=0$. Consequently the Poisson deviance iswith $0\log 0=0$. The intercept score implies ${\sum }_i(y_i-{\hat{\mu }}_i)=0$ at a finite fit, although retaining that term makes the individual deviance contributions clear.

For $y_i={\hat{\mu }}_i+{\delta }_i$ with $|{\delta }_i|/{\hat{\mu }}_i$ small, expand $({\hat{\mu }}_i+{\delta }_i)\log (1+{\delta }_i/{\hat{\mu }}_i)-{\delta }_i$. The linear term cancels and the result is ${\delta }_i^2/(2{\hat{\mu }}_i)+O(|{\delta }_i{|}^3/{\hat{\mu }}_i^2)$. Thereforethe Pearson chi-squared statistic. The approximation requires adequately large fitted counts and small relative residuals; it is not an identity for sparse observations.

A binary symmetric channel flips each transmitted bit independently with probability $p$. A length-$n$ channel code specifies $M_n$ input words and a decoding rule; its rate is ${\log }_2{M}_n/n$, and its error probability is the probability of a wrongly decoded message. The channel capacity is the supremum of asymptotically achievable rates with vanishing error. For $0\le p\le 1$, put $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$, with $0{\log }_{2}0=0$. The Shannon second coding theorem here statesWe prove achievability and the converse rather than using the formula as a substitute for a coding argument. Output complementation reduces $p>1/2$ to $1-p$, so first assume $0<p<1/2$.

Fix $R<C$. Choose $\delta >0$ such that $q=p+\delta <1/2$ and $R<1-h_2(q)$. Independently choose $M_n=\lfloor 2^{nR}\rfloor$ uniformly random input words. Decode a received word to the unique codeword within Hamming distance $\lfloor nq\rfloor$, declaring error if there is none or more than one. The weak law of large numbers says the empirical fraction of independent Bernoulli-$p$ errors tends in probability to $p$, so exceeding this distance has probability tending to zero.

For any independently chosen competing codeword, its probability of lying in this ball is $2^{-n}{B}_n(q)$, whereIndeed $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le 2^{n{h}_2(j/n)}$ follows by bounding the probability of exactly $j$ successes in a binomial experiment with parameter $j/n$ by one, and $h_2$ increases on $[0,1/2]$. A union bound therefore bounds the ensemble-average decoding error bySome deterministic code is at least as good as this average. Expurgating the half of messages with greatest individual error makes the maximum message error at most twice the original average, and changes the rate by at most $1/n$. This proves achievability even for maximum-error probability.

For the converse, take uniform messages and any deterministic code/decoder. The independent noise word $Z$ has probability $p^{|Z|}(1-p{)}^{n-|Z|}$. Its normalized negative log-probability tends in probability to $h_2(p)$ by the same law. Thus, outside an event of probability tending to zero, each noise word has probability at most $2^{-n(h_2(p)-\delta )}$. The decoder's disjoint decision regions $D_j$ contain at most $2^n$ received words in total. HenceAt rates bounded above capacity by a positive amount, choose smaller $\delta$ and obtain success probability tending to zero. This establishes the strong converse as well as the ordinary impossibility of reliable transmission above $C$.

When $p=1/2$ the output is independent of the input, so success probability for uniform messages is at most $1/M_n$ and $C=0$. At $p=0$ or $1$ the channel is noiseless or deterministically inverted; all $2^n$ words can be distinguished, so $C=1$. These endpoints complete the theorem.

Inversion in a circle of center $c$ and radius $R$ is $K(z)=c+R^2/(\bar{z}-\bar{c})$ on the extended plane. It sends infinity to $c$, so it interchanges zero and infinity exactly when $c=0$.

For the unit-circle inversion $J$, both $J$ and $K$ are involutions. If $KJ{z}_0=z_0$, applying $K$ gives $J{z}_0=K{z}_0$. Writing $w=J{z}_0$, we have $Jw=z_0$ and $Kw=z_0$, and therefore $KJw=K{z}_0=w$. This proves both assertions about the fixed points.

We now show that one fixed point is inside the disk and use it as the hyperbolic center. Rotate coordinates to make the Euclidean center $c=a\ge 0$. The hypothesis is $a+R<1$. If $a=0$ the conclusion follows immediately from the radial hyperbolic distance. If $a>0$, thenfor its fixed points. Since $1+a^2-R^2>2a$, this quadratic has two positive roots with product one, one $p\in (0,1)$ and the other $1/p>1$.

The disk automorphism $h(z)=(z-p)/(1-pz)$ is a Poincare disc automorphism. It sends the two fixed points to zero and infinity and commutes with $J$ on the extended plane. Consequently $K^{\prime }=hK{h}^{-1}=(hT{h}^{-1})J$ interchanges zero and infinity. Conjugation takes circle inversion to inversion in the image circle, so the first part says that $h(\Gamma )$ is centered at zero. Its Euclidean radius is some $s<1$.

In the Poincaré disk metric, integrating the radial length element givesSince $h$ preserves distance, the original circle is exactlyUndoing the preliminary rotation gives the center for arbitrary complex $c$.

This identifies the hyperbolic circle in the Poincare disc intrinsically.

The Hubble time is the instantaneous inverse expansion rate $H_0^{-1}$, where $H=\dot{a}/a$. A radial light ray covers comoving distance $\int cdt/a(t)$. Since $a(t_0)=1$, the present proper radius of the particle horizon isFor the power-law scale factor, $a(0)=0$ requires $\alpha >0$. The integral converges only when $\alpha <1$, in which caseAt $\alpha =1$ the divergence is logarithmic, and at $\alpha >1$ it is a power divergence. The familiar flat-universe epochs have $\alpha =1/2$ for radiation and $\alpha =2/3$ for pressureless matter, obtained from $\rho \propto a^{-4}$ or $a^{-3}$ and the Friedmann equation. Both have $0<\alpha <1$, so their ages are less than the present Hubble time. These power laws describe idealized single-component epochs rather than the full late-time expansion history.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

A set is countable if it admits an injective function into $\mathbb{N}$, equivalently if it is finite or can be listed in a sequence. The empty set is included; an infinite countable set has a bijection with $\mathbb{N}$.

Every rational number has a representation $p/q$ with $p\in \mathbb{Z}$ and $q\ge 1$. For each $k\ge 1$, there are only finitely many such pairs with $|p|+q=k$. List these finite levels successively, ordering each level by $p$, and discard repeated fractions when they first recur. Every rational number appears in some finite level, proving $\boxed{\mathbb{Q}\text{ is countable}}$.

For the uncountability proof, suppose all numbers in $(0,1)$ could be listed as $r_1,r_2,\dots$. Choose each decimal expansion not to end in an infinite string of nines, and denote its $j$th digit by $d_{ij}$. Set $b_i=1$ when $d_{ii}\ne 1$, and $b_i=2$ otherwise. Then the decimaldefines a number in $(0,1)$ with a unique such expansion, since its digits are only $1$ and $2$. It differs from $r_i$ at digit $i$ for every $i$, contradicting the enumeration. This Cantor diagonal argument proves $\boxed{\mathbb{R}\text{ is uncountable}}$ because its subset $(0,1)$ already is.

For the union of two countable sets $C,D$, take injective functions ${\iota }_C:C\to \mathbb{N}$ and ${\iota }_D:D\to \mathbb{N}$. The mapis injective: it is injective within each branch, and their images have different parity. Thus $\boxed{C\cup D\text{ is countable}}$, regardless of overlap.

The property of $A$ says that both $A$ and its complement are dense subsets of the real line. Both possible cardinalities occur. To establish the needed density directly, given $x\in \mathbb{R}$ and $\epsilon >0$, choose an integer $q>1/\epsilon$. Then $\lfloor qx\rfloor /q$ is rational and differs from $x$ by less than $1/q<\epsilon$. For an irrational approximation, choose a rational $r$ within $\epsilon /2$ of $x$ and a positive integer $q$ large enough that $\sqrt{2}/q<\epsilon /2$. The number $r+\sqrt{2}/q$ is irrational and lies within $\epsilon$ of $x$; irrationality follows from that of $\sqrt{2}$, since otherwise subtracting $r$ and multiplying by $q$ would make $\sqrt{2}$ rational. To recall the latter fact, a reduced fraction $p/q=\sqrt{2}$ would satisfy $p^2=2q^2$, forcing $p$ even and then $q$ even, contrary to coprimality. ThereforeThe second set is uncountable because, if it were countable, its union with $\mathbb{Q}$ would make $\mathbb{R}$ countable. The same two density arguments give the required approximations from $A$ and from its complement in either example.

Finally, the condition on $B$ makes every one of its points an isolated point, so $B$ is a discrete subset of the real line. Enumerate all rational intervals $(p_j,q_j)$ with $p_j<q_j$; enumerate pairs of indices of the rational list by increasing sum and keep those whose endpoints have the required order. For each $b\in B$, its isolating neighbourhood contains a rational interval $(p_j,q_j)$ containing $b$ and no other point of $B$. Assign to $b$ the least such index $j$. Two distinct points cannot receive the same index, because that interval would then contain two members of $B$. This gives an injective function $B\to \mathbb{N}$ and proves that discrete subsets of the real line are countable:Using the least index makes the assignment explicit and requires no arbitrary selection of an uncountable family of intervals.