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The Fubini theorem for an integrable Borel-measurable function states that, if ${\int }_{{\mathbb{R}}^2}|f|<\infty$, its sections are integrable for almost every coordinate andFor a nonnegative measurable function, Tonelli theorem gives the same equalities with infinity allowed before integrability is established.

Here $f$ is the continuous exponential multiplied by the indicator of a Borel set, so it is Borel-measurable and nonnegative. Tonelli and integration in $x$ giveThis proves integrability. We may now integrate in the reverse order; for $x>0$ the inner integral is $(e^{-ax}-e^{-bx})/x$. Hence the Frullani integral isThe apparent singularity at zero is removable in the integrand's limiting value $b-a$.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree of a holomorphic map is the number of preimages of a regular value. More generally it is the constant sum ${\sum }_{p\in f^{-1}(q)}{e}_p$, where $e_p$ is the local multiplicity. The Riemann-Hurwitz formula is

Negation is well-defined on the complex torus because $-\Lambda =\Lambda$. It is holomorphic with holomorphic inverse itself, so $\psi$ is biholomorphic. Its fixed points satisfy $2z\in \Lambda$. If $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, they are the four distinct classes

Away from these points, choose a small coordinate disk $U$ disjoint from $\psi (U)$. The quotient map is a homeomorphism from $U$ onto $\pi (U)$, so transfer its complex coordinate to that open set of $S^{\prime }$. The transition between two such lifted charts is induced either by identity or by $\psi$, hence is holomorphic. These charts give the required Riemann-surface structure, and $\pi$ is locally a holomorphic bijection there.

The quotient has generic degree two. At a fixed point, a local coordinate centred there changes under $\psi$ as $u\mapsto -u$. The assumed holomorphic quotient map is invariant under this operation, so its local expansion is even and has multiplicity at least two. Since the total degree is two, it is exactly two. There are no other ramification points. The domain torus has genus one, so Riemann-Hurwitz formula gives

If a linear map satisfies $\Vert Tx\Vert \le M\Vert x\Vert$, then $\Vert Tx-T{x}^{\prime }\Vert \le M\Vert x-x^{\prime }\Vert$, so it is continuous. Conversely continuity at zero gives $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ for $x\ne 0$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$. Hence a linear map is continuous exactly when it is bounded.

For the separately continuous bilinear map, put $T_x(y)=F(x,y)$ for $\Vert x\Vert \le 1$. Every $T_x:Y\to Z$ is a bounded linear map by the first result. For each fixed $y$, continuity in the other variable gives $\Vert F(x,y)\Vert \le C_y\Vert x\Vert$, so ${\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert <\infty$. The Uniform boundedness principle therefore gives ${\sup }_{\Vert x\Vert \le 1}\Vert T_x\Vert =M<\infty$, and bilinearity then yields

To spell out the uniform-boundedness step, define the closed subsets $E_m=\{y:{\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert \le m\}$ of the Banach space $Y$. They cover $Y$. The Baire category theorem gives an $E_m$ containing a ball $B(y_0,r)$. For $\Vert h\Vert <r$, both $y_0+h$ and $y_0$ lie in $E_m$, so ${\sup }_x\Vert T_{x}h\Vert \le 2m$. Scaling $h=(r/2)y$ for $\Vert y\Vert \le 1$ bounds all the operator norms by $4m/r$. This proves the claimed uniform estimate without presuming joint continuity.

Use relative homology. Every simplex of $X=B\cup C$ lies in $B$ or $C$, and those lying in both are exactly the simplices of $A$. Consequently inclusion induces a chain-complex isomorphismIt is bijective on the respective bases of simplices and commutes with boundary because it comes from inclusion. Thus

For any pair $(Y,Z)$, the long exact sequence in homology shows that inclusion induces isomorphisms in every degree exactly when $H_j(Y,Z)=0$ for every $j$. To verify both directions, if all relative groups vanish, exactness makes each $H_j(Z)\to H_j(Y)$ injective and surjective. Conversely, if the inclusion maps are isomorphisms, the map from $H_j(Y)$ into the relative group is zero, while the relative connecting map is injective and has zero image because the next inclusion is injective; hence the relative group itself is zero. This includes degree zero, with the negative-degree ordinary group zero.

Apply the criterion first to $(B,A)$ and then to $(X,C)$ using the chain isomorphism. It proves

The factorization form of Kummer-Dedekind theorem is as follows. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial $f$, and $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. ThenThe index condition is essential. All integral bases used below make the index one, so it applies to every prime.

For a squarefree quadratic $d\equiv 2,3(\mathrm{mod}4)$, take $\theta =\sqrt{d}$, $f=X^2-d$ and field discriminant $D=4d$. At an odd prime, a repeated factor occurs exactly when $p\mid d$, in which case $\bar{f}=X^2$ and $p{\mathcal{O}}_K=(p,\sqrt{d}{)}^2$. At $p=2$, reduction is $X^2$ if $d$ is even and $(X+1{)}^2$ if $d$ is odd, so $2$ is totally ramified in both cases. Thus the totally ramified primes are exactly those dividing $4d$.

For $d\equiv 1(\mathrm{mod}4)$, take $\theta =(1+\sqrt{d})/2$, with polynomial $X^2-X+(1-d)/4$ and discriminant $D=d$. At an odd prime it has a double root exactly when $p\mid d$, giving a single linear factor squared and total ramification. At $p=2$, its derivative is $1$, so no repeated factor occurs. Hence again

For the prime cyclotomic field, use ${\Phi }_q(X)=1+X+\cdots +X^{q-1}$. Modulo $q$ this is $(X-1{)}^{q-1}$, so $q{\mathcal{O}}_K=(q,\zeta -1{)}^{q-1}$, proving total ramification. For $p\ne q$, the discriminant has no prime factor $p$, so the reduction is squarefree and $p$ is unramified. There is an actual exponent error in the printed discriminant: the correct value is $(-1{)}^{(q-1)/2}{q}^{q-2}$, not $q^{q-1}$. To check it, at a nontrivial root ${\zeta }^j$,and multiplying these derivatives and using ${\prod }_{j=1}^{q-1}(1-{\zeta }^j)=q$ gives the exponent $q-2$. The prime support is still just $\{q\}$, so the requested ramification conclusion is unaffected.

For $\theta =\sqrt[3]{2}$, the integral basis gives $f=X^3-2$ and $D=-108$. Modulo $2$, $\bar{f}=X^3$; modulo $3$, $\bar{f}=(X+1{)}^3$. Dedekind factorization gives $(2)=(2,\theta {)}^3$ and $(3)=(3,\theta +1{)}^3$. For other primes the polynomial is squarefree because its discriminant is not divisible by the prime. Therefore the equivalence also holds in this cubic field. This is a conclusion for these specific fields, not the false general assertion that every ramified prime in every higher-degree number field is totally ramified.

Extremal graph theory asks how many edges a graph can have while avoiding a prescribed subgraph. Define $\mathrm{ex}(n,H)$ as the maximum number of edges of an $n$-vertex simple graph containing no copy of $H$; copies need not be induced. The extremal graph often reveals the structural reason a forbidden configuration must appear.

A complete proof of Mantel theorem illustrates this principle. In a triangle-free graph, the neighborhoods of the endpoints of any edge are disjoint, so $d(u)+d(v)\le n$. Summing over edges gives ${\sum }_{v}d(v{)}^2\le nm$, where $m$ is the edge count. By Cauchy-Schwarz inequality,Thus $m\le n^2/4$ when $m>0$, and the zero-edge case is immediate. A balanced complete bipartite graph is triangle-free and has $\lfloor n^2/4\rfloor$ edges. ThereforeThe proof links a local forbidden triangle to a global degree inequality.

The general Turan theorem states that the balanced complete $(r-1)$-partite graph $T_{r-1}(n)$ maximizes the edge count among $K_r$-free graphs. Its leading density is $(1-1/(r-1))n^2/2$. A standard symmetrization explanation is to replace a lower-degree vertex by a twin of a nonadjacent higher-degree vertex: no $K_r$ is created and the edge count does not decrease. Iterating at the level of twin classes produces a complete multipartite graph with at most $r-1$ parts. Moving vertices between parts then balances the part sizes and maximizes the number of cross edges. This provides structural intuition beyond the preceding exact triangle proof.

For any fixed graph $H$ with chromatic number $r\ge 2$, the Erdős-Stone theorem isThe lower bound is the Turan construction: an $(r-1)$-partite graph cannot contain $H$. To describe the upper-bound proof, fix an edge-density excess above $1-1/(r-1)$. Apply a regularity partition with an error parameter much smaller than that excess and a positive density threshold. Delete irregular and low-density pairs; the resulting reduced graph still has density above the Turan threshold, so it contains $K_r$. The corresponding $r$ vertex classes form pairwise regular, positive-density pairs. The embedding argument successively chooses vertices with large neighborhoods in every remaining class; regularity discards only a small exceptional set at each choice. With the parameters sufficiently small and $n$ sufficiently large, this embeds a complete $r$-partite graph with $|V(H)|$ vertices in each part, and therefore a properly colored copy of $H$. This explains why chromatic number, rather than the exact shape of $H$, controls the leading quadratic term.

For example $\chi (C_5)=3$, so forbidding a five-cycle still allows asymptotically $n^2/4$ edges, and exceeding that density by any fixed positive fraction eventually forces a five-cycle. For $K_r$ the exact Turan theorem strengthens the asymptotic statement. For bipartite $H$, the theorem gives only $o(n^2)$, leaving an important finer problem: for example $K_{s,t}$-free graphs have upper bounds of order $n^{2-1/s}$ from counting common neighborhoods. Extremal results consequently range from exact finite bounds to asymptotic density thresholds and stability questions about near-extremal structure.

The propositional completeness theorem states that $\Gamma \models \varphi$ if and only if $\Gamma ⊢\varphi$. Soundness follows by checking the axioms against truth valuations and noting that modus ponens rule preserves truth. For completeness, use the deduction theorem for propositional logic: if $\Gamma \cup \{p\}⊢q$, then $\Gamma ⊢p\Rightarrow q$. In the usual negation rules this also says that a contradiction from $\Gamma \cup \{p\}$ proves $\Gamma ⊢\neg p$.

First construct a maximal consistent set extending any consistent set $\Delta$. Enumerate all formulas, possible because the primitive propositions are countable. Starting with ${\Delta }_0=\Delta$, add formula ${\psi }_n$ at step $n$ if consistency is preserved; otherwise add $\neg {\psi }_n$. This latter addition is consistent: if both extensions gave contradictions, the deduction theorem for propositional logic would prove both $\neg {\psi }_n$ and $\neg \neg {\psi }_n$ from ${\Delta }_n$, a contradiction already in ${\Delta }_n$. The union $M$ is consistent because a finite proof uses only finitely many newly added premises. It decides every formula: exactly one of $\psi ,\neg \psi$ belongs to $M$. It is deductively closed, since an omitted derivable formula would have its negation in $M$, contradicting consistency.

Assign a valuation $v(p)=1$ exactly when the primitive proposition $p$ lies in $M$. Structural induction proves the maximal consistent set truth lemma $v(\psi )=1$ if and only if $\psi \in M$. Negation follows from the decision property. For implication, if $\psi \Rightarrow \chi \in M$ and $\psi \in M$, modus ponens rule gives $\chi \in M$. Conversely, if $\chi \in M$, the elementary implication axioms give $\psi \Rightarrow \chi \in M$. If $\psi \notin M$, then $\neg \psi \in M$, and the usual contradiction rule proves $\psi \Rightarrow \chi$. These two alternatives match the truth table. The other connectives, if primitive, are handled by their corresponding rules or defined from negation and implication. Thus every consistent set has a valuation satisfying it.

Now suppose $\Gamma /⊢\varphi$. If $\Gamma \cup \{\neg \varphi \}$ were inconsistent, the deduction theorem for propositional logic would give $\Gamma ⊢\neg \neg \varphi$. Here the third axiom $\neg \neg \varphi \Rightarrow \varphi$ is essential: it would give $\Gamma ⊢\varphi$, a contradiction. Hence that extension is consistent and has a satisfying valuation. This valuation satisfies $\Gamma$ and falsifies $\varphi$, proving the contrapositive of completeness. In this organization of the proof, the explicit classical double-negation elimination occurs in this last consistency step; the construction and truth lemma use the ordinary negation and contradiction rules, not an unmentioned identification of $\neg \neg \varphi$ with $\varphi$.

The propositional compactness theorem says that a set of formulas is satisfiable exactly when each finite subset is satisfiable. If it were not satisfiable, completeness would make it inconsistent. Its finite contradiction proof uses only a finite subset of premises; soundness makes that subset unsatisfiable. This contradicts the assumed finite satisfiability and proves compactness.

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.