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A nonconstant holomorphic map $f:S\to T$ between compact connected Riemann surfaces has degree $d$ equal to the number of points in a fiber counted with local multiplicities. This number is independent of the target point. The Riemann-Hurwitz formula iswhere $e_p$ is the local degree at $p$.

Choose a polynomial $h(s)$ of degree $2g+2$ with distinct roots and take the smooth compactification of $t^2=h(s)$. The affine curve is nonsingular, since simultaneous vanishing of the partial derivatives would require $t=0$ and a repeated root of $h$. At infinity use $u=1/s$, $v=t/s^{g+1}$; the equation becomes $v^2=u^{2g+2}h(1/u)$, which has two smooth points above $u=0$ with nonzero $v$. These charts give a compact hyperelliptic curve, connected by the permitted path-connectedness result for the affine curve.

Projection to the $s$-sphere has degree two and simple ramification at the $2g+2$ roots of $h$, with no ramification at the two points at infinity. The Riemann-Hurwitz formula gives $2g(S)-2=-4+(2g+2)=2g-2$. Hence the surface has genus $g$, proving every nonnegative integer occurs as the genus of a compact connected Riemann surface.

The real Stone-Weierstrass theorem states that a subalgebra $A\subseteq C(K,\mathbb{R})$ on a compact Hausdorff space $K$, containing constants and separating points, is dense in the uniform norm. Let $B$ be its uniform closure. Polynomial approximation to $|t|$ on a bounded interval shows that $f\in B$ implies $|f|\in B$. Therefore $B$ is closed under pointwise maximum and minimum, since $\max (f,g)=(f+g+|f-g|)/2$ and $\min (f,g)=(f+g-|f-g|)/2$.

Fix $h\in C(K)$ and $ϵ>0$. Point separation and constants supply, for each pair $x,y$, a function $f_{xy}\in A$ agreeing with $h$ at both points. For fixed $x$, finitely many open sets where $f_{xy}>h-ϵ$ cover $K$; their corresponding maximum $g_x\in B$ satisfies $g_x>h-ϵ$ everywhere and $g_x(x)=h(x)$. The open sets where $g_x<h+ϵ$, as $x$ varies, cover $K$. Choose finitely many and take their minimum $g\in B$. Then $h-ϵ<g<h+ϵ$ throughout $K$. Approximating $g$ by members of $A$ proves the theorem.

Now let $F$ be the uniform closure of integer-coefficient polynomials on $[a,b]$. It is a closed ring. Since $q={\max }_{[a,b]}|1-2x|<1$, the integer-coefficient polynomials ${\sum }_{j=0}^{N}x(1-2x{)}^j$ converge uniformly to $1/2$, with error at most $b{q}^{N+1}/(1-q)$. Thus $1/2\in F$. Ring operations give every dyadic rational constant, and closure gives every real constant. Since $x\in F$, all real polynomials lie in $F$. The Stone-Weierstrass theorem now yields $\boxed{F=C([a,b])}$, as in integer polynomial approximation away from zero and one.

On $[0,b]$ the answer is no. The constant function $1/2$ cannot be uniformly approximated by integer-coefficient polynomials: at zero every such polynomial has an integer value and therefore error at least $1/2$.

Let $q:S^2\to X$ identify the two points to $p$. Take disjoint small open disks around the original points and let $U$ be their image; $U$ is a contractible wedge of two disks. Let $V=X\setminus \{p\}$, which is homeomorphic to a sphere with two punctures and retracts onto a circle. Then $U\cap V$ consists of two punctured disks, each retracting onto a circle.

In the Mayer–Vietoris sequence, the map $H_1(U\cap V)={\mathbb{Z}}^2\to H_1(U)\oplus H_1(V)=\mathbb{Z}$ has rank one and is surjective: both annular loops map to generators, with a possible orientation sign. Its kernel is $\mathbb{Z}$, so $H_2(X)=\mathbb{Z}$. At degree zero the map ${\mathbb{Z}}^2\to \mathbb{Z}\oplus \mathbb{Z}$ is $(a,b)\mapsto (a+b,-a-b)$, with rank-one kernel. The preceding degree-one map is surjective, so exactness gives $H_1(X)=\mathbb{Z}$. Since $X$ is connected and both pieces have no higher homology,This illustrates homology after identifying two points on a sphere.

Put $N=\lfloor x\rfloor$. Comparing the harmonic number with an integral gives $H_N={\sum }_{n=1}^{N}1/n>{\int }_1^{N+1}dt/t=\log (N+1)>\log x$. By the fundamental theorem of arithmetic, the finite Euler product ${\prod }_{p\le x}(1-p^{-1}{)}^{-1}$ expands as the sum of $1/n$ over positive integers whose prime factors are at most $x$. It therefore includes every term of $H_N$, andThe first strict inequality is the geometric-series bound for the logarithmic remainder; the last uses the telescoping series ${\sum }_{n=2}^{\infty }1/[n(n-1)]=1$. Thus the prime reciprocal lower bound is

On a maximal torus element $\mathrm{diag}(t,t^{-1})$, the SU(2) representation $V_2$ has weights $2,0,-2$. A monomial with respective multiplicities $a,b,c$ has weight $2(a-c)$, givingFor $0\le \ell \le n$, its weight-$2\ell$ multiplicity is $\lfloor (n-\ell )/2\rfloor +1$, and multiplicities are symmetric in $\ell$. An irreducible $V_{2j}$ contributes one at each weight $2j,2j-2,\dots ,-2j$, so subtracting adjacent multiplicities yields the symmetric powers of the three-dimensional SU2 representation decompositionIt contains the trivial representation once when $n$ is even, and never when $n$ is odd. ThereforeThe double cover $SU(2)\to SO(3)$ identifies $V_2$ with the complexified standard three-dimensional representation. It is self-dual, so the same invariant dimensions apply to homogeneous polynomials. The quadratic form $q=x^2+y^2+z^2$ is invariant, and its powers give nonzero invariants in every even degree. Since each such degree has dimension one and each odd degree has dimension zero, every invariant is a polynomial in $q$. There is no polynomial relation on $q$, as seen by setting $y=z=0$. Thus

The chromatic polynomial $p_G(t)$ counts proper vertex colorings with $t$ labeled colors at positive integer $t$. For an edge $e$, the deletion-contraction recurrence for the chromatic polynomial is $p_G=p_{G-e}-p_{G/e}$, where parallel edges in the contracted graph are merged. Induction on edges, starting from the edgeless polynomial $t^n$, shows that the coefficients have alternating signs: the contracted graph has one fewer vertex, so subtracting its alternating polynomial reinforces the same signs. The leading coefficient stays one. The coefficient of $t^{n-1}$ decreases by one per added edge, hence it is $-m$. Thus $p_G(t)={\sum }_{i=0}^n(-1{)}^{n-i}{a}_i{t}^i$, with $a_n=1$, $a_{n-1}=m$ and all $a_i\ge 0$.

For a tree, choose a root, give it any of $t$ colors, and give every other vertex any of the $t-1$ colors different from its parent. This gives $p_G(t)=t(t-1{)}^{n-1}$. Conversely, a graph with this polynomial has $m=n-1$. The chromatic polynomial of a disjoint union is the product of those of its connected components, each divisible by $t$. The given polynomial has a simple zero at zero, forcing a single connected component. A connected graph with $n-1$ edges is a tree. The converse holds.

Finally the polynomial determines the number of vertices, the number of edges and the chromatic number. If it equals that of the Turán graph $T_r(n)$, with $1\le r\le n$, then $G$ has chromatic number $r$, so has no $K_{r+1}$, and has the maximum possible number of edges for such a graph. The equality case in Turán's theorem makes $G$ isomorphic to $T_r(n)$. This is the chromatic polynomial determines a Turán graph property. If $r>n$, the graph is simply complete and its edge count already determines it.

A transitive set $x$ satisfies $z\in y\in x\Rightarrow z\in x$, equivalently $\bigcup x\subseteq x$. If $x$ is transitive and $z\in y\in \bigcup x$, choose $w\in x$ with $y\in w$. Transitivity gives $y\in x$, so $z\in \bigcup x$. Thus $\bigcup x$ is transitive. If $z\in y\in \mathcal{P}(x)$, then $z\in x$ and transitivity implies $z\subseteq x$, whence $z\in \mathcal{P}(x)$. Thus the power set is transitive as well.

The converse for the union is false: take $x=\{\{\varnothing \}\}$. Then $\bigcup x=\{\varnothing \}$ is transitive but $x$ is not. The power-set converse is true: if $z\in y\in x$, then $y\in \{y\}\in \mathcal{P}(x)$ implies $y\in \mathcal{P}(x)$, so $z\in x$.

Define $x_0=x$, $x_{n+1}=\bigcup x_n$ and $\mathrm{TC}(x)={\bigcup }_{n<\omega }{x}_n$. This is a set by replacement and union, contains $x$ as a subset, and is transitive because taking a member advances one stage. Induction shows it is contained in every transitive set containing $x$, proving the transitive closure property.

If the rank of a set $x$ is $\alpha$, thenEvery subset of $x$ has rank at most $\alpha$, and $x$ itself is an element of $\mathcal{P}(x)$, proving the first formula. The second is transitive closure preserves set-theoretic rank. Here the convention is $x\subseteq \mathrm{TC}(x)$, not $x\in \mathrm{TC}(x)$.

For a small Fourier mode $u=U{e}^{ikx+\sigma t}$, $v=V{e}^{ikx+\sigma t}$, the inhibitor equation gives $V=U/(1+k^2)$. The linearization of the cubic reaction at zero is $-rU$. Thus the fast-inhibitor cubic activator growth rate isIt is an even function of $k$, starts at $\sigma (0)=-r<0$, and tends to $-\infty$ as $|k|\to \infty$. In terms of $q=k^2$, its derivative is $-D+\rho /(1+q{)}^2$. If $\rho >D$, the maxima occur at $k=\pm \sqrt{\sqrt{\rho /D}-1}$, withIf $\rho \le D$, the maximum is $-r$ at $k=0$ and all modes decay. For $\rho >D$, the zero state is unstable when $(\sqrt{\rho }-\sqrt{D}{)}^2>r$, and the stability boundary iswhere this curve lies in the allowed parameter region.

Putting $\tilde{u}=1-u$, $\tilde{v}=1-v$ and $\tilde{r}=1-r$ negates both original equations and restores precisely the same form, because $\tilde{u}(\tilde{u}-\tilde{r})(\tilde{u}-1)=-u(u-r)(u-1)$. Therefore the state $u=v=1$ has the corresponding boundary $\boxed{\rho =(\sqrt{1-r}+\sqrt{D}{)}^2}$.

For positive odd coprime integers $m,n$, the Jacobi reciprocity law isFactor $a={\prod }_j{q}_j^{e_j}$. Since $a$ is not a square, some $e_j$ is odd. Choose a quadratic nonresidue modulo that $q_j$, and choose residue one modulo each other $q_i$. The Chinese remainder theorem provides a positive $n$ with these residues and with $n\equiv 1(\mathrm{mod}4)$. Multiplicativity of the Jacobi symbol then gives $(n/a)=-1$.

Suppose only finitely many primes $p_1,\dots ,p_s$ satisfy $(a/p_i)=-1$. They are coprime to $a$. Repeat the preceding Chinese remainder theorem construction, additionally imposing $n\equiv 1(\mathrm{mod}{p}_i)$ at every odd $p_i$ and taking $n\equiv 1(\mathrm{mod}4)$ to avoid two. Since $n\equiv 1(\mathrm{mod}4)$, Jacobi reciprocity gives $(a/n)=(n/a)=-1$. Factoring this Jacobi symbol over the prime factors of $n$ shows that at least one such prime $p$ has $(a/p)=-1$. But none of the listed primes divides $n$, a contradiction. This proves the infinitely many prime quadratic nonresidues of a nonsquare integer assertion.