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A subset is nowhere dense when its closure has empty interior. It is of first category, or meagre, when it is a countable union of nowhere-dense subsets; otherwise it is of second category.

The Baire category theorem says that in a complete metric space every countable intersection of open dense subsets is dense. To prove it, start inside an arbitrary nonempty open set and choose a closed ball of positive radius there and in the first dense open set. Inductively choose a closed ball inside the interior of the preceding ball and the next dense open set, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested closed balls, hence in all the dense open sets and the original open set. Thus a nonempty complete metric space is not meagre in itself.

For $p<r\le \infty$, put $E_m=\{x\in {\ell }^p:\Vert x{\Vert }_p\le m\}$, viewed inside ${\ell }^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the ${\ell }^r$ norm gives coordinatewise convergence, and every finite partial sum of $\sum |x_i{|}^p$ is bounded by $m^p$ in the limit. Taking the supremum of partial sums proves the assertion.

It has empty interior. Given a point in it and any small ${\ell }^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The ${\ell }^r$ distance is $\delta$, while the resulting ${\ell }^p$ norm is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is nowhere dense, andFor $r=p$, including $p=\infty$, the space is Banach and so second category in itself by Baire. There is no case $r>p$ when $p=\infty$.

The Simplicial approximation theorem states that a continuous map between realizations of finite simplicial complexes becomes homotopic to a simplicial map after sufficiently many barycentric subdivisions of the domain. More precisely, a vertex assignment is a simplicial approximation when the image of each open star is contained in the open star of its assigned vertex; sufficiently fine subdivision permits such an assignment. The resulting straight-line homotopy stays in the appropriate target simplices.

The vertices of a barycentric subdivision are the barycentres of nonempty faces. An $n$-simplex has $2^{n+1}-1$ such faces. A top-dimensional subdivided simplex corresponds to a full flag of faces of sizes $1,2,\dots ,n+1$, equivalently to an ordering of the original vertices. ThusTriangulate both copies of the sphere using finite complexes. For each fixed subdivision level there are only finitely many vertex maps to the finite target vertex set, hence only finitely many simplicial maps. Every continuous map is homotopic to one of them at some finite level. A countable union of finite sets is countable, proving at most countably many homotopy classes without needing their classification by degree.

Since $26\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{26}]$ with discriminant $104$. Write $P=(2,ϵ+1)=(2,\sqrt{26})$ and $Q_{\pm }=(5,ϵ\pm 1)$. Reduction of $X^2-26$ modulo two gives a double root, and modulo five gives the roots $\pm 1$. Thus $N(P)=2$, $N(Q_{\pm })=5$, and $(5)=Q_{+}{Q}_{-}$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot 4=2$. The element $ϵ+1=6+\sqrt{26}$ belongs to $P$ and $Q_{+}$ and has field norm ten, so equality of ideal norms givesThe Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every ideal class contains an integral ideal of ideal norm at most five. Three is inert and gives no prime ideal of ideal norm three; ideals of ideal norm four are $P^2$, principal. Thus the class group is generated by $P,Q_{+},Q_{-}$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have field norm $\pm 2$, but $x^2-26y^2=\pm 2$ would give $x^2\equiv 2$ or $11(\mathrm{mod}13)$, neither a quadratic residue. Hence the class number is two.

The unit $ϵ=5+\sqrt{26}$ has field norm $-1$. To show it is fundamental, suppose $1<u<ϵ$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm 1/u$, so $y=(u-u^{\prime })/(2\sqrt{26})$ is a positive integer less than two. Thus $y=1$, and $x^2-26=\pm 1$ forces $x=\pm 5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of ${ϵ}^{-1}$ puts it in $[1,ϵ)$, proving that all units are $\pm {ϵ}^n$.

An element of field norm $\pm 10$ has principal ideal $P{Q}_{+}$ or $P{Q}_{-}$ by prime ideal factorization. The first is generated by $ϵ+1$; the second by $ϵ-1=4+\sqrt{26}$, of field norm $-10$. Two generators of the same principal ideal differ by a unit. Therefore all, and only, the integer solutions areThe two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two field norm signs.

Use the usual positive-definite convention for binary quadratic forms with negative discriminant. Reduction by integral changes of variables of determinant one produces a form $a{x}^2+bxy+c{y}^2$ with $|b|\le a\le c$, and with $b\ge 0$ on the boundary. For completeness, replacing $x$ by $x+ny$ makes $|b|\le a$; if then $c<a$, interchange the variables, reversing one sign to keep determinant one. The new positive leading coefficient is smaller. Repeating terminates and gives a reduced form.

For discriminant $-7$, $4ac-b^2=7$ and the reduction inequalities give $3a^2\le 7$. Thus $a=1$, $b$ is odd and $|b|\le 1$, and $c=2$. The boundary convention chooses $b=1$. Hence there is exactly one positive-definite equivalence class, represented by $x^2+xy+2y^2$.

For an odd prime $p\ne 7$ represented by this form, the identity $4p=(2x+y{)}^2+7y^2$ shows that $-7$ is a square modulo $p$. Here $p\nmid y$, because otherwise $p\mid x$ and $p^2$ would divide the represented number. Conversely, if $-7$ is a square modulo $p$, choose odd $b$ with $b^2\equiv -7(\mathrm{mod}p)$. Then $(p,b,(b^2+7)/(4p))$ is an integral positive-definite form of discriminant $-7$. By the preceding reduction it is equivalent to the given form, so its value $p$ at $(1,0)$ is represented by that form. Quadratic reciprocity gives $(-7/p)=(p/7)$, whose value is one precisely for $p\equiv 1,2,4(\mathrm{mod}7)$. The exceptional primes are represented: $2=Q(0,1)$ and $7=Q(1,-2)$. ThereforeWithout the positive-definite convention, the first printed assertion is false: $-x^2-xy-2y^2$ also has discriminant $-7$ but cannot be equivalent to a positive-definite form, because an invertible change of variables preserves its sign.

Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A real number is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.

If current coordinates lie in a field $K\subseteq \mathbb{R}$, a line-line intersection solves linear equations over $K$, while line-circle or circle-circle intersections reduce to a quadratic equation over $K$. Thus each step requires at most adjoining a square root. Consequently every constructible number lies in a towerConversely field sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive square root by the geometric-mean construction in a semicircle. Hence every real number in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.

In particular a constructible algebraic number has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the Galois group of its normal closure is a finite two-group: a tower's normal closure remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by Galois correspondence.

The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]{2}$, whose polynomial $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos 20^{\circ }$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the unit circle requires side length $\sqrt{\pi }$, impossible because $\pi$ is transcendental whereas constructible numbers are algebraic.

For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\phi (n)$, so it is constructible exactly when this degree is a power of two. Writing the prime factorization of $n$ shows the equivalent conditionIndeed odd prime exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is prime, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic Galois group a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.

Work with finite graphs, as in the usual matching theorem. Hall's condition is $|N(A)|\ge |A|$ for every $A\subseteq X$. It is necessary because a matching injects $A$ into its neighbours.

For sufficiency, induct on $|X|$. If a nonempty proper subset $A$ has $|N(A)|=|A|$, induction matches $A$ to $N(A)$. For $B\subseteq X\setminus A$, Hall's inequality for $A\cup B$ gives $|N(B)\setminus N(A)|\ge |B|$, so induction matches the complement to unused neighbours. Combining the matchings finishes this case. Otherwise every nonempty proper $A$ has $|N(A)|\ge |A|+1$. Choose an edge $xy$, delete its endpoints, and observe that every subset of the remaining $X$ still has at least its own number of neighbours. Induction matches the remainder, and adjoining $xy$ finishes. The one-vertex case and the empty case are immediate.

For $0\le d\le |X|$, the defect version isNecessity follows because at most $d$ vertices of $X$ can be unmatched. For sufficiency add $d$ new vertices to $Y$, each joined to every vertex of $X$. The displayed inequality makes Hall's condition hold in the enlarged graph. Its full matching uses at most $d$ new vertices, giving the required original edges.

To prove the matching-cover equality, let $M$ be a maximum matching and $U$ its unmatched vertices in $X$. Follow alternating paths from $U$, using nonmatching edges from $X$ to $Y$ and matching edges back. Let $Z_X,Z_Y$ be the reachable sets. No reachable $Y$ is unmatched, since that would give an augmenting path and a larger matching. Matching edges give a bijection $Z_Y\leftrightarrow Z_X\setminus U$. The set $C=(X\setminus Z_X)\cup Z_Y$ covers every edge: an edge from a reachable $X$ to an unreachable $Y$ cannot be a nonmatching edge, and its matching edge, if present, is already reachable. Its size isEvery vertex cover has at least $|M|$ vertices, because the matching edges are disjoint. Therefore maximum matching size equals minimum vertex-cover size, as required.