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For $f\in L^{\infty }({\mathbb{R}}^n)$ and $g\in L^1({\mathbb{R}}^n)$, define their convolution byIt is bounded becauseIf ${\tau }_{h}g(z)=g(z+h)$, thenContinuity of translation in $L^1$ makes the right-hand side tend to zero with $h$, uniformly in $x$. Thus the convolution of L infinity and L1 functions is bounded and continuous.

Now let $f=1_A$ and $g(x)=1_A(-x)$. ThenAt zero this equals $|A|>0$. By continuity it remains positive throughout some open neighbourhood $U$ of zero. Positivity at $z$ means that some $y$ satisfies $y\in A$ and $y-z\in A$, so $z\in A-A$. HenceThis is the Steinhaus theorem.

The conclusion also holds when $|A|=\infty$. Since ${\mathbb{R}}^n$ is the union of bounded balls, some measurable subset $A\cap B(0,R)$ has finite positive measure. Its difference set lies inside $A-A$ and already contains a neighbourhood of zero.

The continuous dual space of a real normed vector space $X$ iswith the operator normTwo normed spaces are isometrically isomorphic when there is a bijective linear map between them that preserves norms.

For $y=(y_j)\in {\ell }^{\infty }$, defineThe series is absolutely convergent andso ${\Phi }_y\in ({\ell }^1{)}^{\ast }$ and $\Vert {\Phi }_y\Vert \le \Vert y{\Vert }_{\infty }$. For every $\epsilon >0$, choose $j$ with $|y_j|>\Vert y{\Vert }_{\infty }-\epsilon$ and test on $\mathrm{sgn}(y_j)e_j$. This gives the reverse inequality, hence

Conversely, given $\Phi \in ({\ell }^1{)}^{\ast }$, set $y_j=\Phi (e_j)$. Then $|y_j|\le \Vert \Phi \Vert$, so $y\in {\ell }^{\infty }$. For $x\in {\ell }^1$, its finite partial sums converge to $x$ in norm; continuity of $\Phi$ therefore givesThe representing sequence is unique, and the construction is linear and norm preserving. Thus the duality of l1 and l infinity proves

Choose ${\hat{x}}_i\in p_i^{-1}(x_0)$. By the classification of connected covering spaces and the degree of a connected covering, each cover corresponds to an index-two subgroupAn index-two subgroup is the kernel of a nonzero homomorphism ${\mathbb{Z}}^2\to \mathbb{Z}/2\mathbb{Z}$. There are three:The group $G{L}_2({\mathbb{F}}_2)$ acts transitively on the three nonzero linear functionals on ${\mathbb{F}}_2^2$, and its elementary matrices lift to the integral general linear group. Hence there is $A\in G{L}_2(\mathbb{Z})$ such that

Take $\psi =f_A$. Since ${\psi }_{\ast }(H_1)=H_2$, the lifting criterion for a covering space gives a based liftsuch thatApplying the same argument to $A^{-1}$ gives a lift in the reverse direction. The composites are based lifts of the appropriate identity maps; uniqueness of based lifts makes them identities. Thus $\varphi$ and $\psi$ are homeomorphisms, and the required square commutes. This proves the equivalence of connected double covers of the torus.

Let $A\in G{L}_2(\mathbb{Z})$. If $r-r^{\prime }\in {\mathbb{Z}}^2$, then $A(r-r^{\prime })\in {\mathbb{Z}}^2$, sodefines a continuous map on $X={\mathbb{R}}^2/{\mathbb{Z}}^2$. Since $\det A=\pm 1$, the integral general linear group contains $A^{-1}$ as another integer matrix, and $f_{A^{-1}}$ is the inverse of $f_A$. Hence $f_A$ is a homeomorphism, and $f_A(x_0)=x_0$.

The map $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is the lift of $f_A$ fixing the origin. A loop whose lift ends at $m\in {\mathbb{Z}}^2$ is sent to a loop whose lift ends at $Am$. Under the isomorphism in part (a), the induced map is thereforeThis is precisely the integral linear automorphism of the torus associated with $A$.

The map $p$ is the quotient of ${\mathbb{R}}^2$ by integer translations, andGiven a based loop $\gamma$ at $x_0$, the path lifting theorem gives a unique lift $\tilde{\gamma }$ beginning at $0$. Its endpoint lies in ${\mathbb{Z}}^2$. The homotopy lifting property shows that this endpoint depends only on $[\gamma ]$, so defineLifting a concatenation after translating the second lift by the endpoint of the first shows that $\Phi$ is a group homomorphism. It is surjective because, for every $m\in {\mathbb{Z}}^2$, the path $r\mapsto p(rm)$ is a loop whose lift ends at $m$. If a lift ends at zero, it is a loop in the simply connected space ${\mathbb{R}}^2$ and contracts there; projecting the contraction proves that its original loop is null-homotopic. Thus $\Phi$ is injective, and the lift-endpoint description of the fundamental group of the torus gives

The tower law for field extensions states that for finite field extensionsone hasTo prove it, let $(u_i{)}_{i=1}^m$ be a $K$-basis of $E$ and $(v_j{)}_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as ${\sum }_j{a}_j{v}_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_i{v}_j$ span $L$ over $K$.

Ifthen independence of the $v_j$ over $E$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_i{v}_j)$ is a $K$-basis of $L$, proving the formula.

The sum of the degrees of all faces is $2e$. Each of the $t$ triangular faces contributes three. Because the graph is bridgeless and has no four-cycle, every other face has degree at least five, and therefore

No edge can border two triangular faces: two distinct triangles sharing that edge would have their other two edges form a four-cycle, while the same triangle on both sides would force the connected bridgeless graph to be that triangle, contrary to $n\ge 4$. Thus the $3t$ edge incidences belonging to triangular faces use distinct edges, soCombining the inequalities givesand hence

The Euler formula for a connected planar graph now yieldsso the triangle-pentagon planar edge bound is

Equality is possible. Cut each corner of a dodecahedron through the midpoints of its incident edges. The resulting icosidodecahedral graph haswith every edge incident to one triangular and one pentagonal face. It has no four-cycle, and

Let $G$ have $n$ vertices and $e$ edges. Its complement graph $\overline{G}$ hasedges. If both graphs were planar, the planar graph edge bound would giveEquivalently, $n^2-13n+24\le 0$, whose larger root is $(13+\sqrt{73})/2<11$. Thus an integer $n\ge 11$ is impossible, proving that

Now suppose that $G$ and $\overline{G}$ are both bipartite. In a bipartition $V(G)=A\bigsqcup B$, every pair within $A$ and every pair within $B$ is adjacent in $\overline{G}$. Since a bipartite graph contains no triangle, $|A|,|B|\le 2$, and hence $|V(G)|\le 4$. Equality is attained by the path $P_4$, whose complement is again a copy of $P_4$. Therefore the maximum order is

Assertion (iii) can be false. Let $\kappa ={\aleph }_0$ and $\lambda =2^{{\aleph }_0}$. Part (ii) and the currying law for cardinal exponentiation giveBy Cantor theorem, $2^{\lambda }>\lambda$, and therefore

Assertion (ii) is always true. Since $\kappa$ is infinite and $\kappa \le \lambda$,where the last equality uses ${\lambda }^2=\lambda$. Thus the cardinal exponentiation between two and the exponent gives

Choose sets $K,L,M$ of cardinalities $\kappa ,\lambda ,\mu$. Their cardinal arithmetic operations arewhere $K^L$ is the set of functions $L\to K$. The relation $\kappa \le \lambda$ means that there is an injective function $K\to L$.

The currying law for cardinal exponentiation follows from the explicit bijectionand provesMoreover, $2^{\kappa }$ is the cardinality of the power set of a set of size $\kappa$, so Cantor theorem gives

For completeness, identify each cardinal with its initial ordinal. Suppose that some infinite $\kappa$ violates ${\kappa }^2=\kappa$, and choose the least such cardinal. Well-order the pairs $(\alpha ,\beta )\in {\kappa }^2$ first by $\max \{\alpha ,\beta \}$ and then lexicographically. Every proper initial segment is contained in ${\gamma }^2$ together with finitely many boundary pieces for some $\gamma <\kappa$, and has cardinality below $\kappa$ by minimality. The resulting well-order therefore has cardinality at most $\kappa$. The reverse inequality is immediate from $\alpha \mapsto (\alpha ,0)$, contradicting the choice of $\kappa$. Thus the square of an infinite cardinal satisfies ${\kappa }^2=\kappa$.

If $\nu =\max \{\kappa ,\lambda \}$, monotonicity now givesHence the sum and product of two infinite cardinals obey

Assertion (i) can be false. For any infinite $\lambda$, take $\kappa =(2^{\lambda }{)}^{+}$. Then ${\kappa }^{\lambda }\ge \kappa >2^{\lambda }$, so

Since $1-C=1/2$, the second local hypergeometric solution isUnder $z={\sin }^{2}x$, it must be a constant multiple of the independent elementary solution $\sin (kx)$. As $x\to 0$, both $z^{1/2}$ and $x$ are asymptotic to one another, while $sin(kx)\sim kx$. Matching the normalized hypergeometric factor, whose value at zero is one, givesConsequently the hypergeometric sine identity is

The solution of $y^{\prime \prime }+k^{2}y=0$ with $y(0)=1$ and $y^{\prime }(0)=0$ is $y(x)=\cos (kx)$. Near $x=0$, the map $z={\sin }^{2}x$ has the local inverse $x=\arcsin \sqrt{z}$, so this solution is analytic in $z$, corresponds to exponent zero at $z=0$, and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when $z={\sin }^{2}x$ on the chosen local branch,

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.