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First, finite additivity of a set function gives $f(\varnothing )=0$, since $f(\varnothing )=2f(\varnothing )$. Put $w_s=f(\{s\})$. Decomposing any subset into its singleton elements and using induction gives $f(A)={\sum }_{s\in A}{w}_s$. Fix $s\in S$ and let $r$ be the number of sets $A_i$ containing it. Its total coefficient in the alternating intersection sum is zero if $r=0$, and otherwise isby the binomial theorem. These are exactly the coefficients in $f({\bigcup }_i{A}_i)$. Summing against $w_s$ proves the required inclusion-exclusion principle for the additive set function, including signed weights.

For any $c\in C$, the surjectivity of $g\circ f$ supplies $a\in A$ with $g(f(a))=c$. Therefore $f(a)\in B$ is a preimage of $c$ under $g$, proving surjectivity of $g$.

For the requested counterexample, take $A=C=\{0\}$, $B=\{0,1\}$, $f(0)=0$, and $g(0)=g(1)=0$. The function composition is the identity bijection of the singleton, but $f$ is not surjective and $g$ is not injective.

If $f(a)=f(a^{\prime })$, applying $g$ gives $(g\circ f)(a)=(g\circ f)(a^{\prime })$. The injectivity of $g\circ f$ implies $a=a^{\prime }$, so $f$ is injective.

Suppose $(g\circ f)(a)=(g\circ f)(a^{\prime })$. The injectivity of $g$ implies $f(a)=f(a^{\prime })$, and the injectivity of $f$ then implies $a=a^{\prime }$. Hence $g\circ f$ is injective.

Let $c\in C$. Since $g$ is surjective, choose $b\in B$ with $g(b)=c$. Since $f$ is surjective, choose $a\in A$ with $f(a)=b$. Then $(g\circ f)(a)=g(b)=c$. Thus every element of $C$ has a preimage, proving that $g\circ f$ is surjective.

Take the half-turn about the $z$-axis, represented by the rotation matrixIt is an orthogonal matrix with determinant $1$, and it fixes the $z$-axis while reversing both perpendicular coordinates. Direct substitution gives $Ma=b$, $Mb=a$, $Mc=d$ and $Md=c$. A linear map that permutes the four vertices also preserves their convex hull, which is the solid tetrahedron. Hence this rotation matrix leaves $T$ invariant. Its axis joins the midpoints of the two opposite edges $ab$ and $cd$.

The three known vertices have pairwise squared distances $8$, so the regular tetrahedron must have every squared edge length equal to $8$. Comparing the squared distance from $d$ to $b$ with that from $d$ to $a$ givesEquality forces $x=-1$. Similarly,so $y=-1$. The resulting point has squared distance $8$ from each of the other three vertices. Therefore

Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by $a,b,c$. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by $a,c-b$ and $b,a-c$. Their orthogonality givesHence all three pairwise dot products have the same value $s$. The third pair satisfiesso the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product givesThus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.

Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their dot product isTherefore $|w{|}^2=|a{|}^2-c$. Since $c<|a{|}^2$, this is a strictly positive squared radius. The complete solution isEvery choice of unit vector $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the sphere with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that sphere.

To construct a potential for the exact differential, first integrate its $dx$ coefficient with respect to $x$:Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2y{z}^2+h(z)$. The remaining partial derivative is thenMatching the $dz$ coefficient forces $h^{\prime }=0$. Therefore a potential, valid throughout ${\mathbb{R}}^3$, isDifferentiating this expression recovers all three coefficients, proving that the differential form really is an exact differential.

For an exact differential $Md\theta +Nd\varphi =dF$, equality of mixed partial derivatives requires $M_{\varphi }=N_{\theta }$. Here the two partial derivatives areThey are not identically equal, so the differential form is not a closed differential form and hence is not an exact differential. Equality along special lines such as $\theta =0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two partial derivatives differ.

Yes: every finite group of rotations about the origin in the plane is cyclic. A planar rotation is determined by an angle modulo $2\pi$, and composition adds those angles. If $G$ consists only of the identity, it is a cyclic group. Otherwise, finiteness allows us to choose the smallest positive representative angle $\alpha \in (0,2\pi )$ occurring in $G$.

Take any planar rotation in $G$ with representative angle $\theta \in [0,2\pi )$. Write $\theta =q\alpha +r$, where $q$ is an integer and $0\le r<\alpha$. The composition $R_{\theta }{R}_{\alpha }^{-q}$ belongs to $G$ and has angle $r$. Minimality of $\alpha$ forces $r=0$, so every element of $G$ is a power of $R_{\alpha }$.

Similarly, write $2\pi =m\alpha +r$ with $0\le r<\alpha$. If $r>0$, the planar rotation $R_{\alpha }^{-m}$ has representative angle $r$, a contradiction. Thus $\alpha =2\pi /m$ andThis proves finite groups of planar rotations are cyclic by an elementary angle argument.

A group $G$ is a cyclic group if some element $g\in G$ generates it:For a finite cyclic group, the order of a group element of a generator equals the number of elements of the group.

An example of a finite abelian group that is not a cyclic group is the Klein four-group, realized aswith componentwise addition modulo two. Addition commutes, so this is an abelian group of four elements. Each of its three nonzero elements satisfies $g+g=0$, so the cyclic subgroup it generates has just two elements; the identity generates only itself. No element generates all four elements.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.

Use the product rule for the divergence of the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two dot products cancel. Apply the divergence theorem with outward unit normal vector $n$ to obtain Green's second identity:Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

In Cartesian coordinates, put $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$. Away from the origin the partial derivatives of $\psi =1/r$ areSumming the three second partial derivatives yieldsThus $1/r$ is a harmonic function on the punctured space and solves Laplace equation there. The origin is excluded: the singularity cannot be treated as an ordinary harmonic point.

The 36 ordered outcomes are equiprobable. For $2\le s\le 12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb{P}(A_s)=c_s/36$ and $\mathbb{P}(B_i)=1/6$ for $1\le i\le 6$. Their joint probability is $1/36$ if $1\le s-i\le 6$, and zero otherwise.

If the joint probability is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the independent events condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore within the possible sums and facesIf arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty event, which is trivially independent of every event.

Two independent events satisfy $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. If $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning that $B$ occurred does not change the probability of $A$. The product definition also covers zero-probability events without dividing by zero.

Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\dots ,n-1$. Conditional on that urn, the probability of a blue first draw is $k/(n-1)$, so the law of total probability givesDrawing twice without replacement requires $n\ge 3$. Given $k$, the probability that both draws are blue is $k(k-1)/((n-1)(n-2))$, and henceTaking the ratio defining conditional probability gives the second requested probabilityAlthough a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.

For $\mathbb{P}(B)>0$, the conditional probability is $\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)$. Since the events $B_i$ partition the sample space, their intersections with $A$ are disjoint and exhaust $A$. The law of total probability therefore gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substituting this and $\mathbb{P}(A\cap B_i)=\mathbb{P}(A\mid B_i)\mathbb{P}(B_i)$ into the definition proves Bayes' theorem:The denominator is positive by the given hypothesis on $A$.