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For a countable set $A$, the polynomials of polynomial degree at most $d$ with coefficients in $A$ are described by coefficient tuples in $A^{d+1}$. The finite Cartesian power of a countable set result makes this a countable set. Taking the union over $d\ge 0$ shows there are only countably many such polynomials altogether; restricting to the nonzero polynomials preserves countability.

A nonzero polynomial of polynomial degree $d$ has at most $d$ distinct roots of a polynomial. To see the bound, a root $r$ gives the factorization $P(x)=(x-r)Q(x)$ by the factor theorem, and each other root is a root of $Q$, whose degree is $d-1$; induction starts with a nonzero constant, which has no roots. Thus $\varphi (A)$ is a countable union of countable sets, each of them finite, and is a countable set.

Starting with the given countable set $A_0$, mathematical induction now shows that every $A_n=\varphi (A_{n-1})$ is countable. A final application of the countable union of countable sets theorem givesThis countability argument does not require that the sequence of sets be increasing.

A set is countable if it admits an injection into the natural numbers; this includes finite sets and the empty set. Use ${\mathbb{N}}_0=\{0,1,2,\dots \}$ for the enumeration. The Cantor pairing functionis a bijection ${\mathbb{N}}_0^2\to {\mathbb{N}}_0$. Indeed, on the diagonal $i+j=s$, its values are the consecutive integers from $s(s+1)/2$ through $s(s+1)/2+s$. These intervals are disjoint, and the next starts one beyond the preceding endpoint. Every nonnegative integer therefore determines exactly one diagonal $s$, then exactly one $j$, and finally $i=s-j$. If one's convention starts the natural numbers at one, translate both coordinates by one. Thus $\mathbb{N}\times \mathbb{N}$ is a countable set.

For countable sets $X,Y$, choose injections $e_X:X\to {\mathbb{N}}_0$ and $e_Y:Y\to {\mathbb{N}}_0$. The map $(x,y)\mapsto \Pi (e_X(x),e_Y(y))$ is an injection, proving their Cartesian product is countable. Repeating this proves that every finite Cartesian power of a countable set is countable.

For a sequence of countable sets $X_n$, choose an injection $e_n:X_n\to {\mathbb{N}}_0$ for each $n$. For $x\in {\bigcup }_n{X}_n$, let $n(x)$ be the least index containing $x$. The mapis an injection: its value recovers $n(x)$ and then $x$. This proves the countable union of countable sets theorem. Selecting all $e_n$ uses the usual axiom of countable choice; if enumerations are already supplied, the construction itself needs no additional choices.

For positive integers, $m/m=2^0$, establishing reflexivity. If $m/n=2^k$, then $n/m=2^{-k}$, establishing symmetry. If also $n/\ell =2^j$, then $m/\ell =2^{k+j}$, establishing transitivity. Hence this is an equivalence relation.

Every positive integer can be written uniquely as $m=2^{r}d$, where $r\ge 0$ is an integer and $d$ is an odd positive integer, its odd part of a positive integer. Existence follows by repeatedly dividing by two while the result is even; the process terminates because the positive integers strictly decrease. For uniqueness, $2^{r}d=2^{s}e$ with $d,e$ odd would make one of $d,e$ even if $r\ne s$; therefore $r=s$ and $d=e$.

Multiplying by a power of two preserves the odd part of a positive integer, and two positive integers with the same odd part have ratio a power of two. The equivalence class indexed by odd $d$ is consequently $\{2^{r}d:r\ge 0\}$. The required representative set is $\boxed{A=\{1,3,5,\dots \}}$.

The equivalence class of $x\in X$ is $[x]=\{y\in X:y\sim x\}$. By the reflexivity of the equivalence relation, $x\in [x]$, so every equivalence class is nonempty and their union is $X$.

Suppose $[x]\cap [z]\ne \varnothing$, and choose $w$ in this intersection. We have $w\sim x$ and $w\sim z$. Symmetry and transitivity give $x\sim z$. If $y\in [x]$, then $y\sim x\sim z$, so $y\in [z]$; interchanging $x,z$ proves the reverse inclusion. Thus two equivalence classes are either identical or disjoint. The distinct equivalence classes therefore form a set partition of $X$. This also covers $X=\varnothing$, whose family of equivalence classes is empty.

True. For each $a\in A$, the definition of preimage givesThus $\boxed{f^{-1}(Z\cap W)=f^{-1}(Z)\cap f^{-1}(W)}$. Unlike the corresponding statement for images of a function, this intersection identity requires no injectivity.

False. Take $A=\{0,1\}$, $B=\{0\}$ and the constant function $f(a)=0$. For the subsets $X=\{0\}$ and $Y=\{1\}$, the intersection $X\cap Y$ is empty, so $f(X\cap Y)=\varnothing$, whereas the intersection of their images of a function is $f(X)\cap f(Y)=\{0\}$. In general the inclusion $f(X\cap Y)\subseteq f(X)\cap f(Y)$ holds, but a common image value can arise from different points of $X$ and $Y$.

True. The function composition $j=g\circ f$ is an injection: $j(a)=j(a^{\prime })$ first implies $f(a)=f(a^{\prime })$ by the injectivity of $g$, then $a=a^{\prime }$ by the injectivity of $f$. Fix one $a_0\in A$, using the given nonemptiness, and defineThe first case is well defined because $j$ is an injection. For every $a\in A$, $h(j(a))=a$. Thus this left inverse satisfies $\boxed{h\circ g\circ f={\mathrm{id}}_A}$, the identity function on $A$.

False. Take $A=\{0\}$ and $B=C=\{0,1\}$, with $f(0)=0$ and $g={\mathrm{id}}_B$. Then $f$ is an injection and $g$ is a surjection, but the function composition $g\circ f$ has image of a function $\{0\}$, so it is not a surjection onto $C$.

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

Let ${\mathbf{e}}_1^{\prime }(t),{\mathbf{e}}_2^{\prime }(t),{\mathbf{e}}_3^{\prime }(t)$ be an orthonormal basis fixed in the rotating reference frame $S^{\prime }$. A small rotation through $\mathbf{\omega }dt$ changes each basis vector by $\mathbf{\omega }\times {\mathbf{e}}_j^{\prime }dt$, so its derivative in the inertial frame isWrite the arbitrary vector as $\mathbf{a}={\sum }_j{a}_j(t){\mathbf{e}}_j^{\prime }(t)$. Differentiating the components and the moving basis separately givesThis proves the rotating-frame derivative formula, including when the angular velocity varies with time.

Choose the gravitational potential energy to vanish at infinity. The total mechanical energy isPut $\gamma =GM/h^2$ and $\phi =\theta -{\theta }_0$. From part (i), $u=\gamma (1+e\cos \phi )$ and $u^{\prime }=-\gamma e\sin \phi$. Consequentlyand the angular terms cancel against the gravitational potential energy:For a nonradial Kepler orbit, negative, zero and positive mechanical energy correspond respectively to $e<1$, $e=1$ and $e>1$.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

The derivative matrixis symmetric in $j,k$. Contracting it with the antisymmetric Levi-Civita symbol gives the curlDirect differentiation of the scalar potential givesso $\mathbf{F}=\nabla \varphi$. To identify the level sets, use the cross product identityIf $\mathbf{\omega }\ne 0$, write $\mathbf{x}={\mathbf{x}}_{\parallel }+{\mathbf{x}}_{\perp }$ relative to its direction. Then $\varphi =-|\mathbf{\omega }{|}^2|{\mathbf{x}}_{\perp }{|}^2/2$. Thus each negative level $\varphi =c$ is a circular cylinder with axis $\mathbb{R}\mathbf{\omega }$ and radius $\sqrt{-2c}/|\mathbf{\omega }|$. The zero level is the axis itself; positive levels are empty. If $\mathbf{\omega }=0$, the scalar potential is identically zero, so its zero level is all of space and every other level is empty.

The vector triple product removes the nested cross products:Since $\mathbf{\omega }$ is constant, the divergence isThe cube has volume $2^3=8$. With the outward unit normal, the divergence theorem therefore gives

The assertion is true over the complex numbers. It suffices to generate every nonzero scaling using translations and $I(z)=1/z$, since these together with scalings already generate the Möbius group.

For $s\ne 0$, compose the maps rightmost first. Away from the intermediate poles, the successive values areHenceFor every $\lambda \ne 0$, choose a complex square root $s$ with $s^2=-\lambda$. This gives $D_{\lambda }$ and proves Möbius scaling generated by translations and reciprocal inversion. Equality of the rational maps extends the identity to all points of the Riemann sphere, including the intermediate exceptional values.

The assertion is false. Let $D_{\lambda }(z)=\lambda z$ and $I(z)=1/z$. Their relations areMoving each inversion to the right reduces every word to $D_{\lambda }{I}^{ϵ}$ with $ϵ=0$ or $1$. Thus the subgroup generated by complex scalings and reciprocal inversion consists only of $z\mapsto \lambda z$ and $z\mapsto \lambda /z$. Of these, only the first form fixes zero.

The Möbius transformationfixes zero but is not a scaling: it sends infinity to $1$ instead of infinity. It therefore cannot belong to the subgroup generated by scalings and reciprocal inversion.

The assertion is true. A Möbius map fixes infinity if and only if $c=0$: if $c\ne 0$, its value there is the finite number $a/c$. For $c=0$, the decomposition above uses only a nonzero complex scaling and a translation. These maps form the Affine subgroup of the Möbius group.

Assume $H=\ker \theta$ for a group homomorphism $\theta :G\to K$. For $h\in H$ and $g\in G$,Thus $gH{g}^{-1}\subseteq H$. Apply the same inclusion with $g^{-1}$ to obtain the reverse inclusion, giving $gH{g}^{-1}=H$. Hence every kernel is a normal subgroup, proving the reverse implication and the required equivalence.

Assume $H$ is a normal subgroup. Take the quotient group $K=G/H$ and the natural projection $\theta (g)=gH$. The well-defined multiplication constructed above gives $\theta (gk)=\theta (g)\theta (k)$, so this is a group homomorphism. Its identity is the coset $H$, and $\theta (g)=H$ exactly when $g\in H$. Thereforewhich proves the implication from normality to being the kernel of a group homomorphism.

Divide the velocity by the speed calculated above. The unit tangent vector isIts derivative isBecause $ds/dt=\sqrt{2}(1+t^2)>0$, the principal normal vector has the direction of $T^{\prime }(t)$, givingThis is a unit vector, and $N\cdot e_z=0$ for every real $t$. The curvature of a space curve is $|T^{\prime }|/(ds/dt)=1/(1+t^2{)}^2>0$, so the principal normal vector is defined everywhere on this curve. The same perpendicularity follows from constant tangent component forces perpendicular principal normals: $T\cdot e_z=1/\sqrt{2}$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.