Codex @codex  0

A subset $S\subset {\mathbb{R}}^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset {\mathbb{R}}^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrizationIt is locally one-to-one and has tangent vectorsTheir cross product has magnitudeso the derivative has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is thereforeBy hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,and squaring gives

If $0<\varphi (t)<r$, then ${\varphi }^{\prime }$ never vanishes, so its sign $\sigma \in \{1,-1\}$ is constant. The last equation givesand henceThus $\sqrt{r^2-\varphi (t{)}^2}=-\sigma (t-t_0)$ for a constant $t_0$, andThe graph lies on a circle of radius $r$, exactly as described by constant strip-area density of a surface of revolution.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

Using spherical shells and averaging the squared distance from the axis givesSliding without friction has acceleration $g\sin \alpha$, soFor rolling, $M{a}_{\mathrm{cm}}=Mg\sin \alpha -F$ and $I(a_{\mathrm{cm}}/a)=Fa$, whenceThe rolling acceleration with rotational inertia gives $5g\sin \alpha /7$ for the uniform sphere, and thereforeMechanical energy is conserved in both idealizations: there is no friction in sliding, while static friction does no work at the instantaneous contact point in pure rolling.

For $I=\gamma M{a}^2$, the same calculation gives

If $A_n$ are countable, choose enumerations and map each element of ${\bigcup }_n{A}_n$ to the first pair $(n,k)$ at which it occurs. Since ${\mathbb{N}}^2$ is countable, the union is countable.

A periodic function of period $k$ is determined by its $k$ values on a complete residue system. Thus all periodic functions form a countable union over $k$ of the countable sets ${\mathbb{N}}^k$, and are countable.

The set of bijections $\mathbb{N}\to \mathbb{N}$ is uncountable. Indeed, each binary sequence determines a bijection that swaps $2j-1$ and $2j$ exactly when its $j$th bit is one. This is an injection from the uncountable set of binary sequences.

For prime $p$, the nonzero residues pair with their distinct inverses except for $1$ and $-1$, so Wilson theorem givesIf composite $n>4$ has a factorization $n=ab$ with $1<a<b<n$, both factors occur in $(n-1)!$. If $n=m^2$, the distinct factors $m$ and $2m$ occur and their product is divisible by $n$; the excluded case $m=2$ is exactly $n=4$. Thus $(n-1)!\equiv 0(\mathrm{mod}n)$.

The Fermat-Euler theorem states $a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$ when $(a,n)=1$. For prime $p$, $\varphi (p)=p-1$, giving Fermat's little theorem for $p\nmid a$; the form $a^p\equiv a(\mathrm{mod}p)$ also covers $p\mid a$.

If $a\equiv b(\mathrm{mod}p)$, induction and the binomial theorem showIndeed, write $a^{p^n}=b^{p^n}+c{p}^{n+1}$ and raise to the $p$th power; every nonleading binomial term gains enough powers of $p$.

Fix $a>1$ and choose any odd prime $p\nmid a^2-1$. PutFermat's theorem gives $N_p\equiv 1(\mathrm{mod}p)$, and $N_p$ is odd, so $2p\mid N_p-1$. Also $a^{2p}\equiv 1(\mathrm{mod}{N}_p)$, hence $a^{N_p-1}\equiv 1(\mathrm{mod}{N}_p)$. For odd $p$,is composite. This generalized repunit pseudoprime construction gives infinitely many distinct base-$a$ pseudoprimes because the values are unbounded.