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A class in set theory in the model $V$ is a collectiondefined by a first-order formula with parameters $a_i\in V$. A set-theoretic class function is a definable class relation $F$ for which every input in its domain has exactly one output. Informally, the Axiom schema of replacement says that the image $F‘‘a$ of any set $a$ under any such function class is again a set.

Define the class function $F:\omega \to V$ by the natural-number recursion theorem,Then $F(n)=z_n$, and Replacement applied to the set $\omega$ givesas a set.

Call a set small when it injects into some $z_n$. Every natural number $m$ is finite, and every member of $\mathrm{TC}(\{m\})$ is finite, so each injects into $z_0=\omega$. Hence $m\in \mathbf{HS}$. The set $\omega$ itself injects into $z_0$, and its hereditary members are natural numbers, so $\omega \in \mathbf{HS}$.

We next prove $z_n\in \mathbf{HS}$ by mathematical induction. The case $z_0=\omega$ was just proved. If $z_n\in \mathbf{HS}$ and $u\in z_{n+1}=\mathcal{P}(z_n)$, then $u\subseteq z_n$, so inclusion injects $u$ into $z_n$ and makes $u$ small. Every set below $u$ in its transitive closure is already below $z_n$ and is small by the induction hypothesis. The set $z_{n+1}$ itself injects into $z_{n+1}$. Thus every member of $\mathrm{TC}(\{z_{n+1}\})$ is small.

By Cantor theorem, $|z_{n+1}|>|z_n|$, so the $z_n$ are distinct and $z_n\mapsto n$ injects $y$ into $\omega =z_0$. Thus $y$ is small. Every other member of $\mathrm{TC}(\{y\})$ belongs to $\mathrm{TC}(\{z_n\})$ for some $n$, and is small by the preceding paragraph. Therefore $y\in \mathbf{HS}$. This proves the finite-power-set hereditary-small construction.

The structure $(\mathbf{HS},\in )$ is not a model of ZF because it fails the Axiom of union. If $\bigcup y$ were small, it would inject into some $z_m$. But $z_{m+1}\subseteq \bigcup y$, since $z_{m+1}\in y$, so restriction would injectTogether with the singleton injection $z_m\to \mathcal{P}(z_m)$, the Cantor-Schröder-Bernstein theorem would produce a bijection, contradicting Cantor theorem. Hence $\bigcup y$ is not small and therefore does not belong to $\mathbf{HS}$. Since $y\in \mathbf{HS}$ has no union inside the class, the Union axiom fails.

A deterministic finite automaton is a tuple $(Q,\Sigma ,\delta ,q_0,F)$ with finite state set $Q$, finite input alphabet $\Sigma$, transition function $\delta :Q\times \Sigma \to Q$, initial state $q_0$, and accepting-state set $F\subseteq Q$. It accepts a word when the extended transition function of a deterministic finite automaton carries $q_0$ to a state in $F$. A regular language is a language accepted by some deterministic finite automaton.

The pumping lemma for regular languages says that if a regular language is accepted by an automaton with $N$ states, every accepted word $w$ of length at least $N$ has a decomposition $w=xyz$ with $|xy|\le N$, $|y|>0$, and $x{y}^{i}z$ in the language for every $i\ge 0$. Indeed, among the $N+1$ states visited before and after the first $N$ symbols, two are equal by the pigeonhole principle. The intervening nonempty word $y$ labels a loop, which may be traversed any number of times without changing the final accepting state.

In base two, the powers of two have representations $10^n$, so their language is the regular expression $10^{\ast }$ and is regular.

Suppose instead that their base-ten representations formed a regular language. Apply the pumping lemma to a sufficiently long decimal power of two $xyz$, and put $d=|y|>0$. Pumping gives decimal integers $a_i$ represented by $x{y}^{i}z$, all powers of two. A direct place-value calculation shows thatfor a constant $C$ independent of $i$. The lengths, and hence the exponents in $a_i=2^{n_i}$, tend to infinity. Dividing by $a_i$ givesFor large $i$, the right side has absolute value less than one while the left side is an integer, so it must vanish. This would make a power of two equal to $10^d=2^d{5}^d$, impossible for $d>0$ by unique prime factorization. Thus the decimal powers of two are not a regular language.

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are