A class is a collection of sets specified by a formula, possibly with set parameters. A class is a set when one set has exactly those members.

Let $(P,\le )$ be a nonempty partially ordered set. If every chain in a partial order $C\subseteq P$ has an upper bound in $P$, then $P$ has a maximal element. The hypothesis meansThe upper bound may depend on $C$ and need not belong to $C$. A least upper bound is not required. The conclusion asserts that some $m\in P$ satisfies $m\le x⟹x=m$; it does not assert a greatest element, which would satisfy $x\le m$ for every $x\in P$.^[ref]

Here a chain in a partial order may be the empty set. Every member of $P$ is an upper bound of the empty chain in a partial order, so including it in the hypothesis already forces $P$ to be nonempty. If “chain” means only nonempty chains in a partial order, the separate nonempty-$P$ assumption is essential: the empty poset otherwise satisfies the hypothesis vacuously and has no maximal element. These are two equivalent conventions for the theorem.

The strength of Zorn's lemma is that compatible partial constructions can have a completed, unextendible member without an explicit procedure for finding it. The next examples separate the upper bound calculation from the reason a maximal element has the desired property. All chains in a partial order must be covered, including uncountable ones, rather than just increasing sequences. The condition is sufficient, not necessary: a poset can have a maximal element and still have an unbounded chain in a partial order elsewhere.

A standard model of set theory interprets membership by the ambient membership relation restricted to its domain. A transitive model additionally has a transitive set or class as its domain. Standardness alone does not impose transitivity.

A family of subsets of $X$ is independent when every finite choice of distinct members and their complements has nonempty intersection. Write $A^1=A$ and $A^0=X\setminus A$ in the displayed condition. Every assignment of zero or one to the family then gives a collection with the finite intersection property, which can be extended to an ultrafilter.

Two sets with exactly the same members are equal. In a transitive model, all actual members of its sets are available internally, so ambient extensionality gives the internal axiom as well.

For any two sets there is a set containing exactly those two members. Together with other ZFC axioms it permits ordinary ordered-pair coding. Countable-subset closure implies this axiom for the structure of hereditarily countable sets.

Objects have sorts $0,1,\dots$, with membership allowed only from sort $i$ to sort $i+1$. A set predicate distinguishes sets from atoms; atoms have no members, extensionality applies to sets, and every well-typed formula defines a set in the next sort by comprehension. First-order models may have only a specified collection of subsets, as in Henkin semantics. For a full finite-type model, take $D_i=V_{r_i}$ at widely separated increasing ranks, represent sets at level $i+1$ by $\mathcal{P}(D_i)\subseteq D_{i+1}$, and treat the remaining objects as atoms under typed membership.

New Foundations uses ordinary extensionality and comprehension restricted to stratified formulas. Its atom-permitting weakening is New Foundations with urelements.

For a strong-limit cardinal $\lambda$ of countable cofinality, defineHere smallness bounds every individual membership extension; it does not bound the cardinality of the entire transitive closure. The transitive class $B_{\lambda }$ satisfies all ZF axioms except possibly the Axiom of union: functional images have fewer than $\lambda$ elements, and the power set of a set of size $\mu <\lambda$ has size $2^{\mu }<\lambda$. These facts give Axiom schema of replacement and Axiom of power set; pairs and definable subsets also remain locally small. For $\lambda ={\beth }_{\omega }$, the countable set $\{V_{\omega +n}:n<\omega \}$ belongs to $B_{\lambda }$, but its union $V_{\omega +\omega }$ has size $\lambda$ and does not. This gives an explicit relative-independence model for the Axiom of union.

An urelement is an object permitted alongside sets, with no members, distinguished from the empty set by an atom or set predicate. Extensionality is restricted to sets, so distinct atoms need not be equal merely because they have no members. An atom is different from a Quine atom, which is a set with a self-member. In simple typed set theory with atoms, each sort may contain urelements in addition to the objects representing subsets of the preceding sort.

Unrestricted formation of a set $R=\{x:x\notin x\}$ contradicts its own membership condition: substituting $R$ for $x$ gives the displayed equivalence. Thus a predicate cannot always define a set of all objects satisfying it. Restricted set comprehension forms a subset of a previously given set, avoiding this argument. The contradiction does not assert that a set whose elements happen not to contain themselves is impossible; it rules out the universal collection with exactly the specified membership condition.

For nonnegative integers $n,k$, the Stirling number of the second kindcounts the set partitions of an $n$-element finite set into exactly $k$ unlabelled blocks of a set partition. Set $S(0,0)=1$, $S(n,0)=0$ for $n>0$, and $S(n,k)=0$ for $k>n$. Also $S(n,1)=S(n,n)=1$ for $n\ge 1$. Grassl and Levin develop these counts through explicit set partitions and a recurrence relation. ^[ref]

Insert one distinguished element. For $n\ge 1$ and $1\le k\le n$, the Stirling numbers of the second kind obeyProof outline. Inspect the block of a set partition containing $n$. If it contains another element, removing $n$ leaves a set partition into $k$ blocks of a set partition; conversely, insert $n$ into any of those $k$ blocks of a set partition. If $n$ is alone, remove its singleton set and obtain a set partition into $k-1$ blocks of a set partition. The cases cannot overlap, and deletion reverses each construction. This explains the factor $k$ even though the blocks of a set partition have no names. ^[ref]

For $n=4,k=2$, start with the three two-block set partitions of $\{1,2,3\}$. Inserting $4$ into either block of a set partition gives six possibilities. The remaining possibility is $\{\{1,2,3\},\{4\}\}$, obtained by making $4$ a singleton set. Hence $S(4,2)=2\cdot 3+1=7$. The seven set partitions in the diagram comprise four with sizes $3+1$ and three with sizes $2+2$; this grouping by sizes independently gives $4+3=7$.

Three different counting questions. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ selects one $k$-element subset. The Stirling number of the second kind $S(n,k)$ groups all $n$ distinct elements into $k$ nonempty unnamed blocks of a set partition. The power $k^n$ counts all functions to a specified $k$-element finite set, allowing some destination labels to go unused. At $n=4,k=2$, these counts are respectively $6$, $7$, and $16$. Choosing two people, making two unnamed nonempty teams, and assigning each person a red or blue label are different tasks.

When the destinations have labels. For fixed finite sets $X,Y$ of sizes $n,k$, respectively, the number of surjective functions $X\to Y$ isHere every element of $Y$ must be used. Each surjective function gives a set partition of $X$ into its nonempty preimages $f^{-1}(\{y\})$. Conversely, assign the $k$ distinct destination labels to the $k$ blocks of a set partition in $k!$ ways, where $k!$ is a factorial. For two labelled teams of four people this gives $2!\cdot 7=14$: of the $2^4=16$ assignments, exactly two use only one label. Dividing arbitrary function counts by $k!$ fails in general because unused labels can be permuted without changing a function. For example, $3^4/3!=13.5$ cannot count set partitions. The empty-to-empty function is a surjection, consistent with $0!S(0,0)=1$. Stanley gives the bijection underlying this qualification.^[ref]

For example, the inclusion-exclusion principle counts the surjections from four elements to three labelled destinations as $3^4-3\cdot 2^4+3\cdot 1^4=36$. Thus $S(4,3)=36/3!=6$. An independent check chooses the only two-element block of a set partition in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways; the other two elements form singleton sets.

Why these numbers also occur in algebra. Classifying all functions to an $m$-element finite set by the size $k$ of their image givesFor a fixed set partition, the falling factorial $(m{)}_k$ counts assignments of distinct labels to its $k$ blocks of a set partition, and is zero when $k>m$. At $n=3,m=4$, this reads $64=1\cdot 4+3\cdot 12+1\cdot 24$. Replacing $m$ by $x$ yields a polynomial identity: both sides agree at every nonnegative integer. Thus the Stirling numbers of the second kind translate powers into falling factorials. Grouping by equal values turns a counting rule into an algebraic change of coordinates. Summing $S(n,k)$ without destination labels instead gives the Bell number.

A set partition $\mathcal{Q}$ refines $\mathcal{P}$ if every cell of $\mathcal{Q}$ is contained in a cell of $\mathcal{P}$. Thus $\mathcal{Q}$ retains every separation made by $\mathcal{P}$ and may make additional separations. Refinement of a set partition is reflexive and transitive. If each partition refines the other, their cells coincide, so refinement of a set partition is a partial order on partitions. For finite partitions the nonempty intersections of cells form their common refinement of a set partition. In graph regularity arguments, subdividing cells and treating discarded vertices as singletons preserves refinement of a set partition; merging discarded vertices into one cell need not. The refinement variance identity for regularity energy consequently proves that this singleton treatment cannot decrease the equitable regularity energy.

A block of a set partition is one of its nonempty subsets. Distinct blocks of a set partition have empty intersection, and their union is the original set. For $\pi =\{\{1,3\},\{2\}\}$, the blocks of a set partition are $\{1,3\}$ and the singleton set $\{2\}$. The indices in a written list of blocks of a set partition do not label or order the set partition.

For the nondimensional simple pendulum $\ddot{x}=-\sin x$, the separatrix has energy ${\dot{x}}^2/2+1-\cos x=2$. Starting at $x=0$ with velocity $2$, its time to angle $a<\pi$ is $\log (\sec (a/2)+\tan (a/2))$. The saddle at angle $\pi$ is approached only as time tends to infinity.

// Target: classical-mechanics.bigb

A probabilistic model assigning distinct keys independent bucket indices with a uniform distribution on a finite set. It supplies an expected value bound for separate chaining, not a guarantee for every deterministic hash function or workload.

For a neutral sample of size $n\ge 2$ under the infinite sites mutation model, the Watterson estimator is ${\hat{\theta }}_W=S/a_{n-1}$, where $a_{n-1}={\sum }_{i=1}^{n-1}1/i$. Since the coalescent has $ES=\theta a_{n-1}$, this estimator is unbiased. It matches the observed segregating-site count to its expected value, but need not maximize the exact segregating-site likelihood under the neutral coalescent.

The Maykut-Untersteiner model is a seasonally forced one-dimensional sea ice thermodynamics model with snow and saline ice layers. It couples radiative and turbulent surface fluxes, heat conduction, interior shortwave absorption, brine-dependent effective heat capacity and conductivity, snowfall, and a basal Stefan condition driven by prescribed ocean heat flux. Its annual sea-ice thermodynamic equilibrium is a periodic cycle with zero net yearly thickness change. The snow insulation and summer ice protection mechanisms compete, so additional snow need not reduce annual equilibrium thickness. The transient temperature profile need not be linear.

When seawater freezes, most salt remains in liquid, increasing salinity and density and potentially driving convection.

The water-equivalent thickness expresses an ice mass per area as a depth of water at a declared reference density. With $m_i={\rho }_i{h}_i$ per area, $h_w=m_i/{\rho }_{w,\mathrm{ref}}$. Ice at $900\mathrm{kg}{\mathrm{m}}^{-3}$ and thickness $2.44\mathrm{m}$ has about $2.20\mathrm{m}$ water equivalent at a fresh-water reference density of $1000\mathrm{kg}{\mathrm{m}}^{-3}$. The depth of seawater consumed at $1025\mathrm{kg}{\mathrm{m}}^{-3}$ is instead about $2.14\mathrm{m}$. A reference depth must not replace ice thickness in an ice-volume calculation or the liquid mass removed in brine rejection.