For $n\ge 2$, the intersection of all pairwise intersections $A_i\cap A_j$ is ${\bigcap }_i{A}_i$. Apply the union bound to the complements of the pairwise intersections and then take the complement to obtain the displayed lower bound. A sum exceeding $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-1$ therefore forces positive common probability without any independence assumption.

A failure-probability budget assigns nonnegative numbers ${\delta }_i$ to a finite or countable family of failure events $F_i$ in a common probability space, with verified bounds $\mathbb{P}(F_i)\le {\delta }_i$. The union bound guarantees that the probability of any failure is at most $\min \{1,{\sum }_i{\delta }_i\}$. By De Morgan's laws and the complement rule of probability, the probability that every requirement succeeds obeysThus ${\sum }_i{\delta }_i\le \delta$ guarantees simultaneous success with probability at least $1-\delta$, even when all failure events depend on the same observations. The numbers ${\delta }_i$ are upper allowances; adding them is not a calculation of the exact failure probability.

For three requirements, give each a failure-probability budget ${\delta }_i=1/20$. The total budget is $3/20$, so simultaneous success has probability at least $17/20$. On a sample space of twenty equally likely outcomes, let $F_1=\{1\}$, $F_2=\{2\}$, and $F_3=\{3\}$. These disjoint events attain the budget and the success bound. If instead all three failure events equal $\{1\}$, their individual probabilities are unchanged, but exact failure has probability $1/20$ and simultaneous success has probability $19/20$. This second model shows how overlap can leave the guarantee conservative.

For $n$ requirements, an equal allocation ${\delta }_i=\delta /n$ is convenient; unequal allocations also work when some requirements are harder to satisfy. For infinitely many requirements indexed by $i\ge 1$, the allocation ${\delta }_i=\delta /2^i$ uses total budget $\delta$ by the geometric series. A fixed positive allowance for every requirement instead gives an infinite sum and no positive simultaneous-success guarantee from the union bound. The individual probability bounds must cover all the failure events being combined: omitting possible failure modes leaves the resulting guarantee about the listed requirements only. The same addition of error allowances underlies the Bonferroni correction in simultaneous hypothesis tests.

For a finite family, the random variable $N={\sum }_i{1}_{A_i}$ counts the events that occur. Linearity of expectation gives $\mathbb{E}N={\sum }_i\mathbb{P}(A_i)$, while $\mathbb{P}(N\ge 1)$ is the exact union probability. Their difference isAn outcome in exactly $k\ge 1$ events contributes $k-1$ extra counts, weighted by its probability. This nonnegative difference vanishes exactly in equality in Boole's inequality. For two events it is $\mathbb{P}(A\cap B)$, as the inclusion-exclusion principle for probabilities shows. For more events, the Bonferroni inequalities use pairwise and higher intersection information to bound the exact union probability; blindly subtracting pairwise intersections from the sum gives a lower bound, rather than a corrected upper bound.

Severe overlap can make the union bound arbitrarily pessimistic. Let all $m$ events be the same event $A$ with probability $1/m$, where $m\ge 1$ is an integer. A concrete model is the discrete uniform distribution on $\{1,\dots ,m\}$ with $A=\{1\}$. The exact union probability is $1/m$, whereas the sum and capped bound are both one. The ratio is $m$: adding more names for the same event does not create more ways to fail. The identity above describes the gap in the uncapped sum; capping at one can reduce that gap.

For finitely many events, or a countable family with finite sum of individual probabilities, equality in the uncapped union bound holds exactly when every pairwise intersection has probability zero. Equivalently, at most one event occurs almost surely. Pairwise disjoint events give equality, but literal disjointness is stronger than necessary: shared outcomes of probability zero do not change either side. For the capped bound, if the sum exceeds one, equality instead means that the union is an almost sure event; overlap can then have positive probability.

The cap and the disjoint case are both sharp. If the individual probabilities sum to at most one, put the events on separate pieces of a sample space to attain their sum. The twelve-outcome disjoint example does this with two events of probability $1/3$. If the sum exceeds one, a covering union can attain the cap: on the same twelve-outcome sample space, $A=\{1,\dots ,8\}$ and $B=\{5,\dots ,12\}$ have sum $4/3$ and exact union probability one. These are sharpness examples, not an assertion that every overlapping family attains the bound. Positive-probability disjoint events are also not independent events, because their joint probability is zero while the product of their individual probabilities is positive.

For isentropic spherical accretion with this adiabatic index, define $\lambda =\dot{M}{c}_{\infty }^3/(4\pi G^2{M}^2{\rho }_{\infty })$ and $x=r{c}_{\infty }^2/(GM)$. The steady-flow equation is $\sqrt{\lambda }[{\mathcal{M}}^{3/2}/2+3{\mathcal{M}}^{-1/2}/2]=1+3x/2$. The bracket has minimum two at Mach one. Reaching arbitrarily small positive radii therefore requires $\lambda \le 1/4$. For smaller rates the outer-rest branch remains subsonic; at the maximal rate it approaches Mach one only as $r\to 0$, not at a finite sonic point. Supersonic outer branches fail the specified resting finite-density condition, and larger rates terminate at a positive minimum radius. An additional inner boundary prescription is needed to select a unique rate among the globally defined subsonic solutions.

For the induced square lattice subgraph in $[0,m]\times [0,n]$, $h_p(m,n)$ is the probability of an open graph path from its left boundary to its right boundary under independent bond percolation of density $p$. Width and height count lattice spacings. Dual bond percolation exchanges failure with a top-to-bottom dual crossing of dimensions $(m-1)\times (n+1)$, with the usual exterior boundary vertices.

Every real sequence has a subsequence converging to an extended real number. A bounded sequence has a real convergent subsequence by the Bolzano-Weierstrass theorem. If it is unbounded above, every tail is unbounded above, so choose increasing indices $n_j$ with $a_{n_j}>j$. These terms tend to positive infinity. If it is unbounded below, use $a_{n_j}<-j$.

The strings in Boltzmann entropy of a two-box occupancy are complete microstates only within the binary toy model. They retain a coarse-grained variable for each particle and omit its exact position within a box and its momentum. An actual classical gas instead occupies continuous phase space; obtaining a dimensionless count from a phase space volume requires a specified cell convention and the appropriate treatment of particle identity. A quantum state count must respect the allowed quantum states and quantum statistics. Thus $2^N$ is not a general count of microstates for a gas obeying quantum ideal-gas statistics. ^[ref]

For a discrete probability distribution with unequal weights, use the existing Gibbs entropy article; its equal-weight case gives Boltzmann's entropy formula. Shannon entropy uses the corresponding information measure, with the logarithm base setting its units. A count alone cannot replace the specified probability distribution, as the Boltzmann distribution examples demonstrate. ^[ref]

Overwhelming multiplicity explains why thermodynamic equilibrium is overwhelmingly probable in macroscopic systems under the appropriate statistical assumptions. It does not make low-entropy thermal fluctuations logically impossible. Nor does counting alone derive irreversible macroscopic evolution: one also needs dynamics, an initial preparation, and a reason the system samples the allowed microstates on the timescale of interest. ^[ref]

Historically, Boltzmann's 1877 paper counted microscopic arrangements, called complexions, and related the natural logarithm of their number to entropy. The Sharp and Matschinsky translation appeared in Entropy 17 (2015), pages 1971–2009, DOI 10.3390/e17041971. Its notation differs from ours: Boltzmann's permutability measure $\Omega$ is already logarithmic, whereas here $\Omega$ is the count. ^[ref] Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) developed the statistical ensemble framework; Chapter XIV compares several statistical definitions of entropy with thermodynamics. ^{[ref] [ref]}

For two subsystems with independently specified macrostates, suppose every allowed microstate of $A$ can be paired with every allowed microstate of $B$. Their allowed set is a Cartesian product. With equal probability for each pair, Boltzmann's entropy formula givesFor example, take two independent copies of the toy model, each with two labels choosing between its two boxes. Each copy has four configurations and Boltzmann entropy $k_B\ln 4$. Combining their independent labels gives sixteen configurations and Boltzmann entropy $k_B\ln 16=2k_B\ln 4$. This explains why entropy adds even though microscopic counts multiply. ^[ref]

The product assumption matters. A constraint shared between the subsystems can forbid some pairs; for example, fixing only their combined energy generally requires summing over its possible divisions. See entropy maximization under thermal contact for how thermal equilibrium selects the dominant division. Statistical independence cannot be inferred merely from drawing a boundary between two parts.

In Boltzmann entropy of a two-box occupancy, impose an initial restriction that all labels are left. There is one allowed coarse microstate. Remove that restriction, allowing all occupancy values: the allowed set is now the union of every occupancy macrostate, containing $2^N$ strings. If the final description assigns equal probability to this whole set, its Boltzmann entropy change isThe counting identity is the binomial theorem. For twenty labels, $\Delta S=20k_B\ln 2\approx 13.862944k_B$. This final description fixes neither box occupancy; it is coarser than the particular macrostate $k=10$, whose Boltzmann entropy is $k_B\ln 184756$. The two counts answer different questions, so substituting the balanced macrostate for the full allowed set would change the calculation.

For four labels, the initial all-left string is 0000; removing the restriction makes all sixteen strings available. The $k=2$ column has only six strings. Accordingly, the whole allowed set has Boltzmann entropy $k_B\ln 16=4k_B\ln 2$, while the balanced macrostate has Boltzmann entropy $k_B\ln 6$.

Removing a barrier changes the allowed set immediately, but it does not instantaneously randomize the actual probability distribution. Motion and equilibration must make those configurations accessible. The counting result illustrates the configurational contribution to free expansion; it does not describe the relaxation time or every contribution to the entropy of a real gas.

Take $N=20$ distinguishable, labeled particles and two boxes of equal volume. Retain only whether each particle is in the left or right box: use a bit $0$ for left and $1$ for right. Each length-20 string is one coarse microstate of this toy model. Assume the labels make independent side choices with equal probability, so the $2^{20}=1048576$ strings are equiprobable. Equal volumes motivate these weights for noninteracting particles with no preference for either box; equal volumes alone would not establish them for an arbitrary interacting gas.

Now retain only the number $k$ in the left box as the macrostate. Choosing which $k$ labels are left gives a binomial coefficient, soThe factorials count assignments to labels; the probability of a macrostate sums the probabilities of its constituent microstates. This is the binomial distribution with parameters $20$ and $1/2$. The same binomial coefficient appears in the counting of distinguishable two-level systems, although here the macrostate records occupancy rather than energy. ^[ref]

The largest binomial coefficient occurs at $k=10$, the balanced macrostate. All-left means $k=20$ and supplies just one string. ThusThere are $184756$ balanced strings for every all-left string, even though each individual string has the same probability. Exact balance has about $17.6\%$ of the total probability; equilibrium reasoning concerns the many macrostates near balance too, rather than certainty of exactly ten particles on each side. The diagram compares the counts with their natural logarithms.

Four fair, independent coins give a smaller worked example with the same binomial distribution. If $k$ is the number of heads, the five macrostates $k=0,1,2,3,4$ contain respectively $1,4,6,4,1$ microstates. Their probabilities are respectively $1/16,4/16,6/16,4/16,1/16$. In particular, two heads is one macrostate containing six ordered coin sequences, with Boltzmann entropy $k_B\ln 6$ in this counting analogy. A prescribed ordered sequence has one realization. The next diagram enumerates the same sixteen strings as left/right assignments.

For zero scalar potential, the static field equations are $D_i{D}_i\Phi =0$ and $D_i{F}_{ij}=[\Phi ,D_j\Phi ]$. If $B_i=-D_i\Phi$, the gauge-theory Bianchi identity gives the first equation. For the second,using $F_{ik}={ϵ}_{ik\ell }{B}_{\ell }$ and ${ϵ}_{ijk}{ϵ}_{ik\ell }=-2{\delta }_{j\ell }$. Thus the first-order equations imply the full second-order equations locally, independently of the radial ansatz.

For the trace inner product and adjoint covariant derivative, compatibility gives $\Delta |\Phi {|}^2=2|D\Phi {|}^2+2⟨\Phi ,D_i{D}_i\Phi ⟩$. The gauge-theory Bianchi identity and $B_i=\pm D_i\Phi$ imply $D_i{D}_i\Phi =0$. Therefore $\Delta |\Phi {|}^2=2|D\Phi {|}^2$, equal to the energy density $|B{|}^2+|D\Phi {|}^2$ on a Bogomolny-Prasad-Sommerfield monopole. If the energy density is instead defined with an overall factor $1/2$, this Laplacian is twice that density.

The sign $\sigma =\pm 1$ distinguishes an inverted quartic potential from a double-well potential. The origin has determinant $\lambda$ and trace $\kappa$. Nonzero equilibria obey $q^2=\lambda /\sigma$, with determinant $-2\lambda$ and trace $\kappa -2\lambda$. Its energy $p^2/2+\lambda q^2/2-\sigma q^4/4$ has derivative $(\kappa -2\sigma q^2)p^2$. The origin's Hopf bifurcation is supercritical for $\sigma =1$ and subcritical for $\sigma =-1$. In the latter case the nonzero equilibria have supercritical Hopf bifurcations at $\kappa =2\lambda <0$.

// Target: dynamical-systems.bigb

For $\lambda >0$, the equilibrium points are $x=\pm \sqrt{\lambda }$, $y=0$. The positive branch is a saddle; the negative branch has a subcritical Hopf bifurcation at $\mu =\sqrt{\lambda }$. A homoclinic orbit occurs at $\mu =(5/7)\sqrt{\lambda }+O(\lambda )$ near the double-zero point. Between this homoclinic curve and the Hopf curve, an unstable periodic orbit surrounds a stable equilibrium point. The factor $5/7$ follows from a Melnikov energy-balance method, using the integrals of $v^2$ and $u{v}^2$ along the leading conservative saddle loop.

// Destination: dynamical-systems.bigb

For $\lambda >0$, the equilibrium points are the origin and $(s,s^2)$ with $s=(1\pm \sqrt{1+4\mu })/2$. The nonzero branches have Jacobian trace $s-\lambda$ and determinant $\lambda s(2s-1)$. They have a saddle-node bifurcation at $\mu =-1/4$, $s=1/2$; the origin exchanges stability with the lower branch at the transcritical bifurcation $\mu =0$. A Hopf bifurcation occurs on the upper branch at $\mu =\lambda (\lambda -1)$ for $\lambda >1/2$. The point $(\mu ,\lambda )=(-1/4,1/2)$ has a double zero eigenvalue and a nontrivial Jordan block. Its generic unfolding includes a saddle-loop curve and the intervening stable periodic orbit region.

For finite sums of compactly supported convolution squares, $\hat{p},\hat{q}\ge 0$. Insert the Bochner theorem representation of $q$ into $p\ast q$ and apply the Fubini theorem: its representing measure is $\hat{p}{\mu }_q$. Reversing $p,q$ and using uniqueness gives the displayed identity. It makes the local quotients ${\mu }_p/\hat{p}$ consistent wherever $\hat{p}>0$.

A positive-definite function $r$ on the discrete group $\mathbb{Z}$ is the sequence of Fourier coefficients of a unique finite positive Borel measure on the circle, with total mass $r(0)$. Positivity makes the finite sums ${\sum }_{|n|<N}(1-|n|/N)r(n)e^{-2\pi int}$ nonnegative. Their measures converge weakly to the required measure; uniqueness follows from density of trigonometric polynomials in continuous functions.

An ordinary time integral of random variables taken in the L2 space of the underlying probability space. Under strong measurability and ${\int }_a^b(\mathbb{E}|X_t{|}^2{)}^{1/2}dt<\infty$, it exists as a Bochner integral; for a mean-square continuous process it is also the limit of Riemann sums in that space. If the process is centered and the covariance is integrable, Fubini's theorem givesFor a Gaussian process, the integral is a Gaussian random variable, by approximating it in mean square by finite linear combinations of jointly Gaussian random variables.

For a transitive finite permutation group, suppose each point stabilizer has orbits of a common size $r$ on the other points. If a nontrivial block system exists, all two-point stabilizers are trivial. Indeed for block size $b$, $r\mid n-1$ gives $\gcd (r,b)=1$. An orbit of a block not containing $\alpha$ has size at most $br$ and divisible by $br$, so its b point-orbits are disjoint. A stabilizer of $\alpha$ and a point of that block fixes the whole block. Interchanging the two points also fixes the block of $\alpha$. Equal two-point-stabilizer orders then show that each two-point stabilizer fixes every point, and faithfulness makes it trivial.