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In ZF, let $K,L$ be nonempty. If $K\times L$ injects into $K\bigsqcup L$, then there is either an injection or a surjection from $K$ to $L$.

Indeed, extend the inverse of the given injection to a surjection $p:K\bigsqcup L\to K\times L$ by sending points outside its range to one fixed pair $(k_{\ast },l_{\ast })$. If the second coordinate of $p$ restricted to the $K$-summand covers $L$, it is the required surjection. Otherwise choose $l_0$ that it misses. Every $(k,l_0)$ then has its preimage in the $L$-summand. Except possibly for the default pair, that preimage is unique, so these preimages inject $K$ into $L$. In the exceptional case they inject $K\setminus \{k_{\ast }\}$ into $L$; either their image is all of $L$, yielding a surjection from $K$, or an omitted point extends the map to an injection from $K$.

For an oriented closed connected $d$-manifold, the fundamental class is the unique generator of $H_d(M;\mathbb{Z})$ that restricts to the orientation generator in every local homology group.

The leading term ${\omega }^{\delta }n$ is the highest-exponent term in Cantor normal form; the remaining tail is strictly below ${\omega }^{\delta }$.

The Belinfante-Rosenfeld construction adds the divergence of a spin-current superpotential to the canonical stress-energy tensor, producing a symmetric conserved tensor with the same integrated four-momentum under suitable boundary conditions.

For a Poisson mean $\mu$, the natural parameter is $\theta =\log \mu$, so the canonical link is the logarithmic link $g(\mu )=\log \mu$.

If a nondecreasing oscillation quantity satisfiesfor $0<\rho \le R$, with $\gamma >\beta$, then iteration gives $\Phi (\rho )\le C(\rho /R{)}^{\beta }\Phi (R)+CB{\rho }^{\beta }$.

An operator $U$ on a Hilbert space is a partial isometry when it is isometric on $(\ker U{)}^{\perp }$. Equivalently, $U^{\ast }U$ is the orthogonal projection onto this initial space.

An element $a$ of a C-star algebra is positive when $a=b^{\ast }b$ for some $b$, equivalently when $a=a^{\ast }$ and $\sigma (a)\subseteq [0,\infty )$. Continuous functional calculus gives it a unique positive square root $a^{1/2}$.

If $(U,V)$ is $C$-relevant to $(X,Y)$ and $U_1,U_2,V_1,V_2$ are mutually independent copies, then $(U_1+U_2,V_1+V_2)$ is $2C$-relevant to $(X,Y)$. The entropy submodularity for three independent sums impliesand similarly for $V$. The Entropic Ruzsa triangle inequality boundswhich proves the claim after addition.

For distinct accessible subset states, choose a state in their symmetric difference. A word taking it to the unique final state accepts from one subset; uniqueness of the starting state prevents acceptance from the other. Thus the accessible part of the subset automaton is irreducible.

Let $W$ be continuous with $W_0=0$. Ifis a martingale for every real $c$, then conditional expectations solveThus $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and distributed as $N(0,t-s)$, so $W$ is Brownian motion.

For suitable $f$,defines the Brownian transition semigroup. Its generator identity is

Each Gaussian-convolution operator is a rescaled Weierstrass transform; the semigroup combines all times.

For the Cameron-Martin space of Wiener measure path $h(s)=\mu s$, the Cameron-Martin theorem says that translation of Brownian motion by this linear function changes its law through time $t$ by the density

A Brownian motion with drift $\mu$ is $X_t=W_t+\mu t$ for a standard Brownian motion $W$.

If $W$ is Brownian motion, thenis also Brownian motion. Its covariance is $\min (s,t)$, and continuity at zero follows from $W_u/u\to 0$ almost surely.

A centered Gaussian process with almost surely continuous sample paths and covariance $\mathbb{E}[B_s{B}_t]=\min (s,t)$ is standard Brownian motion. Here Gaussian process means that every finite-dimensional distribution is a multivariate normal distribution; having separate normal distributions at each time would not suffice. This characterization is stated in Pitman, Basic Properties of Brownian Motion, Definition 15.2.

To compute the Brownian covariance kernel directly, suppose $0\le s\le t$. Since $B_t=B_s+(B_t-B_s)$ and the added Brownian increment is centered and independent of $B_s$,Symmetry gives $\min (s,t)$ for arbitrary nonnegative times. For instance, $\mathrm{Cov}(B_1,B_4)=1$ and their correlation is $1/\sqrt{1\cdot 4}=1/2$. Positions are therefore related even though disjoint Brownian increments are independent: both positions contain the displacement accumulated during the first unit of time. Conversely, taking differences of $\min (s,t)$ gives zero covariance for disjoint increments; their joint multivariate normal distribution turns this into independence.

For every real $c$,is a martingale. Conditional expectation factors at time $s$ because the Gaussian increment $W_t-W_s$ is independent of the past and has exponential moment $e^{c^2(t-s)/2}$.

A standard Brownian bridge is the centered Gaussian process $B(t)=W(t)-tW(1)$ on $[0,1]$. Its covariance is $\min (s,t)-st$ and it satisfies $B(0)=B(1)=0$.

For a domain $D$, the Brownian exit time is $T_D=\inf \{t\ge 0:B_t\notin D\}$. It is a stopping time, and it has finite expectation when $D$ is bounded.

Brownian motion in ${\mathbb{R}}^d$ is transient for $d\ge 3$: its distance from the origin tends to infinity almost surely. In dimension three, the positive local martingale $|B_t{|}^{-1}$ and Brownian scaling give a short proof.

It is a dimension-dependent long-time property of Brownian motion.