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An element $\perp$ of a partially ordered set $P$ is a least element if $\perp \le x$ for every $x\in P$. A least element is unique: two such elements are below each other, so antisymmetry makes them equal.

Use metric signature $(+---)$ and the curvature sign fixed by the stated Newtonian limit. In flat spacetime the source-free Maxwell equations are ${\partial }_a{F}^{ab}=0$ and ${\partial }_{[a}{F}_{bc]}=0$. The stress-energy tensor in mixed form is $T^a{}_b=F^{ac}{F}_{cb}+{\delta }_b^a{F}_{cd}{F}^{cd}/4$. Its divergence isContracting the cyclic Maxwell equation with $F^{ac}$ gives $F^{ac}{\partial }_a{F}_{cb}=-{\partial }_b(F_{cd}{F}^{cd})/4$. The first term also vanishes, proving $\boxed{{\partial }_a{T}^{ab}=0}$. The component $b=0$ expresses local electromagnetic energy conservation, and the spatial components express momentum conservation. Integrating converts them into the usual energy and momentum flux balances.

In curved spacetime the tensor equations areThey agree with the flat equations in coordinates of a local inertial frame at a point, where the connection vanishes, and covariance then extends them to arbitrary coordinates. Antisymmetry makes the homogeneous equation equivalent to $dF=0$. Metric compatibility permits the same contraction proof, now giving $\boxed{{\nabla }_a{T}^{ab}=0}$. In general,which need not vanish. Covariant local balance remains valid; ordinary coordinate components do not provide a universal conserved global energy-momentum vector in a curved geometry. Additional spacetime symmetries can furnish conserved currents. A locally inertial frame removes the connection at a point, not the curvature or the global obstruction.

The alternating array with fixed component ${ϵ}_{0123}=1$ is appropriate in oriented inertial special-relativistic coordinates, whose proper Lorentz transformations have determinant one. It is not itself a covariant tensor under general coordinate changes. For an oriented Lorentzian metric, replace it by the metric metric volume tensor ${\epsilon }_{abcd}=\sqrt{-\det g}[abcd]$. The genuine dual isThis is the Hodge star; the determinant factor, orientation and raising conventions are essential. Even in flat spacetime arbitrary curvilinear coordinates require the metric volume tensor, so the distinction is about coordinate covariance rather than curvature alone.

The isolated equation for $R_{00}$ singles out a coordinate component and cannot be a generally covariant law. Replacing it by a proportionality $R_{ab}\propto T_{ab}$ still fails: the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}={\nabla }_{b}R/2$, whereas stress-energy conservation requires ${\nabla }^a{T}_{ab}=0$. Such a law would force $R$ constant and hence impose an unphysical restriction on a varying relativistic dust density. Missing units or a coupling constant in the proposed proportionality would not repair this structural problem.

The Einstein tensor $G_{ab}=R_{ab}-R{g}_{ab}/2$ is divergence-free. Thereforeis consistent with covariant stress-energy conservation in the source's curvature convention. Its trace gives $R=8\pi GT$ and hence $R_{ab}=-8\pi G(T_{ab}-T{g}_{ab}/2)$. For slowly moving relativistic dust with $V^a{V}_a=1$, $T_{00}\simeq \rho$, $T\simeq \rho$, $g_{00}\simeq 1$, so $R_{00}\simeq -4\pi G\rho$, exactly matching $-\Delta \varphi$ and Poisson's equation.

Finally, electromagnetic stress-energy is trace-free: $T^a{}_a=F^{ac}{F}_{ca}+F_{cd}{F}^{cd}=0$ by antisymmetry. The traced field equation then gives $\boxed{R=0}$ for a purely electromagnetic source. This does not say $R_{ab}=0$ or that spacetime is flat; its trace-free Ricci curvature can still be nonzero.

A binary relation $R$ on $A$ is a subset of $A\times A$. Its properties are quantified over elements of $A$:

Thus an equivalence relation is a binary relation that is simultaneously reflexive, symmetric and transitive. The equivalence class of $a$ is the set of all elements related to it:Antisymmetry permits diagonal pairs and does not mean that a binary relation fails to be a symmetric relation.

The interval containment order compares the lower endpoints in reverse and the upper endpoints in their ordinary order. The smallest containing integer interval uses the smaller lower endpoint and the larger upper endpoint.Endpoint comparison immediately proves reflexivity, transitivity, and antisymmetry on interval values. If two separately allocated instances have the same endpoints, they denote the same abstract interval; object identity is not the mathematical equality used by this partial order.

The returned interval contains both operands. Every other containing interval has a lower endpoint at most both original lower endpoints and an upper endpoint at least both original upper endpoints, so it also contains the returned hull. HenceThis least upper bound need not equal the ordinary union: the hull of $[1,2]$ and $[5,6]$ is $[1,6]$, including the gap.

For each unordered pair $\{a,b\}$ with $a\ne b$, antisymmetry allows three choices: neither direction, only $(a,b)$, or only $(b,a)$. This contributes $3^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. The diagonal is unrestricted by antisymmetry, but being not irreflexive requires a nonempty subset of the $n$ diagonal pairs. There are $2^n-1$ such choices. HenceThis allows some diagonal pairs to be absent; “not irreflexive” does not require a reflexive relation.

In the partially ordered set $(A,⪯)$, the assertion $m=\mathrm{lub}(S)$ means bothThe first condition makes $m$ an upper bound; the second makes it the least upper bound. Both conditions are necessary, and antisymmetry makes such an element unique.

This is the pointwise order on a function space. Each of the three partial order axioms follows from the corresponding axiom in $B$.

For reflexivity, $f(x){⪯}_{B}f(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}f$. For antisymmetry, if $f{⪯}_{\mathcal{F}}g$ and $g{⪯}_{\mathcal{F}}f$, then $f(x)=g(x)$ for every $x\in A$; equality of functions therefore gives $f=g$. For transitivity, if $f{⪯}_{\mathcal{F}}g{⪯}_{\mathcal{F}}h$, then $f(x){⪯}_{B}g(x){⪯}_{B}h(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}h$.

Thus $(\mathcal{F},{⪯}_{\mathcal{F}})$ is a partially ordered set. The argument also covers an empty domain $A$, since then there is exactly one function $A\to B$.

A preorder is reflexive and transitive: $xRx$ for every element, and $xRy,yRz$ imply $xRz$. A partial order additionally requires antisymmetry: $xRy$ and $yRx$ imply $x=y$. It need not compare every pair of distinct elements.

Given a partially ordered set $B$ and a set $A$, the pointwise order on the function space of functions $A\to B$ compares their values at every argument. It is a partial order: reflexivity and transitivity hold at every argument, and antisymmetry follows because two functions equal at every argument are equal. Even if $B$ has a total order, this function space need not: the two functions on $\{0,1\}$ with values $(0,1)$ and $(1,0)$ are incomparable when $B=\mathbb{R}$.

Reachability in a directed graph, allowing length-zero paths, is a preorder: zero steps give reflexivity and concatenation gives transitivity. Distinct mutually reachable vertices violate antisymmetry. Its equivalence classes under mutual reachability are the strongly connected components; the induced reachability relation on those classes is a partial order. Equivalently, contracting the classes gives the acyclic condensation graph.

The least reflexive relation and transitive relation containing $R$. Here $R^0$ is the identity relation. Membership means that a finite sequence of zero or more $R$-steps connects the ordered pair. Adding paths can create reverse connections that were absent in $R$, so initial antisymmetry need not be preserved.