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No. Let $S$ be the displayed infinite axiom set. Any finite subset $S_0\subseteq S$ mentions only finitely many of the two infinitude requirements. Choose $N\ge 1$ at least as large as every index mentioned. A finite equivalence relation with exactly $N$ singleton classes and $N$ doubleton classes satisfies $S_0$, but is not a first-order model of the complete theory.

If a finite axiomatization existed, let $\sigma$ be the conjunction of its axioms. Then $S\models \sigma$, so the compactness theorem gives a finite $S_0\subseteq S$ with $S_0\models \sigma$. Every first-order model of $S_0$ would then satisfy the full first-order theory, contradicting the finite first-order model above. This is the finite subaxiomatization from an equivalent finite theory argument: finite axiomatizability would force one finite fragment of the explicit axioms to suffice.

WriteThese are first-order formulas in the original language, not new predicate symbols. An axiomatization consists of the equivalence relation axioms for $E$, the boundand, for every $n\ge 1$, the two sentencesThe first scheme supplies infinitely many singleton classes. The second supplies infinitely many distinct non-singleton classes; the size bound makes these doubletons.

Every countable first-order model has countably infinitely many classes of each size. A bijection between singleton classes, and another between doubleton classes followed by a bijection inside each matched pair, gives an isomorphism between any two countable first-order models. To infer completeness, suppose a sentence had different truth values in two arbitrary first-order models. The Downward Lowenheim-Skolem theorem supplies countable elementary substructures of those first-order models, which would still disagree on the sentence, contradicting the isomorphism just constructed. Thus all first-order models satisfy the same sentences. Since the original first-order structure satisfies the axioms, these axioms form a complete axiomatization of its first-order theory. This is the theory of infinitely many singleton and doubleton equivalence classes.

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

For each continuous map to a Hausdorff target, equality of its values is reflexive, symmetric and transitive. The relation $R$ requires all these equalities simultaneously. Explicitly, $f(x)=f(x)$ gives reflexivity; $f(x)=f(y)$ implies $f(y)=f(x)$; and $f(x)=f(y)=f(z)$ gives $f(x)=f(z)$. These arguments apply to every map in the defining family, proving $R$ is an equivalence relation.

Use the equivalence relation of belonging to the same fibre:The map $\bar{f}:X/\mathcal{R}\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the quotient topology: for open $V\subseteq Y$, $q^{-1}({\bar{f}}^{-1}(V))=f^{-1}(V)$ is open.

The quotient space is compact as the continuous image of compact $X$, and $Y$ is Hausdorff. The compact-to-Hausdorff continuous bijection theorem makes $\bar{f}$ a homeomorphism. Indeed, images of closed subsets of a compact space are compact and hence closed in a Hausdorff space, so the inverse is continuous.

Use the equivalence relation of belonging to the same fibre:The map $\bar{f}:X/\mathcal{R}\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the quotient topology: for open $V\subseteq Y$, $q^{-1}({\bar{f}}^{-1}(V))=f^{-1}(V)$ is open.

The quotient space is compact as the continuous image of compact $X$, and $Y$ is Hausdorff. The compact-to-Hausdorff continuous bijection theorem makes $\bar{f}$ a homeomorphism. Indeed, images of closed subsets of a compact space are compact and hence closed in a Hausdorff space, so the inverse is continuous.

Conjugacy is an equivalence relation: the identity element gives reflexivity, conjugation by $g^{-1}$ reverses conjugation by $g$, and composing conjugations gives transitivity. Its equivalence classes therefore partition the finite set $G$. Counting this disjoint union givesSeparating singleton classes from the others gives the class equation $|G|=|Z(G)|+{\sum }_{|C_i|>1}|C_i|$.

The relation is reflexive because the exceptional set for $f=f$ is empty, and symmetric because the exceptional sets for $f=g$ and $g=f$ agree. If $f=g$ off a null set $N_1$ and $g=h$ off a null set $N_2$, then $f=h$ off $N_1\cup N_2$, which is null by part (b). Hence it is transitive as well, so equality almost everywhere is an equivalence relation.

The relation is reflexive because the exceptional set for $f=f$ is empty, and symmetric because the exceptional sets for $f=g$ and $g=f$ agree. If $f=g$ off a null set $N_1$ and $g=h$ off a null set $N_2$, then $f=h$ off $N_1\cup N_2$, which is null by part (b). Hence it is transitive as well, so equality almost everywhere is an equivalence relation.

The crucial distinction is between an observer-independent binary relation of absolute simultaneity and a relation defined relative to an inertial frame. The relativity of simultaneity removes the former from the usual structure of special relativity; it does not make the latter meaningless or subjective.

For two spacetime events separated by $(\Delta t,\Delta x)$, a Lorentz boost of velocity $v$ givesIf $\Delta t=0$ and $\Delta x\ne 0$, then $\Delta t^{\prime }\ne 0$ for a nonzero boost along their separation. More generally, spacelike separated events can be simultaneous in some inertial frame, and their time order can reverse between inertial frames. Distinct timelike separated or null separated events cannot be simultaneous in any inertial frame; their causal order is preserved by time-orientation-preserving Lorentz transformations. These are precise geometric facts, independent of anyone's beliefs.

In Minkowski simultaneity relative to an inertial observer, the observer is represented by a constant future-directed four-velocity $u$, normalized by $\eta (u,u)=-c^2$ in metric signature $(-,+,+,+)$. ThenFor a fixed $u$, this is a perfectly well-defined equivalence relation. It can be specified without coordinates. What the Minkowski metric does not supply is a uniquely preferred $u$ and hence a uniquely preferred simultaneity foliation. This is the strongest defensible interpretation of the claim by Tim Maudlin: there is no additional fundamental, frame-independent simultaneity relation in the bare theory.

Calling it merely “relative to an observer” can mislead in two ways. First, distant simultaneity does not depend on when an image is received: travel time of the light signal must be accounted for. Second, two observers at the same spacetime event but with different four-velocities assign different simultaneity hyperplanes; the dependence is on state of motion, rather than consciousness. Einstein synchronisation operationalizes the standard frame assignment by setting a reflected pulse's remote reading to the midpoint of its sending and returning readings.^[ref]

A material system can pick out a useful rest frame, for example a gas in thermal equilibrium. That adds physical information beyond the Minkowski metric; it does not provide a universal absolute present. Accordingly, Maudlin's denial is justified if it denies a primitive observer-independent relation, but is too strong if read as denying every objective relation $P{\sim }_{u}Q$. The indexed relation and its measurable consequences remain part of the physics.

Writefor the quotient map defining the mapping cylinder, and put $j(x)=q(x,0)$. Since the equivalence relation only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an injective continuous map.

Define the homotopyIt is a continuous map, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. HenceEven if $f$ identifies many points, the free end of the mapping cylinder retains a separate copy of every point of $X$.

The set $X/\mathcal{R}$ consists of the equivalence classes $[x]$ for $\mathcal{R}$. The canonical surjection $q:X\to X/\mathcal{R}$, $q(x)=[x]$, defines its quotient topology:The quotient space includes this topology, not just the collection of equivalence classes.

Regard the circle as $S^1=\{z\in \mathbb{C}:|z|=1\}$ and use the continuous surjectionDefine $z\mathcal{R}w$ precisely when $h(z)=h(w)$. This is an equivalence relation, being equality of function values. Equivalently $w=z$ or $w=\overline{z}$, so the construction is the circle modulo reflection.

Since $h$ is constant on each equivalence class, it induces a bijectionIt is continuous by the universal property of the quotient topology, because $\overline{h}\circ q=h$ is continuous. The circle is compact by the Heine-Borel theorem, and its quotient space is a continuous image of a compact space, hence compact. The closed interval is Hausdorff. Part (a)(ii) therefore provesGeometrically, the equivalence relation folds the upper and lower semicircles together, retaining the two real endpoints separately.

Use the equivalence relation induced by a function:Equality in $Y$ supplies reflexivity, symmetry, and transitivity, so this is an equivalence relation. Define $\bar{f}:X/\sim \to Y$ by $\bar{f}([x])=f(x)$. It is well-defined; it is injective by the definition of $\sim$ and surjective because $f$ is surjective.

For an open set $W\subseteq Y$,is open by continuity of $f$. The quotient topology therefore makes ${\bar{f}}^{-1}(W)$ open, proving that $\bar{f}$ is continuous.

The continuous image of a compact space theorem applied to $p$ makes $X/\sim$ compact. If $C$ is closed in $X/\sim$, it is compact, and $\bar{f}(C)$ is compact in the Hausdorff space $Y$, hence closed. Thus $\bar{f}$ is a closed map. Its inverse function is continuous, since its preimage of $C$ is precisely $\bar{f}(C)$. We have proved that $\bar{f}$ is a homeomorphism, including the compact-to-Hausdorff continuous bijection theorem used in the argument.

The “same age” binary relation is a Euclidean relation, with age understood as one consistently measured value at a fixed time. If $x$ has the same age as both $y$ and $z$, the ages of $y$ and $z$ are equal. HenceThis is another instance of equality of the values of a function inducing an equivalence relation.

Suppose $R$ is an equivalence relation, and $xRy$, $xRz$. Since it is a symmetric relation, $yRx$. Since it is a transitive relation, $yRx$ together with $xRz$ implies $yRz$. This proves

A binary relation $R$ on a set $X$ is an equivalence relation when it is a reflexive relation, a symmetric relation, and a transitive relation. In first-order logic, these three requirements are

On the nonempty set $X=\{a\}$ take the empty binary relation $R=\varnothing$. There are no triples for which $xRy\wedge xRz$ holds. The defining implication is therefore true for every triple by vacuous truth, making $R$ a Euclidean relation.

But $aRa$ is false, so $R$ is not a reflexive relation and hence is not an equivalence relation:

The answer is no, even when both original binary relations are equivalence relations. On $X=\{a,b,c\}$, let $R_1$ have equivalence classes $\{a,b\},\{c\}$ and $R_2$ have equivalence classes $\{a\},\{b,c\}$. Each is a transitive relation and, by the earlier proof, a Euclidean relation.

Use exactly the paper's convention for composition of relations: $xRy$ means $x{R}_{1}z$ and $z{R}_{2}y$ for some $z$. The possible successors areIndeed, from $a$ or $b$ the first step can reach $a$ or $b$, whose $R_2$ successors together exhaust $X$; from $c$ the first step must stay at $c$, whose $R_2$ successors are only $b,c$. Consequently $cRb$ and $bRa$, but not $cRa$. Therefore

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 relation is reflexive because $a-a=2\cdot 0$. It is symmetric because $a-b=2k$ implies $b-a=2(-k)$. It is transitive because $a-b=2k$ and $b-c=2\ell$ imply $a-c=2(k+\ell )$. It is therefore an equivalence relation, namely congruence modulo an integer for modulus two. Its two equivalence classes consist of the even integers and the odd integers.