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Give ${\mathbb{R}}^2$ the lexicographic order and the order topology. Every point $(a,b)$ has an interval neighbourhood between $(a,b-\epsilon )$ and $(a,b+\epsilon )$, lying entirely in the vertical line $\{a\}\times \mathbb{R}$. These intervals give each such line its usual real topology, and every line is open. Thus this space is a disjoint union of uncountably many open copies of the real line and is not a second-countable space: any basis of a topology must supply a different member inside each of these disjoint nonempty open sets.

If $\mathcal{B}$ is a countable topological basis of $X$, then $\{B\cap Y:B\in \mathcal{B}\}$ is a countable topological basis of the subspace $Y$. Indeed every subspace-open neighborhood is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its point and $U$. This proves the first assertion.

Urysohn's metrisation theorem: every second-countable space that is also a T3 space is a metrizable space. Here regularity of a topological space allows a neighborhood of a point to be shrunk to an open neighborhood whose closure is still inside the original one.

We first prove the separation facts needed in a complete proof. Second countability implies the Lindelöf property: for each basic set contained in some member of an open cover choose one such member; these countably many members cover. In particular closed subspaces are Lindelöf. For disjoint closed sets $F,G$, regularity of a topological space and this property give open covers $F\subseteq {\bigcup }_n{U}_n$, $G\subseteq {\bigcup }_n{V}_n$ with ${\overline{U}}_n\cap G=\varnothing$ and ${\overline{V}}_n\cap F=\varnothing$. DefineThese sets are open and contain $F,G$ respectively. They are disjoint: if a point belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the closure of the other covering set. Thus the space is a normal space.

For completeness, Urysohn's lemma supplies continuous separating functions in a normal topological space. Its construction is as follows. Given disjoint closed $F,G$, choose open sets $O_r$ for dyadic $r\in [0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and ${\overline{O}}_r\subseteq O_s$ whenever $r<s$. Start by shrinking a neighborhood of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the closed-neighborhood shrinking in a normal space. DefineThen $f=0$ on $F$, $f=1$ on $G$, and $0\le f\le 1$. The identities $\{f<a\}={\bigcup }_{r<a}{O}_r$ and $\{f>a\}={\bigcup }_{a<r<1}(X\setminus {\overline{O}}_r)$ prove continuity for $0<a<1$, with the endpoint cases immediate.

Let $\{B_n\}$ be a countable topological basis. For each pair $(n,m)$ with ${\overline{B}}_n\subseteq B_m$, choose a continuous function $f_{nm}:X\to [0,1]$ equal to one on ${\overline{B}}_n$ and zero outside $B_m$. These form a countable family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a neighborhood of $x$ inside $B_m$, and then choose $B_n$ inside that smaller neighborhood. ThusConsequently these functions generate the original topology. They also separate distinct points: use the $T_1$ open set $X\setminus \{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto (f_j(x){)}_j$ is a topological embedding into the countable cube. Pulling back its metric, or directly usinggives a compatible metric. The fact that the functions distinguish distinct points proves positivity for distinct points, the triangle inequality is termwise, and the uniformly bounded tails show that this metric induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially metrizable spaces. This proves the theorem.

For the application, a separable metric space has a countable topological basis consisting of balls with rational positive radii about a countable dense set, and it is a T3 space. The given product theorems make a countable product of these spaces second countable and a T3 space. Urysohn's theorem makes it a metrizable space. Finally every second countable space is separable: choose a point from each nonempty basic open set; the resulting countable set meets every nonempty open set. Hence the product is separable and a metrizable space.

For a fixed nonzero second coordinate $c$, the rational expression defining $h(x,c)$ is continuous on the half-line where $xc>0$. On the opposite half-line it is constant. At $x=0$ the ratio tends to $1$, agreeing with the value on the other side. If $c=0$ the entire restriction is constant. By symmetry the restrictions with the first coordinate fixed are continuous too. Hence $h$ is separately continuous, and so is $h_t(x,y)=h(x-t,y-t)$ for every real $t$.

On the diagonal $D$, $h_t(t,t)=1$, while $h_t(s,s)=0$ for $s\ne t$. ThereforeEach diagonal singleton is open in the induced topology; arbitrary unions show that $D$ has the discrete topology.

If $X$ were metrizable, its countable dense subset from the preceding part would make it second countable. Indeed balls with rational radii about dense points contain a countable basis after the usual smaller-ball argument. Every subspace of a second-countable space is second countable. A discrete second-countable space is countable, since each singleton must itself occur as a basis member. But $D$ is uncountable. This contradiction proves $X$ is not metrizable.

The Lie group $PSL(2,\mathbb{C})$ is a second-countable space. For every point of its discrete subgroup, choose the first basic neighbourhood isolating that point; different elements require different such neighbourhoods. This injects $G$ into a countable set, proving countability.

A discrete subgroup is closed and meets compact sets in finitely many points. Indeed, distinct subgroup elements accumulating anywhere would give quotients converging to identity, contradicting an identity neighbourhood containing no other subgroup element. To apply this, use the positive Hermitian determinant-one model of hyperbolic space. At the base point $I$, the image under a determinant-one representative $g$ is $g{g}^{\ast }$ andThis distance formula follows by unitary diagonalization: eigenvalues are $e^t,e^{-t}$ and the axial hyperbolic displacement is $|t|$. Thus bounded displacement bounds all matrix entries. The determinant-one condition is closed, so the resulting set is compact; the quotient by $\{I,-I\}$ is compact too. Conjugation gives the same conclusion at any base point. These are the bounded displacement sets of hyperbolic isometries are compact facts.

Hence, for each $R$, only finitely many $g\in G$ satisfy $d(p,gp)\le R$. Any enumeration of an infinite group without repetitions therefore has $d(p,g_{n}p)\to \infty$. A finite discrete group is a harmless exception to the wording: it has no infinite enumeration of distinct elements, and proper discontinuity is immediate.

For a compact set $C$, put $R={\max }_{x\in C}d(p,x)$. If $gC\cap C\ne \varnothing$, choose $x,gx\in C$. The triangle inequality gives $d(p,gp)\le d(p,gx)+d(x,p)\le 2R$. Only finitely many such $g$ exist, proving properly discontinuous group action. Conversely, if the subgroup is not discrete, there are distinct $g_n\to e$. A closed ball about $p$ then meets its images under infinitely many $g_n$, violating this compact-set definition. Thus discreteness is equivalent to proper discontinuity. Here proper discontinuity allows finite point stabilizers; a stronger convention requiring freeness would exclude discrete elliptic subgroups and is not the equivalence intended.

A basis of a topology is a family $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, whenever $x\in U$ and $U$ is open, there is $B\in \mathcal{B}$ with $x\in B\subseteq U$. For a metric space, all open balls form a basis. For the product topology on $X\times Y$, the rectangles $U\times V$ with $U,V$ open in their respective factors form a basis; rectangles made from factor bases also suffice.

If $C\subseteq X$ and $D\subseteq Y$ are countable dense subsets, then $C\times D$ is countable. Every nonempty basic rectangle meets it, since $U\cap C$ and $V\cap D$ are nonempty. Every nonempty open set contains such a rectangle, proving that a product of two separable spaces is separable.

Suppose a metric space has a countable dense set $C$. The balls $B(c,r)$ with $c\in C$ and positive rational $r$ form a countable basis. Indeed, given $x\in U$ open, choose $ϵ>0$ with $B(x,ϵ)\subseteq U$, then choose $c\in C$ with $d(c,x)<ϵ/3$ and rational $r$ satisfyingThe triangle inequality gives $x\in B(c,r)\subseteq B(x,ϵ)\subseteq U$. Conversely, from any countable basis select one point of each nonempty member. These points form a countable dense subset, because every nonempty open set contains a basis member. This direction holds for arbitrary topological spaces. Thus a metric space is separable exactly when it is a second-countable space. Intersecting a countable basis with a subspace gives a countable basis for its subspace topology; selecting points as above shows that every subspace of a separable metric space is separable.

For the Sorgenfrey line, $\mathbb{Q}$ meets every nonempty interval $[a,b)$, so it is a countable dense subset. In the Sorgenfrey plane, however, for any $x\in \mathbb{R}$ and $ϵ>0$,Indeed, $(t,-t)$ in that rectangle requires both $t\ge x$ and $t\le x$. Every singleton is therefore open in the subspace topology on $Y$, which is the discrete topology. A dense subset of a discrete space must contain every point. Since $Y$ is in bijection with the uncountable set $\mathbb{R}$, it has no countable dense subset. Hence $X$ is separable and $Y$ is not separable, illustrating why the metric assumption in the inheritance statement matters.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

Put $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$. The invertible linear map $x\mapsto u=Dx$ takes the ellipsoid to the unit sphere $S^n$. Write $u=(v,s)$ with $v\in {\mathbb{R}}^n$, and let $N=(0,1)$ and $S=(0,-1)$ be its two poles. The following two stereographic projections define manifold charts on the ellipsoid:Their ranges are ${\mathbb{R}}^n$. With $r^2=|w{|}^2$, their inverses, followed by $D^{-1}$, areThese are continuous and smooth functions, and directly satisfy the defining constraint. On the overlap, the smooth transition map isIt is its own inverse and is smooth away from zero. The two domains are open in the subspace topology and cover the ellipsoid. Its subspace topology is inherited from Euclidean space, so the ellipsoid is Hausdorff and has a second-countable space topology. Thus these charts form a smooth atlas and make it an $n$-dimensional smooth manifold, with an explicit diffeomorphism to $S^n$. For the degenerate dimension $n=0$, each chart consists of a single point with range ${\mathbb{R}}^0$, and their overlap is empty; the same conclusion holds.

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis countable, since the Cartesian product of two countable sets is countable. It is a basis of a topology for the product topology: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

A smooth manifold of dimension $n$ is a Hausdorff second-countable space covered by manifold charts ${\phi }_i:U_i\to V_i\subseteq {\mathbb{R}}^n$ whose smooth transition mapsare smooth functions with smooth inverses. Such charts form a smooth atlas. Adjoining every compatible manifold chart gives the maximal smooth atlas; different compatible atlases describe the same smooth structure.^[ref]

The circle is a one-dimensional smooth manifold. On the open subset obtained by removing one point, an angular coordinate chart identifies a sufficiently short arc with an open interval; overlapping angular coordinates differ by a constant multiple of $2\pi$. No single coordinate chart covers the whole circle, since a nonempty compact circle cannot be homeomorphic to an open interval.^[ref]

The smooth atlas makes derivatives, tangent vectors and differential forms independent of a chosen coordinate chart. Charts into ${\mathbb{R}}^n$ describe manifolds without boundary; a manifold with boundary additionally permits charts into a closed half-space.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

If $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$ is invertible, $x\mapsto Dx$ sends the ellipsoid ${\sum }_i{a}_i^2{x}_i^2=1$ to the unit sphere. On $u=(v,s)\in S^n$, the two stereographic projections $v/(1-s)$ and $v/(1+s)$ omit opposite poles and have inverses $(2w/(1+|w{|}^2),\pm (|w{|}^2-1)/(1+|w{|}^2))$, with the sign chosen for the appropriate pole. Their overlap transition is $w\mapsto w/|w{|}^2$, smooth and involutive on ${\mathbb{R}}^n\setminus \{0\}$. Composing their inverses with $D^{-1}$ gives a two-chart smooth atlas of the ellipsoid. Its subspace topology is Hausdorff and has a second-countable space topology, so it is a smooth manifold. For $n=0$ the overlap is empty and each chart is a point.

Every second-countable space that is also a T3 space is a metrizable space. First regularity of a topological space and the Lindelöf property yield a normal space: cover each of two disjoint closed sets by countably many open sets whose closures avoid the other, and subtract the first $n$ opposite closures from the $n$th neighborhood. The resulting unions are disjoint open neighborhoods. Urysohn's lemma then gives, for each pair of basic sets ${\overline{B}}_n\subseteq B_m$, a continuous function equal to one on the smaller closure and zero outside the larger set. There are countably many such pairs. The shrinking form of regularity of a topological space shows these functions generate the topology, and $T_1$ makes them separate points. Their coordinate map embeds the space into the countable cube, with compatible metric $d(x,y)={\sum }_j{2}^{-j}|f_j(x)-f_j(y)|$. The bounded tails prove compatibility with the product topology. Thus a countable base converts separation axioms into a numerical distance.