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If $A$ has a smallest element $a_0$, any open cover has a member containing $a_0$. This member is either the whole line or a ray $(c,\infty )$ with $c<a_0$, so it already contains all of $A$. Thus a single member suffices and $A$ is compact.

If $A$ has no smallest element, the family $\{(a,\infty ):a\in A\}$ covers $A$: each point has another point of $A$ below it. A finite subfamily has union $({\min }_j{a}_j,\infty )$, which misses its smallest listed endpoint, itself a point of $A$. No finite subcover exists. Therefore a nonempty subset is compact exactly when it has a smallest element in this topology.

The family contains the whole space and the empty set. A finite intersection of nontrivial members is $({\max }_j{a}_j,\infty )$. A nonempty union of rays is $({\inf }_j{a}_j,\infty )$ when the infimum is finite, and is the whole line when the infimum is $-\infty$. Empty unions and unions involving the whole space are already included. Hence $\tau$ is a topology. This is the upper-ray topology on the real line.

A topology $\mathcal{T}$ on $X$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets. A subset $F\subseteq X$ is closed precisely when $X\setminus F$ is open.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

The cosmological principle gives spatial homogeneity and cosmological isotropy for the large-scale spatial slices. About any chosen comoving observer, spherical coordinates therefore separate a radial direction from the angular directions. In particular,is the metric tensor on the unit sphere. Its appearance expresses cosmological isotropy, rather than a privileged centre of the universe. The factor $(1-K{r}^2{)}^{-1}$ permits the radial geometry to have nonzero spatial curvature of an FLRW universe.

The scale factor $a(t)$ changes every spatial separation between fixed comoving coordinates by the same multiplicative factor. For example, the radial proper distance in cosmology between fixed radial labels at a fixed cosmic time isExpansion or contraction through this single function preserves spatial homogeneity and cosmological isotropy; direction-dependent scale factors would generally spoil the latter. The Friedmann-Lemaître-Robertson-Walker metric specifies this possible evolution geometrically; a dynamical theory is needed to determine $a(t)$.

The constant $K$ fixes the sign and normalization of the spatial curvature of an FLRW universe. The spatial scalar curvature is ${}^{(3)}\mathcal{R}=6K/a^2(t)$. With dimensionless $r$ and a suitably normalized scale factor, the familiar choices are $K=+1,0,-1$ for spherical, flat, and hyperbolic spatial geometry. A rescaling of $r$ changes the normalization of $K$ and $a$ together; its sign has invariant geometric significance. The local metric tensor alone does not fix global topology.

The coordinates share a cosmological, comoving construction: $r,\theta ,\varphi$ label the worldlines of the smooth cosmic fluid, and $t$ is the common cosmic time measured as proper time along those worldlines. Observers following that fluid keep their spatial labels while the scale factor changes their physical separations. The coordinate $t$ itself increases; it is not another constant spatial label. Thus the question's final phrase is understood as asking for their common adaptation to the cosmological fluid, rather than suggesting that all four coordinates are time independent or have the same units.

Use the Monge representation $\mathbf{r}(x,y)=(x,y,h(x,y))$ for a single-valued surface without overhangs. Its area element is $dA=\sqrt{1+|\nabla h{|}^2}dxdy$; at small slopes, $dA\simeq [1+|\nabla h{|}^2/2]dxdy$ and twice the mean curvature is $2\mathcal{H}\simeq {\nabla }^{2}h$, up to orientation sign. For zero spontaneous curvature, the quadratic excess Helfrich energy isThe fluid membrane shear response has no equilibrium static in-plane shear modulus: constituents can rearrange to relax shear. Bending and isotropic tension still cost energy; viscous shear stresses can matter during motion. A Gaussian curvature term is omitted under the stated approximation; for fixed closed-surface topology its integral is constant, but boundaries and topology changes require care. The small-slope Monge expression describes weak deformations, not an entire fully wrapped sphere.^[ref]

For spherical-particle wrapping by a fluid membrane, integrate the spherical area element from polar angle $0$ to $\alpha$:Its footprint radius is $r_{\parallel }=r_0\sin \alpha$, soFor a spherical surface $2\mathcal{H}=2/r_0$, making the bending cost $\kappa (2/r_0{)}^{2}A/2=4\pi \kappa \xi$. adhesion contributes $-2\pi w{r}_0^2\xi$, and tension contributes $\sigma \Delta A$. ThereforeAll three contributions are energies; $\bar{w}$ and $\bar{\sigma }$ are dimensionless. This is the specified bound-patch model: deformation of the free membrane and the neck is omitted.

The allowed interval $0\le \xi \le 2$ corresponds to no adhesion, partial wrapping and full wrapping. At $\xi =2$, the cap covers a whole sphere and its footprint shrinks to zero. This is a closed neck in the geometric limit. It does not by itself prove membrane scission: creating a detached enclosure changes topology and involves neck physics absent from this energy.

For positive membrane tension, $E^{\prime \prime }/(\pi \kappa )=2\bar{\sigma }>0$. Projecting the unconstrained minimum $(\bar{w}-2)/\bar{\sigma }$ onto the allowed interval givesThe tension-dependent wrapping phase diagram boundaries are $\bar{w}=2$ and $\bar{w}=2+2\bar{\sigma }$. At zero membrane tension, the energy is linear: the minimum jumps from $0$ to $2$ as $\bar{w}$ crosses $2$, while all indentation values are degenerate at $\bar{w}=2$. Negative membrane tension is outside the stable-flat-membrane interpretation used here. Increasing membrane tension makes the excess area required for wrapping more expensive, broadening the partially wrapped region.

TODO: user-deferred biological application. The final request for biological relevance is intentionally left unsolved; all preceding generic geometry and membrane mechanics are included.

In gel sieving of synthetic polymers, gel electrophoresis combines drift under an electric field with size- and conformation-dependent obstruction by a porous matrix. In weak-field free solution, force balance gives $v=\mu E$ and, in a simple effective-charge/effective-friction model, $\mu =q_{\mathrm{eff}}/\zeta$. The mobility is motion relative to the medium, so fluid motion from electro-osmosis must be accounted for separately. This is the operational mobility convention, not a claim that the bare structural charge is always the effective driving charge.^[ref]

For a chain with free-draining polymer friction with approximately constant effective electric charge per segment, both charge and friction scale with $N$. The free-solution mobility can therefore become nearly independent of length, making size separation poor. A gel adds steric constraints. Coils smaller than the mesh can pass relatively easily; larger coils require deformation and pore-to-pore rearrangement. A schematic weak-sieving law is $\mu (c_g)\simeq \mu (0)e^{-K_r{c}_g}$, where matrix concentration $c_g$ and retardation coefficient $K_r$ are defined by a calibration; it is not a universal size law.

When a long flexible charged polymer is confined to a tube of available paths, reptation supplies another regime. An additive-friction model gives contour diffusion $D_c\propto N^{-1}$ and renewal time $\tau \sim N^2/D_c\propto N^3$. For ideal large-scale chain statistics, $R^2\propto N$, so centre-of-mass diffusion scales as $N^{-2}$. With total effective driving charge proportional to $N$, weak-field mobility can then scale as $N^{-1}$. This assumes screened/ideal large-scale conformations and linear response; strong-field alignment, trapping and nonideal matrices can alter the scaling.

Operationally, load narrow zones of independently characterized synthetic standards and sample into a controlled gel, apply a known field and detect migration positions after a measured time. Compare with a free-solution measurement, vary matrix concentration and field, and control temperature, ionic strength, pH and matrix charge. Larger gel pores or lower matrix concentration usually reduce sieving within a given calibrated material system. Excessive field causes Joule heating and can change conformation and mobility. adsorption or electro-osmotic drift can mimic a size effect; standards should match charge chemistry and topology as well as nominal mass.

Band centres move as $\mu Et$ while ordinary diffusion gives ${\sigma }_x^2(t)={\sigma }_x^2(0)+2Dt$. Resolution depends on separation of centres relative to band widths, not travel distance alone. A neutral tracer helps identify bulk flow, and multiple concentrations test adsorption or overloading. A mobility or band position does not uniquely determine size without a regime-specific calibration. TODO: biomolecular examples, sample preparation and biological applications are user deferred; the account here concerns synthetic charged polymers only.

Solutions to all three questions of the topology and groups course of the University of Oxford Part B examination, sat on 31 May 2023. Source: original released PDF, bundle pages 221–223; paper pages 1–3. The Mathematics and Statistics version is labelled B3.4. All eighteen parts are treated below. Question 3(a)(iii), as printed, needs an additional hypothesis such as local path connectedness; its solution includes both the intended argument and a counterexample to the unrestricted assertion.

Throughout Question 1, $I=[0,1]$ has its usual topology, and the mapping cylinder uses the paper's convention that $(x,1)$ is attached to $f(x)$.

Assume $p$ has a common finite degree of a covering map $d$. Declare the lifted vertices to be $p^{-1}(V(X))$, so there are exactly $dv(X)$ of them. Each open edge is homeomorphic to $(0,1)$, a simply connected interval. The restriction of the covering map above it is a disjoint union of $d$ copies of that open edge.

These copies acquire endpoints as follows. Lift the parametrizing path $[0,1]\to X$ of each oriented edge, starting at each of the $d$ points over its initial vertex. The path lifting theorem gives a unique lift all the way to $1$. Its endpoint lies over the terminal vertex, and the interiors of these lifted paths are exactly the $d$ open edges already described. This also handles a graph loop, whose two endpoints may coincide downstairs or upstairs. Local covering sheets above a small star at a vertex show that the lifted incidence relations have precisely the original topology. Thus the graph structure lifted through a covering map is a finite topological graph withSubtracting gives the Euler characteristic under a finite covering formulaThe formula holds even when $\tilde{X}$ is disconnected. If a disconnected base is allowed to have degrees $d_C$ that vary with its connected components $C$, the precise replacement isThe printed formula with a single $d$ uses the constant-degree convention explained in part (a)(ii).

Source paper: Mathematics Part A FOI release, member FOI-20210708-1 - Data/4324.pdf. The original cover identifies Topology, A5, formal code 4324, Wednesday 25 June 2014, 9.30am to 11.00am. All three source pages are readable.

A proper closed subset in the cofinite topology is finite, so cannot contain the infinite subset $A$. Its only closed superset is therefore the whole space:In particular, every infinite subset is dense in this topology.

The quotient set $X/\mathcal{R}$ consists of the equivalence classes. Let $p(x)=[x]$. Equip that set with the quotient topologyPreimages preserve the empty set, whole set, arbitrary unions and finite intersections, so this is a topology. The canonical projection is continuous and surjective, and the displayed equivalence specifies the quotient space completely.

The product topology has basisOpen subsets of the product are arbitrary unions of these rectangles. The rectangles cover the product and their finite intersections are rectangles, so they satisfy the basis axioms. Equivalently, this is the smallest topology making both coordinate projections continuous.

An abstract simplicial complex $K=(V,\Sigma )$ consists of a vertex set $V$ and a collection $\Sigma$ of finite nonempty subsets containing every singleton and every nonempty subset of each member. A member with $n+1$ vertices is an $n$-simplex.

Its topological realization glues a standard geometric simplex for each member along common faces. A concrete model iswith finite support and the weak topology of a simplicial realization: a set is closed precisely when its intersection with every simplex is closed. This agrees with the quotient topology of the gluing construction. Specifying this topology matters for complexes with infinitely many vertices.

Use the basis of a topology consisting of open rectangles:It covers $X\times Y$, since $X\times Y\in \mathcal{B}$. Whenever a point lies in two basis rectangles, their intersection is another basis rectangle containing it:These are precisely the two basis of a topology axioms. Arbitrary unions of the rectangles therefore form a topology, the product topology. To check the finite-intersection axiom directly, distribute the intersection of two unions over all pairs of rectangles and use the displayed formula. The empty union gives the empty set.

The product topology on the Cartesian product $X\times Y$ has a basis of a topology consisting of all rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the smallest topology making the two coordinate projections continuous.

Take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, use the Hausdorff condition in $X$ to find disjoint open neighbourhoods $U_1,U_2$. Then $U_1\times Y$ and $U_2\times Y$ are disjoint open neighbourhoods of the two points. If $x_1=x_2$, then $y_1\ne y_2$; disjoint open neighbourhoods $V_1,V_2$ in $Y$ give $X\times V_1$ and $X\times V_2$. Thus $X\times Y$ is Hausdorff, as established by the forward implication of the Hausdorff property of a nonempty product. An empty Cartesian product is automatically Hausdorff.

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.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

Original solutions to all three questions of the University of Oxford Part A topology examination, sat on 25 June 2021. Source: officially released examination PDF, parent PDF pages 94–96; printed paper pages 1–3. All 25 answer parts are covered, including every alternative. Only the best two questions counted towards the examination mark.

The two conditions for a basis of a topology both hold. First, every real number $x$ belongs to $[x,x+1)$, so the proposed basic intervals cover the real line. Second, ifthen $\max (a,c)\le x<\min (b,d)$, andis itself one of the proposed basic intervals. Thus every point in an intersection of two basic sets has a basic neighbourhood contained in that intersection.

Consequently the arbitrary unions of members of $\mathcal{B}$, including the empty union, form a topology $\mathcal{T}$. The basis of a topology conditions give closure under finite intersections; arbitrary unions are built into the definition. This is the lower limit topology, also called the Sorgenfrey line.