Past exam of the options II course of the University of Oxford / 2012 / E (Topology)

The closure of $A$ is the intersection of every closed subset containing it:The collection is nonempty because it contains $X$. Arbitrary intersections of closed sets are closed, so this is the smallest closed set containing $A$.

Every subset is contained in its closure, giving $\overline{A}\subseteq \overline{\overline{A}}$. On the other hand, $\overline{A}$ is already closed by its definition as an intersection of closed sets. It is therefore one of the closed supersets used when taking its own closure, so $\overline{\overline{A}}\subseteq \overline{A}$. HenceThis is the idempotence of the closure operation.

If $x\notin \overline{A}$, some closed superset $F$ of $A$ misses $x$. Its complement is an open set containing $x$ and disjoint from $A$. Conversely, if an open set $U$ contains $x$ and misses $A$, the closed set $X\setminus U$ contains $A$ and misses $x$, so $x\notin \overline{A}$.

Negating this equivalence gives the neighbourhood characterization

An accumulation point of $A$ is a point $x\in X$ such that every open neighbourhood of $x$ contains a point of $A$ different from $x$:The point itself need not belong to $A$. Requiring a different point distinguishes accumulation from mere membership in the closure.

In a metric space,The sequence may equivalently be required to have pairwise distinct terms. The metric hypothesis makes a countable sequence sufficient; arbitrary topological spaces need not have this sequential characterization.

Take $A_1={\mathbb{Q}}^2$. Every open disk contains infinitely many points with both coordinates rational, so every point of the plane is an accumulation point of $A_1$. Every disk also contains infinitely many points with an irrational first coordinate, and these lie in the complement. ThusThe infinitely-many-points observation ensures that one can avoid the centre of the disk itself in both cases.

TakeIts only accumulation point is $(0,0)$, which is absent from $A_2$. Every listed point is isolated, so none belongs to $A_2^{\prime }$. ConsequentlyFor completeness, away from the origin a sufficiently small disk meets only finitely many of the listed points and can be shrunk to exclude all but its centre, proving there are no further accumulation points.

Use three nested sequences:For fixed $n,m$, letting $k\to \infty$ produces the point $(1/n+1/(n^{2}m),0)$. Letting $m\to \infty$ produces $(1/n,0)$, and letting $n\to \infty$ produces $(0,0)$. Accordingly, putThe successive derived sets areTo see that no limits are missing, take any convergent sequence of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to \infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint intervals $[1/n,1/n+1/n^2]$.

The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a three-stage accumulation tower.

The quotient topology is defined byPreimages commute with arbitrary unions and finite intersections and take the empty and whole sets to the corresponding sets, so this definition indeed gives a topology.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

If $g$ is continuous, so is the composition $g\circ p$. Conversely, assume the composition is continuous. For every open set $V\subseteq Y$,is open in $X$. The quotient definition then says that $g^{-1}(V)$ is open in $X/\mathcal{R}$. Hence $g$ is continuous andThis is the universal property of the quotient topology.

Let two distinct quotient points correspond to two equivalence classes. By hypothesis those classes lie in disjoint open saturated sets $U,V\subseteq X$. Saturation gives $p^{-1}(p(U))=U$, so the quotient definition makes $p(U)$ open, and likewise $p(V)$.

The image sets contain the two quotient points and are disjoint. Indeed, a class meeting both $U$ and $V$ would be contained in each because both sets are saturated, contradicting their disjointness. Thus the quotient is Hausdorff.

Define $\tilde{f}:X/\mathcal{F}\to Y$ by $\tilde{f}([x])=f(x)$. Its definition is independent of the representative because the classes are precisely the fibres of $f$. It is onto since $f$ is onto, and injective because two points with equal $f$-values lie in the same fibre.

The relation $\tilde{f}\circ p=f$ and part (iii) imply continuity of $\tilde{f}$. Since $X$ is compact and $p$ is continuous, $X/\mathcal{F}$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore

Every iterate $g^n(x,y)=(x+n,(-1{)}^{n}y)$ is an isometry of the strip with its Euclidean subspace metric. Let $a,b$ lie in distinct orbits. Since the first-coordinate difference in $a-g^{n}b$ grows without bound as $|n|\to \infty$, the positive distances $d(a,g^{n}b)$ have a positive minimum $\delta$. Only finitely many indices can give a distance below any fixed bound, and none gives zero.

Choose $0<\epsilon <\delta /3$. Saturate the relative open balls by settingThese are open saturated neighbourhoods of the two orbits. Any two centres $g^{n}a,g^{m}b$ are at least $\delta$ apart, since their distance equals $d(a,g^{m-n}b)$. Their radius-$\epsilon$ balls therefore cannot intersect. Hence $U\cap V=\varnothing$, and part (a)(iv), an instance of positive orbit separation implies a Hausdorff quotient, proves

Let $D=[0,1]\times [-1,1]$ and $q=p\circ i:D\to S/{\mathcal{R}}^{\prime }$. This is continuous. It is onto: given $(x,y)\in S$, choose an integer shift bringing $x$ into $[0,1]$, with the corresponding possible sign change in $y$; the resulting representative lies in $D$.

Two representatives in $D$ are in the same orbit only if their first coordinates differ by an integer. An interior point can therefore be equivalent only to itself. Distinct boundary representatives have first coordinates zero and one, and the iterate $g$ identifies exactly $(0,y)$ with $(1,-y)$. Thus the fibres of $q$ are precisely the equivalence classes defining the Möbius band $M$.

The rectangle $D$ is compact, and the target is Hausdorff by part (i). Applying part (a)(v) to $q$ therefore yieldsThe compactness argument supplies continuity of the inverse without requiring a discontinuous choice of one representative per orbit.