Past exam of the topology course of the University of Oxford / 2015 / 2

A topological space is compact if every open cover has a finite subcover:There is no Hausdorff requirement in this definition.

Let $(V_i)$ be an open cover of $Y$. By continuity, the inverse images $f^{-1}(V_i)$ are open in $X$; by surjectivity, they cover $X$. Compactness supplies finitely many covering $X$. Applying $f$ and using $f(X)=Y$ shows that the corresponding finitely many $V_i$ cover $Y$.

Thus a continuous surjective function from a compact space has compact target. This is the continuous image of a compact space theorem.

For the canonical projection of a quotient set $p$, defineThe inverse images of the empty set and whole target are $\varnothing$ and $X$. Inverse images commute with arbitrary unions and finite intersections, so the topology axioms follow from those of $\mathcal{T}$. This is the quotient topology, and it makes $p$ continuous by definition.

The crucial criterion is an equivalence: $U$ is open precisely when $p^{-1}(U)$ is open. This is stronger than continuity alone and is what makes the next part work.

If $g$ is continuous, then so is $g\circ p$, as a composition of continuous functions. Conversely, suppose $g\circ p$ is continuous. For an open subset $W\subseteq Z$,is open in $X$. The quotient topology criterion therefore says $g^{-1}(W)$ is open in $X/\mathcal{R}$. This proves continuity of $g$.This is the universal property of the quotient topology.

Since $K$ misses $A$, the open set $U=X\setminus K$ contains the whole collapsed equivalence class $A$. Every other equivalence class is a singleton, so $U$ is saturated under the saturation of a subset operation:Its preimage is open, so the quotient topology givesThe hypothesis $K\cap A=\varnothing$ is exactly what prevents the saturation of a subset from adding missing points of the collapsed class.

Use $F:[0,1]\to S^1$, $F(t)=e^{2\pi it}$. This is continuous, identifies the two endpoints, and is otherwise injective. Its fibres are exactly the equivalence classes for collapsing a subset to a point with $A=\{0,1\}$. Hence it factors uniquely through a bijectionBy the universal property of the quotient topology, $\overline{F}$ is continuous. The interval is compact, so its quotient $X_A$ is compact. The unit circle is Hausdorff. The compact-to-Hausdorff continuous bijection theorem now shows that $\overline{F}$ is a homeomorphism:Indeed the theorem makes the inverse continuous because images of closed subsets of a compact space are compact and hence closed in a Hausdorff target.

Every quotient class has a representative in $[0,1]$: the collapsed class has representatives $0,1$, and the others are the points of $(0,1)$. Thus $p{|}_{[0,1]}$ is a continuous surjection from a compact interval onto $X_A$. It follows that $X_A$ is compact, despite the noncompact original space.

Define the continuous mapThe formulas agree at both junctions, and its fibres are exactly the required equivalence classes. It induces a continuous bijection $X_A\to S^1$, which is a homeomorphism by compactness and the Hausdorff condition on the circle.The quotient notation means that the displayed entire subset becomes one point.

Collapsing the integers gives a Hausdorff space. For two distinct noninteger points, choose disjoint small open intervals around them that avoid $\mathbb{Z}$. Each interval is saturated under the saturation of a subset operation, so its image is an open neighbourhood of its class. For a noninteger $x$ and the collapsed class, choose $0<r<\mathrm{dist}(x,\mathbb{Z})$. The disjoint open setsare saturated; $U$ contains $x$ and $V$ contains all integers. Their images separate the two quotient points. These exhaust all pairs of distinct classes.

Collapsing the rationals does not give a Hausdorff space. The preimage of the singleton collapsed point is $\mathbb{Q}$, which is not closed in $\mathbb{R}$. Therefore that quotient singleton is not closed, whereas every singleton of a Hausdorff space is closed.Equivalently, every nonempty saturated open subset of $\mathbb{R}$ meets the dense set $\mathbb{Q}$ and therefore contains it all. Every nonempty quotient open set contains the collapsed point, making separation from any other quotient point impossible.

No countable subset works. For any countable set $A\subseteq \mathbb{R}$, choose $b_n\in (n,n+1/2)\setminus A$: each nonempty interval is uncountable. The sequence tends to infinity, so every tailis closed in $\mathbb{R}$ and disjoint from $A$. By part (i), $p(\mathbb{R}\setminus K_n)$ is open in $X_A$. These sets increase and cover $X_A$, since no point belongs to every tail. But any finite selection is contained in its largest member, which still misses every quotient point $p(b_k)$ in that member's remaining tail. The $b_k$ represent distinct singleton classes.

This is an open cover without a finite subcover, provingThe argument does not assume that $A$ is closed, bounded, or discrete. It is the noncompactness after collapsing a countable real subset obstruction.