Past exam of the topology course of the University of Oxford / 2017 / 2

A topological space is Hausdorff when any two distinct points have disjoint open neighbourhoods.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of $Z$. Thus $Z$ is a hyperconnected space.

If the continuous map $f$ took distinct values at $s,t\in Z$, choose disjoint open neighbourhoods $G,H$ of $f(s),f(t)$ in the Hausdorff space $Y$. By continuity, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty disjoint open sets of $Z$, a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.

A homeomorphism is a bijection for which both the function and its inverse are continuous maps.

Let $f:X\to Y$ satisfy the given assumptions. For a closed subset $F$ of $X$, the closed subset of a compact space theorem makes $F$ compact. The continuous image of a compact space theorem makes $f(F)$ compact. Because $Y$ is Hausdorff, its compact subset $f(F)$ is closed. Thus $f$ is a closed map.

For $g=f^{-1}:Y\to X$, the inverse image of a closed set $F\subseteq X$ iswhich is closed in $Y$. The closed-set criterion for continuity therefore makes $g$ continuous. Hence $f$ is a homeomorphism, proving the compact-to-Hausdorff continuous bijection theorem.

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.

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.

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.

Write $Q=(Z\times I)/\mathcal{R}$. Since $Q$ is homeomorphic to the circle, it is Hausdorff. For fixed $t\in I$, the functionis continuous, being a coordinate slice of the quotient map. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and setThen $g$ is continuous andSince the quotient map $p$ is a surjection, so is $g$. This is Hausdorff factorization through a cofinite product factor: a continuous map into a Hausdorff space cannot distinguish the points of the coordinate with the cofinite topology.

If $g$ were injective, it would be a continuous bijection from the compact closed interval $I$ to the Hausdorff space $Q$. Part (a)(ii) would make it a homeomorphism, contradicting the fact that the circle and interval are not homeomorphic.

To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting topological subspace is separated by the nonempty relatively open sets $[0,1/2)$ and $(1/2,1]$, so it is not connected. The circle with any one point removed is connected: after a rotation, it is the image under a continuous map of the connected open interval $(0,2\pi )$ under $t\mapsto (\cos t,\sin t)$. A homeomorphism and its restriction preserve connectedness, giving a contradiction.

Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. ThenIn particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.