Past exam of the options II course of the University of Oxford / 2009 / E (Topology)

A subset $U$ of a metric space is open if every $x\in U$ has a radius $r>0$ with the open ball $B_d(x,r)\subseteq U$.

Continuity at a point means: for every $\epsilon >0$ there exists $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ implies $d_Y(f(x),f(x^{\prime }))<\epsilon$.

The function is continuous if it is continuous at every point of $X$. Equivalently, the preimage of each open set in $Y$ is open in $X$.

Nonnegativity and symmetry follow from $d$, and $d_a(x,y)=0$ with $x\ne y$ would force both $x=a$ and $y=a$, a contradiction. For the triangle inequality, when the intermediate point $z$ is different from both endpoints,If $z$ equals an endpoint the inequality is immediate, and if $x=y$ its left target is zero. These cover every case, so $d_a$ is a metric.

This is the star metric centered at a point.

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$, so no such $x$ lies in the strict open ball. Its centre always does. HenceThus every point other than $a$ is isolated.

The singleton calculation shows that every map out of $(X,d_a)$ is continuous at each point other than $a$: use the domain ball consisting only of that point. Thus global continuity in this metric is equivalent to continuity at $a$, proving $[\alpha ]⟺[\beta ]$.

For every $x$, including $x=a$, one has $d_a(a,x)=d(a,x)$. Therefore the open balls centred at $a$ are identical for $d_a$ and $d$. The epsilon-delta definitions of continuity at $a$ are identical, giving $[\beta ]⟺[\gamma ]$. Hence all three conditions are equivalent.

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

If $A$ is open and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\epsilon )\subseteq A$, this is a $d_a$-neighbourhood of $a$, while every other point of $A$ has its open singleton inside $A$. Hence $A$ is open. Thus the stated ball-containment condition is necessary and sufficient.

Every point other than $a$ is isolated, so no such point can enter the closure unless it was already in $B$. The only possible extra point is $a$, and it lies in the closure precisely when every $d$-ball about $a$ meets $B$. Therefore the three answers, in the printed order, areIn the middle case, a ball of radius below $d(a,B)$ misses $B$; in the last case the infimum zero makes every positive-radius ball meet $B$. If $B$ is empty its closure is empty as usual, with the convention $d(a,\varnothing )=+\infty$.

A subset is connected when its subspace topology admits no separation into two disjoint nonempty open sets. Equivalently, its only subsets that are both open and closed in that subspace are the empty set and the entire subset.

A separation of $f(X)$ would pull back under the continuous map to two disjoint nonempty open sets covering $X$: nonemptiness follows because their points belong to the image. This contradicts connectedness of $X$. Hence a continuous image of a connected space is connected.

The underlying set of the quotient space is the set of equivalence classes $X/R$. With quotient map $q(x)=[x]$, its quotient topology is defined by declaring $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. This makes $q$ a continuous surjection and is the finest topology with that property.

Let $C$ be any equivalence class. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the subspace topology on $C$. Its connectedness makes $C\cap A$ either empty or all of $C$. Thus an equivalence class meeting $A$ lies wholly in $A$. Hence $A$ is saturated, a union of whole equivalence classes.

If $X$ is connected, its continuous quotient image is connected by part (a). Conversely, suppose $X$ is disconnected. It has a nonempty proper subset $A$ both open and closed. Part (i) makes $A$ and its complement saturated, so $q^{-1}(q(A))=A$ and $q^{-1}(q(X\setminus A))=X\setminus A$. The quotient topology therefore makes these disjoint nonempty images open, and they cover $X/R$. Thus the quotient is disconnected. This proves the two spaces are connected simultaneously.

The difference map $(z,w)\mapsto z-w$ is continuous, and $X$ is its preimage of the open set $\mathbb{C}\setminus \{0\}$. Thus $X$ is open in the product.

The class of $(z,w)$ is $C_z=\{z\}\times (\mathbb{C}\setminus \{z\})$. A punctured plane is path-connected: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is connected.

The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is continuous as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-topology basis gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq \pi (U)$. Hence $\pi (U)$ is open.

The map is constant exactly on the equivalence classes, so it induces a bijection $\overline{\pi }:X/R\to \mathbb{C}$. It is continuous by the defining property of the quotient topology. It is also open: for quotient-open $V$, $\overline{\pi }(V)=\pi (q^{-1}(V))$ is open. ThusThe plane is connected, and the classes are connected, so part (b) now yields $X$ is connected.