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

The family $\mathcal{F}$ of closed sets contains $\varnothing$ and $X$, is closed under arbitrary intersection, and is closed under finite union. Conversely these axioms define a topology by declaring the complements of the members of $\mathcal{F}$ to be the open sets.

The map is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ in $Y$. Equivalently the preimage of every closed set in $Y$ is closed in $X$, since preimages commute with complementation.

Convergence of a sequence to $\ell$ means that every open neighbourhood $U$ of $\ell$ contains all sufficiently late terms: there is $N$ such that $x_n\in U$ for all $n\ge N$. The threshold may depend on $U$.

Both $\varnothing$ and $X$ satisfy the condition. For an arbitrary intersection of members, either every member contains $a$, in which case the intersection contains $a$, or one does not, in which case that member is countable and so is the intersection. The empty intersection is $X$.

For a finite union, either a member contains $a$, so the union does too, or all members omit $a$ and are countable, so their finite union is countable. Thus the closed set axioms hold and define a topology. This is a Fortissimo space: every point other than $a$ is isolated, while the open neighbourhoods of $a$ have countable complement.

Let $S=\{x:f(x)\ne f(a)\}$, which is countable. For any closed set $C\subseteq \mathbb{R}$, consider its preimage. If $f(a)\in C$, then $a\in f^{-1}(C)$ and that preimage is closed by definition. If $f(a)\notin C$, then $f^{-1}(C)\subseteq S$, so it is countable and again closed. Therefore every closed set has a closed preimage, proving continuity.

Set $S=\{x_n:n\in \mathbb{N},x_n\ne \ell \}$. This is a countable set, hence a closed set in the stated topology, and $\ell \notin S$. Consequently $X\setminus S$ is an open neighbourhood of $\ell$. By convergence of a sequence, all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is eventually constant, and only finitely many indices have $x_n\ne \ell$.

A subset $A$ is compact when every open cover of $A$ has a finite subcover. This may be stated using sets open in the ambient space or using the subspace topology on $A$; the definitions agree because every subspace-open set is the intersection of $A$ with an ambient-open set.

Let $(U_i{)}_{i\in I}$ be an open cover of $f(X)$, with the sets open in $Y$. The preimages $f^{-1}(U_i)$ form an open cover of $X$, because $f$ is continuous. By compactness, finitely many of these preimages cover $X$. Their corresponding sets $U_i$ cover $f(X)$, providing a finite subcover. Hence the continuous image is compact.

The family contains the empty set and the entire real line. The intersection of two proper nonempty members isso it is closed under finite intersection. For an arbitrary nonempty family of proper rays, put $c={\inf }_i{a}_i$. If the thresholds are unbounded below, the union is $\mathbb{R}$; otherwise the union is $(c,\infty )$. In the latter case $c$ itself is in no ray, and every point greater than $c$ lies above some threshold, proving the equality. Empty unions and unions involving $\mathbb{R}$ cause no difficulty. Thus arbitrary unions also belong to the family, establishing the topology axioms.

If $m=\min A$, any open cover contains a member containing $m$. Such a member is either $\mathbb{R}$ or a ray $(a,\infty )$ with $a<m$. It therefore contains every element of $A$, so that single member is already a subcover.

Conversely suppose $A$ is nonempty and has no smallest element. Thenis an open cover of $A$: for any $x\in A$, the lack of a minimum gives an $a\in A$ below $x$. A nonempty finite subcollection has union $(a_0,\infty )$, where $a_0$ is the smallest of its thresholds. It misses $a_0\in A$. The empty subcollection cannot cover nonempty $A$ either. Hence there is no finite subcover, and $A$ is not compact. Thus compactness is equivalent to having a smallest element.