Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / E (Topology)

The closed sets contain the empty set and the whole space, are closed under arbitrary intersections, and are closed under finite unions. These conditions are equivalent, by complements, to the open-set axioms of a topology.

A map is continuous when the inverse image of every open set is open. Equivalently, the inverse image of every closed set is closed.

The definition of convergence of a sequence in a topological space to $\ell$ is that every open neighborhood $U$ of $\ell$ contains all sufficiently late terms: there exists $N$ such that $x_n\in U$ for every $n\ge N$.

The empty set is countable, and the whole space contains $a$, so both belong to the proposed closed family. A finite union contains $a$ if any member does; otherwise it is a finite union of countable sets and remains countable. For an arbitrary intersection, if every member contains $a$, so does the intersection. Otherwise some member excludes $a$ and is countable, and the intersection is a subset of it, hence countable. Thus all closed-set topology axioms hold.

Let $E=\{x:f(x)\ne f(a)\}$ be countable. For a closed subset $C$ of the real line, if $f(a)\in C$, its inverse image contains $a$ and is closed by definition. If $f(a)\notin C$, its inverse image lies inside $E$ and is countable, hence is again closed. Every closed inverse image is therefore closed, proving $f$ is continuous.

The set $S=\{x_n:x_n\ne \ell \}$ is countable, hence closed, and does not contain $\ell$. Its complement is an open neighborhood of $\ell$. Convergence therefore places every sufficiently late term outside $S$, which is possible only if those terms equal $\ell$. Thus the sequence is eventually constant at its limit, and the set of exceptional indices is finite. This conclusion concerns repeated indices as well as distinct values.

A subset is compact if every family of open sets covering it has a finite subfamily still covering it. Equivalently, its subspace topology has the finite-subcover property.

Let $\{U_i\}$ be an open cover of $f(X)$. By continuity the inverse images $f^{-1}(U_i)$ are open and cover $X$. Compactness provides finitely many of them covering $X$, and the corresponding $U_i$ then cover $f(X)$. Thus a continuous image of a compact space is compact.

The family contains the whole space and the empty set. A finite intersection of nontrivial members is $({\max }_j{a}_j,\infty )$. A nonempty union of rays is $({\inf }_j{a}_j,\infty )$ when the infimum is finite, and is the whole line when the infimum is $-\infty$. Empty unions and unions involving the whole space are already included. Hence $\tau$ is a topology. This is the upper-ray topology on the real line.

If $A$ has a smallest element $a_0$, any open cover has a member containing $a_0$. This member is either the whole line or a ray $(c,\infty )$ with $c<a_0$, so it already contains all of $A$. Thus a single member suffices and $A$ is compact.

If $A$ has no smallest element, the family $\{(a,\infty ):a\in A\}$ covers $A$: each point has another point of $A$ below it. A finite subfamily has union $({\min }_j{a}_j,\infty )$, which misses its smallest listed endpoint, itself a point of $A$. No finite subcover exists. Therefore a nonempty subset is compact exactly when it has a smallest element in this topology.