Past exam of the options I course of the University of Oxford / 2008 / E (Topology)

A map of topological spaces is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ of $Y$. Equivalently, the preimage of every closed subset of $Y$ is closed in $X$.

For every $\epsilon >0$ there must be a $\delta >0$ such thatThe choice of $\delta$ may depend on $x_0$ and $\epsilon$. This is the epsilon-delta definition of continuity.

A basis for a topology is a family of open sets such that every open set is a union of members of the family. Equivalently, for every point in an open set there is a basis member containing that point and contained in the open set.

For a basis $\mathcal{B}$ on $Y$, continuity is equivalent to $f^{-1}(B)$ being open for every $B\in \mathcal{B}$. Necessity follows because basis sets are open. Conversely, taking preimages of a union of basis sets gives a union of open preimages, so every open set has an open preimage.

Each interval $(a,b)$ is open in the usual metric: if $x\in (a,b)$, the ball with radius $\min (x-a,b-x)/2$ is contained in it. Conversely, if $U$ is open and $x\in U$, the metric definition supplies $r>0$ with $(x-r,x+r)\subseteq U$. Thus every open set is a union of open intervals, which proves they form a basis.

If $f$ is continuous, the sets in question are preimages of the open rays $(-\infty ,\lambda )$ and $(\lambda ,\infty )$, so they are open. Conversely, suppose all those sets are open. For every $a<b$,is open as the intersection of two open sets. Part (i) and the basis criterion of part (b) then prove $f$ is continuous.

A subset is connected when it has no separation into two disjoint nonempty relatively open sets. It is path-connected when every pair of its points can be joined by a continuous path from $[0,1]$ lying in the subset.

Path-connectedness implies connectedness; the reverse fails in general, for example for the topologist's sine curve. They are equivalent for open subsets of every Euclidean space ${\mathbb{R}}^n$. More generally this holds for open subsets of locally path-connected spaces.

If $f(A)$ were separated by two disjoint nonempty relatively open sets, their preimages under the restriction $f{|}_A$ would be disjoint nonempty relatively open sets covering $A$. Continuity makes them open in $A$, contradicting its connectedness. Therefore $f(A)$ is connected.

The circle is path-connected, using circular arcs, and hence connected. Its continuous image is connected by part (ii). A connected subset of the real line is an interval: if it contained values on opposite sides of a missing value $c$, intersection with $(-\infty ,c)$ and $(c,\infty )$ would separate it. Thus $f(C)$ is an interval. It is also compact, so it is a closed bounded interval, possibly reduced to one point.