Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 4

In the subspace topology on $S$, a subset $A$ is open in $S$ precisely when $A=S\cap O$ for some open set $O\subseteq \mathbb{C}$. A subset $B$ is closed in $S$ precisely when $S\setminus B$ is open in $S$, equivalently $B=S\cap F$ for some closed set $F\subseteq \mathbb{C}$.

A compact space is one for which every open cover has a finite subcover. Here the cover may use relatively open subsets of $S$, or equivalently ambient open subsets covering it. The Heine-Borel theorem states thatClosedness and boundedness here refer to the ambient complex plane, not merely closedness in $S$ itself.

A connected space $S$ cannot be expressed as $S=A\cup B$ with $A,B$ disjoint nonempty relatively open sets. Equivalently, the only subsets simultaneously relatively open and closed are $\varnothing$ and $S$. A decomposition into such $A,B$ is a separation of the topological space.

A path-connected space $S$ has, for every pair $z_0,z_1\in S$, a continuous path $\gamma :[0,1]\to S$ with $\gamma (0)=z_0$ and $\gamma (1)=z_1$. Every path-connected space is a connected space, because a separation of the target would pull a joining path apart and contradict connectedness of an interval.

The integers are unbounded in the complex plane, so the Heine-Borel theorem shows they are not compact. They are also not connected: $\{0\}$ and $\mathbb{Z}\setminus \{0\}$ are disjoint nonempty relatively open sets covering $\mathbb{Z}$. Each integer is isolated, so the induced subspace topology is discrete.

The disc is a closed set and is bounded in the complex plane, so it is compact by the Heine-Borel theorem. It is convex: the triangle inequality gives $|(1-t)z_0+t{z}_1|\le 2$ for all $t\in [0,1]$ when both endpoints lie in it. Straight segments therefore supply paths, so it is path-connected and hence connected.

The union is bounded but is not a closed set: the points $-1+1/n$, $n\ge 2$, belong to the open disc and tend to $-1$, which lies in neither disc. Thus it is not compact by the Heine-Borel theorem.

Nevertheless it is connected, indeed path-connected. Every point of the open disc joins $0$ by a straight segment. The segment from $0$ to $1$ stays in the union, since its points before the endpoint belong to the open disc and the endpoint $1$ belongs to the closed disc. Every point of that closed disc joins $1$ inside it by convexity. Concatenating these paths joins any two points of the union. The fact that the two discs are disjoint does not imply disconnection: their contact point belongs to one of them. This illustrates connectedness at a one-sided contact point.