Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 4F (Metric and Topological Spaces)

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently its only subsets that are both open and closed are the empty set and the whole space. For a subspace topology, open means relatively open within that subspace. A path-connected space is connected, since a separation would pull back along a path to a separation of an interval.

Use the continuous slope map $(x,y)\mapsto x/y$ on the given set, where $y>0$. The attained ratios are positive rational numbers. Choosing the irrational cut $\sqrt{2}$, the two subsetsare disjoint, relatively open, and cover the entire set. Both are nonempty: ratios one and two are attained inside the square. This irrational-cut separation of rational slopes is a separation, soThe individual rational-slope rays are connected, but their common origin is excluded; their separate connectivity does not imply connectivity of their union.

The set is a rational comb with a connecting spine: it contains the whole vertical segment $x=0$, and every horizontal segment at rational height. Any point with $x>0$ has rational height and can move horizontally to the vertical segment without leaving the set. It can then move vertically to $(0,0)$. Points already on the vertical segment use only the second step. Concatenating these polygonal paths connects any two points. ConsequentlyNo claim about a horizontal segment at irrational height is needed: at such a height, the only point in the set lies on the connecting spine.