Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 4F (Metric and Topological Spaces)

True. Fix a point $x_0$ of a nonempty connected open set and let $P$ be its path-connected component. Each point of the open set has a small Euclidean ball contained in it. A ball is convex and path-connected, so each such ball lies entirely in one path component. All path components are therefore open, and the complement of $P$ is open as well. If the complement were nonempty, these two open sets would disconnect the original set. Thus $P$ is the whole set. In fact the joining paths may be chosen polygonal.

True. For nonempty connected spaces $X,Y$, choose $x_0\in X$, $y_0\in Y$. For each $x\in X$, the subspace $(X\times \{y_0\})\cup (\{x\}\times Y)$ is connected: its two connected pieces meet at $(x,y_0)$. The union of these subspaces over $x$ is all of $X\times Y$, and every piece contains the common connected subset $X\times \{y_0\}$. Their union is therefore connected. The assertion is also true if a factor is empty, since the empty product space is connected. This proves product of connected spaces.

False. The rational numbers $\mathbb{Q}$ with their subspace topology from $\mathbb{R}$ are Hausdorff. If a subset contains two distinct rationals $a<b$, choose an irrational $c$ between them; intersecting the subset with $(-\infty ,c)$ and $(c,\infty )$ gives a separation into two nonempty relatively open sets. Thus every nonempty connected subset of $\mathbb{Q}$ is a singleton, so it is totally disconnected. Yet no rational is isolated, because every interval around it contains another rational. Therefore $\mathbb{Q}$ is not a discrete space. Hausdorffness and total disconnectedness do not imply discreteness.