Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 4A (Metric and Topological Spaces)

Every point has arbitrarily small path-connected open neighbourhoods. Indeed, in a coordinate chart homeomorphic to ${\mathbb{R}}^n$, take the inverse image of a sufficiently small open Euclidean ball. The ball is convex, hence path-connected, and its inverse image is path-connected by the homeomorphism. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a locally path-connected space.

Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a path. For $x\in A$, a path-connected open neighbourhood $U_x$ is contained in $A$: concatenate a path from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is open. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a path to $x$. Hence $X\setminus A$ is also open. Since $X$ is a connected space and $A$ is nonempty, $A=X$. This proves $X$ is path-connected, including the general principle that connected locally path-connected spaces are path-connected. The empty-space case, if allowed, is vacuous.

Every nonempty open set in the given topology contains $1$. Suppose a continuous map $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset \mathbb{R}$ containing these respective values. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty open sets in $\mathbb{N}$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus every such continuous map is constant. Conversely, constant maps are continuous. This is the continuous real maps from the initial-segment topology argument, using the Hausdorff separation of distinct real points.