Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 10G (Analysis and Topology)

A space is connected when it has no separation into two nonempty disjoint open sets; the connected subspaces of $\mathbb{R}$ are exactly the intervals. A space is path connected if every two points are joined by a continuous path. A separation would pull back along a path to a separation of $[0,1]$, so path connectedness implies connectedness.

Each of the $n$ interval components of $I_n$ must map into one of the $m$ components of $I_m$. This assignment is constant on every connected component of the uniform function space. Conversely, for a fixed assignment, straight-line interpolation within each target interval gives paths between maps. Hence

Take the topologist's sine curveIt is closed and bounded, hence compact. The oscillating graph is connected and accumulates on every point of the vertical segment, so its closure $S$ is connected. No path joins the graph to the vertical segment, as continuity of the first coordinate and the unbounded oscillation of $\sin (1/x)$ would fail at the first time the path reaches $x=0$. Thus $S$ is not path connected.

Constant maps identify a copy of $S$ inside $C([0,1],S)$. If the function space were path connected, evaluating a path of functions at $0$ would give a path in $S$ between any two constant values. Since $S$ is not path connected,

The constant maps form a connected subspace homeomorphic to $S$. Every $f\in C([0,1],S)$ is joined to the constant map $f(0)$ byUniform continuity of $f$ makes $s\mapsto H(s)$ continuous in the uniform metric. Thus the function space is a union of connected sets, each meeting the connected set of constant maps, and so