Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 10G / b

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