Does every connected metric space , with more than one point , contains a path connected subset with more than one point ?

Question

Does every connected metric space , with more than one point , contains a path connected subset with more than one point ? Is there any additional condition imposing which on the mother space will guarantee the existence of such non-trivial path connected subset ? I know that every connected metric space , with more than one point , contains a proper connected subset with more than one point ; but I cannot make any headway if we want the subset to be path connected . Please help . Thanks in advance

Answer 1

I refer you to an old answer of mine on Mathoverflow Is every locally connected subset of Euclidean space Rⁿ locally path connected. This demonstrates that the plane can be partitioned into a pair of connected (and locally connected) subsets which contain no path connected subspaces of more than one point.

We can construct much simpler examples in the plane, as the graph of a (non-continuous) function $f\colon\mathbb{R}\to\mathbb{R}$. For example, consider the topologist's sine curve $$ f(x)=\begin{cases} \sin(1/x),&{\rm if}x > 0,\\ 0,&{\rm if}x\le0. \end{cases} $$ Although this is not continuous at $0$, it does have a connected graph $\{(x,f(x))\colon x\in\mathbb{R}\}\subseteq\mathbb{R}^2$. Now, let $x_1,x_2,\ldots$ be a sequence dense in the reals. For example, take it to be an enumeration of the rationals, and set $$ g(x)=\sum_{n=1}^\infty2^{-n}f(x-x_n) $$ As $f$ is bounded by $1$, this sum converges uniformly. It is not difficult to show that $g$ also has a connected graph $G=\{(x,g(x))\colon x\in\mathbb{R}\}$, and is discontinuous at each $x_n$. The fact that $g$ is discontinuous on a dense subset of the reals means that no two points in $G$ are connected, in $G$, by a continuous curve.