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

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

• How about the totally path disconnected (but still connected) spaces? topology.jdabbs.com/spaces/127 Commented Apr 4, 2016 at 6:45
• @Santiago Isn't the line segment from $(0,0)$ to $(1,1)$ a path-connected subset with more than one point? (And am I crazy or does the ¬ sign in front of "Totally Path Disconnected" mean that this space is not totally path disconnected? Otherwise whoever designed that website should rethink it.) Commented Apr 4, 2016 at 6:51
• @Najib Could be that the link was changed, now it points to the/a pseudo-arc, which the site asserts is totally path-disconnected. But, as it points out, the assertion was added without a proof. Commented Apr 6, 2016 at 13:06
• @DanielFischer The link did change, it used to point to the infinite broom. Commented Apr 6, 2016 at 13:30
• @NajibIdrissi : So , we have got a counter-example ?
– user228169
Commented Apr 7, 2016 at 7:05

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.

• Why is $A$ connected ?
– user228169
Commented Apr 8, 2016 at 15:24
• In my linked answer, I showed that the intersection of $A$ with rectangles of the form $(x_0,x_1)\times(y_0,y_1)$ are connected. This is enough to show that $A$ is both connected and locally connected. Commented Apr 10, 2016 at 5:33
• It is possible to construct simpler examples using the graphs of (non-continuous) functions $f\colon\mathbb{R}\to\mathbb{R}$. I can add this to my answer later. Commented Apr 11, 2016 at 8:27