# 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 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.) 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. Apr 6, 2016 at 13:06
• @DanielFischer The link did change, it used to point to the infinite broom. Apr 6, 2016 at 13:30
• @NajibIdrissi : So , we have got a counter-example ?
– user228169
Apr 7, 2016 at 7:05

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 ?
• 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. 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. Apr 11, 2016 at 8:27