Hi
Do you know an exemple of connected space but path connected in no open subset (except perhaps the whole space) ?
Definitions I use is:
$\displaystyle X$ connected iff for all $\displaystyle U,V$ disjoint open subsets in $\displaystyle X,\ X=U\cup V\Rightarrow (U=\emptyset,\ V=X)$ or $\displaystyle (U=X,\ V=\emptyset)$
$\displaystyle X$ is path connected iff between any two points $\displaystyle x,y$ in $\displaystyle X$ there is always a path. (a continuous map $\displaystyle \sigma$ from $\displaystyle [0,1]$ to $\displaystyle X$ such that $\displaystyle \sigma(0)=x$ and $\displaystyle \sigma(1)=y$)
Why not only considering
$\displaystyle
\mathbb{Q}\times\mathbb{R},
$
which has the same nature with the first quadrant of
$\displaystyle
(\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\}).
$
It is clear that this set is not path connected.
Because the mapping $\displaystyle \sigma:[0,1]\to\mathbb{Q}\times\mathbb{R}$ will have discontinuity while matching $\displaystyle \sigma(0)=(q_{0},r_{0})$ and $\displaystyle \sigma(1)=(q_{1},r_{1})$ for some $\displaystyle P_{0}=(q_{0},r_{0})$ and $\displaystyle P_{1}=(q_{1},r_{1})$, for instance when $\displaystyle q_{0}\neq q_{1}$, and in any open subset you may find such points (may be sketching a graphic would help for this).
In this case, you will always have $\displaystyle \sigma(t)\in\mathbb{Q}^{c}\times\mathbb{R}$ for some $\displaystyle t\in[0,1]$.
About the connectivity, I am not sure for this set.
You're right it's not path-connected, but unfortunately it's not connected: you can write this space as a disjoint union of non trivial open sets, for instance:
$\displaystyle \mathbb{Q}\times\mathbb{R}= ((\mathbb{Q}\cap ]-\infty ,\pi [)\times\mathbb{R}) \cup ((\mathbb{Q}\cap]\pi ,+\infty [)\times\mathbb{R})$
(this sets are open because they are intersections with open subsets in $\displaystyle \mathbb{R}^2$, and we're working with subspace topology)
But the space I was looking for may be more complicated: it has to be connected, but, instead of not path connected, not path-connected in any open subset (i.e. if you want never locally path-connected), and I'm not sure that ($\displaystyle X$ not path-connected)$\displaystyle \Rightarrow$($\displaystyle X$ not path-connected in any open subset)...
Okay, please see here Connected and Path Connected
I guess this will help you much, and indicates that your assertion is true.