Could anyone tell me if that's a wrong exemple (but I think it works): in
Why not only considering
which has the same nature with the first quadrant of
It is clear that this set is not path connected.
Because the mapping will have discontinuity while matching and for some and , for instance when , 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 for some .
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:
(this sets are open because they are intersections with open subsets in , 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 ( not path-connected) ( 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.