connectednessgeneral-topology
I think the statement is true that two connected subspaces of a connected topological space is connected, and there are two different situations here to be discussed.
First of all, when the intersection is empty set, do we consider it as connected or not?
For the other situation, if the intersection is not empty, how could I get the conclusion?
Others have mentioned the topologist's sine curve, that's the canonical example. But I like this one, though it is the same idea:
This is the image of the parametric curve $\gamma(t)=\langle (1+1/t)\cos t,(1+1/t)\sin t\rangle,$ along with the unit circle. There is no path joining a point on the curve with a point on the circle. Yet, the space is the closure of the connected set $\gamma((1,\infty))$, so it is connected.
The space between a circle of radius 2 centered at 0 and a
circle of radius 1 centered at 1 is simply connected, regular
open. The point of tangency is a counter example.
Is the property you'd like, E is regular open with a
simply connected closure? Here we are considering every
disk D. If you want just one disk, then if E is bounded,
any disk containing E will suffice.
Best Answer
The empty set is connected (trivially), because we cannot write it as a union of non-empty (!) disjoint open sets..
But the intersection of two connected sets need not be connected at all. Consider $C = \{(x,y): x^2 + y^2 = 1\}$, which is the unit circle (connected) and $D = \{(x,y): (x-1)^2 + y^2 = 1 \}$, the circle of radius 1 around $(1,0)$, also connected. Their intersection is $\{(\frac{1}{2},\frac{1}{2}\sqrt{3}),(\frac{1}{2},-\frac{1}{2}\sqrt{3})\}$, which is a two point set in the plane, hence disconnected.