Points of a connected open subset joined by a curve Let $\Omega$ be a connected open subset of $\Bbb C$. Is it necessarily true
that any two points of $\Omega$ can be joined by a non-selfintersecting curve, that
is, an injective continuous map $\gamma:[0,1]\to \Omega$?
Thanks in advance!
 A: The answer is no.  The property that you're describing is called path-connectedness.  Path-connectedness of a space always implies connectedness but not the other way around.  The most classic example of this is the topologist's sine curve defined by $T = T_1 \cup T_2$ where $T_1, T_2 \subseteq \mathbb{R}^2$ are given by 
$$
T_1 \;\; =\;\; \left \{ \left ( x, \sin \left . \frac{1}{x}  \right ) \; \right | x \in (0, \infty) \right \}
$$
and 
$$
T_2 \;\; =\;\; \{(0,y) \; | \; -1 \leq y \leq 1\}.
$$
$T_1$ and $T_2$ are each connected themselves since one is the graph of a continuous image of a connected space, and the other is just an interval.  If $T$ were disconnected we could write it as $T = U\cup V$ with $U$ and $V$ nonempty, disjoint, open subsets of $T$.  Necessarily $T_2$ must be contained in one of them, say $U$, but $U$ necessarily contains points of $T_1$ (why?) so since $T_1$ is connected we must have that $T_1 \cup T_2 = T \subseteq U$ which is a contradiction.  
Can you construct an argument for why path-connectedness doesn't follow for $T$?
