# Path-connected iff connected on extended complex plane

The following is true for an open set $D \subset \mathbb{C}$:

$D$ is connected $\iff$ $D$ is path-connected.

The latter means there is a continuous function $\gamma: [0,1] \to D$ for $z_0, z_1 \in D$ so we have $\gamma(0)=z_0$ and $\gamma(1)=z_1$.

Now, I want to consider the statement on the extended complex plane $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$.

I have read that $\hat{\mathbb{C}}$ is a connected topological manifold which is locally path-connected and thus path-connected.

I must admit that I don't feel comfortable with this statement, probably because I'm not that familiar with these topological terms. Maybe one of you can clarify it or even show the statement for $\hat{\mathbb{C}}$ in a more direct way.

If $z_0, z_1 \in D \subset \hat{\mathbb{C}}, z_1=\infty$, then $\gamma(t)=(1-t) z_0 + \frac{t}{1-t}, t\in[0,1]$ is a continuous function so that $z_0$ and $z_1$ are path-connected on $D$?

Help is appreciated!

• The claim is false even for subsets of $\mathbb{C}$. The topologist's sine curve is a standard counterexample. Commented Nov 22, 2011 at 19:01
• Sorry, an important point is that D is an open subset. Edited. Commented Nov 22, 2011 at 19:09

• You can get rid of compactness by observing that in a locally path-connected space, the set $C_x$ of points that can be connected to a certain fixed point $x$ is closed and open. Commented Nov 22, 2011 at 23:45