Open and connected set in metric space

Question

In a normed space, we know that if a set is open and connected, it is path connected. Is it true for general metric space or general topological space?

Answer 1

It is not true. The typical counterexample is the "topologist's sine curve". $$\{(t,\sin(1/t):t>0\}\cup \{(0,y):-1\le y\le 1\}.$$

Answer 2

For general topological spaces the answer is no. See here for a nice counterexample, the union of the graph of $\sin(1/x)$ and $\{0\}$:

http://topospaces.subwiki.org/wiki/Connected_not_implies_path-connected

If your space is locally path-connected, i.e. every point has a path-connected neighborhood, then connected implies path connected. This particularly holds for normed vector spaces, since any open ball around a point is path-connected.

So for metric spaces, connectedness and path-connectedness are equivalent, but for general topological spaces not.