# Example for the set that is open connected but not path-connected in $R^n$

Connected set may not be path-connected.

Certainly,there are many examples to show it is true, such as the closure of topologist's sine curve.

More examples can be found in the following questions posted in this website:

Is there a topological group that is connected but not path-connected?

Show that this set is connected but not path connected

Another example of a connected but non path connected set

connected but not path connected?

However, all examples are close connected sets.

So is there any open connected sets that is not path-connected? If not , how to prove the theorem?

If $A$ is an open subset of $\mathbb R^n$, then it is easy to see that every path-connected component of $A$ is also open. Therefore, if $A$ has at least two path-connected components, then it is a disjoint union of at least two nonempty open sets, and therefore is not connected.
• If not in $R^n$ , could such set exist? I mean topology may be very weird. – Syuizen Oct 11 '15 at 11:33
• @Syuizen: Sure -- just take a known connected-but-not-path-connected set to be the entire topological space, and let $A$ be all of it... – Henning Makholm Oct 11 '15 at 11:35