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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?

Thanks for your time.

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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.

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  • $\begingroup$ Oh, I get it. Connected + locally path-connected = path-connected, right? $\endgroup$ – Syuizen Oct 11 '15 at 11:31
  • $\begingroup$ @Syuizen: Yes, that sounds right. $\endgroup$ – Henning Makholm Oct 11 '15 at 11:32
  • $\begingroup$ If not in $R^n$ , could such set exist? I mean topology may be very weird. $\endgroup$ – Syuizen Oct 11 '15 at 11:33
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    $\begingroup$ @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... $\endgroup$ – Henning Makholm Oct 11 '15 at 11:35
  • $\begingroup$ Uh, subspace topology works. Thank you. I thought that I could find the set in some well-known topology space, but failed. $\endgroup$ – Syuizen Oct 11 '15 at 11:41

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