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There is a theorem that:A space is locally connected iff each connected components of an open set is open.

But recently I had seen to prove That each connected component is closed. Connected Components are Closed

Then how can the connected component of an open set be open if it is a locally connected space ? It will be contradiction to the statement that connected components is closed.

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  • $\begingroup$ Think about $\mathbb{R} \setminus \{0\}$. What are the connected components? $\endgroup$ – Carl Mummert Mar 6 at 2:12
  • $\begingroup$ In fact, it is true in general that any connected component is clopen (it is both open and closed). If you want a proof, you should add your definition of a "connected component" first. $\endgroup$ – stressed out Mar 6 at 13:41
  • $\begingroup$ sets are not doors, they can be open and closed at the same time... $\endgroup$ – Henno Brandsma Mar 6 at 17:36
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A subset being closed doesn't preclude that subset from being open. For a simple example, every discrete space is locally connected, and every subset of a discrete space—in particular the singleton sets (which are the connected components)—is both open and closed.

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