I would like to prove that the connected components of an open set are open.
Take $U$ an open set in a space $X$ and $U=\cup_\alpha C_\alpha$ where the $C_\alpha$ are the connected components.
Suppose that there exists a $C_\alpha$ which is closed. Then ${C_\alpha}^c$ is open and ${C_\alpha}^c\cap U$ is open. However $$U= U\cap X = U\cap(C_\alpha\cup{C_\alpha}^c) = C_\alpha\cup ({C_\alpha}^c\cap U)$$
which is neither open nor closed. So we have a contradiction.
Is this proof correct?
Best Answer
A space $X$ is locally connected iff a component of an open set is open. See here e.g.
So this proof is not correct. Consider the rationals $\mathbb{Q}$, then all components are singletons which are never open in $\mathbb{Q}$ (nor in any of its open sets).
Logical fallacies: "not open" does not imply "closed" (sets aren't doors). Components are always closed, so the complement of a component is a union of closed sets, namely all other components,but unions of closed sets need not be closed, only finite unions are. So the complement need not be closed, etc.