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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?

user405156
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    There's a problem. If $C_\alpha$ is not open, why must that imply that it is closed? Note that being not open and being closed are not the same. Therefore, you need to see what conclusions you can draw from non-openness, or go in for a direct argument. – Sarvesh Ravichandran Iyer Jan 01 '18 at 11:59
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    What if no such open $U$ exists (unless it is empty)? The proof cannot be correct since connected components are not necessarily open. For that you need extra conditions like: their number is finite (they are closed), or the space is locally connected. – drhab Jan 01 '18 at 12:03
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    If we consider as our topological space the open set itself, then each of its connected components is both open and closed in that topology – QED Jan 01 '18 at 12:06
  • can this be proved in the case of a. manifold? I would like to have this property for the open set of a map – user405156 Jan 01 '18 at 13:13

2 Answers2

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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.

red_trumpet
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Henno Brandsma
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  • can this be proved in the case of a. manifold? I would like to have this property for the open set of a map. – user405156 Jan 01 '18 at 12:57
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    @tomak A manifold is locally Euclidean, so in particular locally connected. So there it does hold, but with a different proof. – Henno Brandsma Jan 01 '18 at 13:20
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Connected components are open if X is locally connected.
Q is not locally connected, I = { x in Q : 0 < x < 1 } is
open subset and its connected components are the singletons
which are closed and not open.

William Elliot
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    can this be proved in the case of a. manifold? I would like to have this property for the open set of a map – user405156 Jan 01 '18 at 13:13