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I am trying to prove that if $C$ is a connected component of $X$, then $C$ is closed. Here is my attempt:

Let $C$ be a connected component of $X$. Then $\overline{C} \supseteq C$ must be connected as well, and since every connected subspace intersects one, and only one, connected component (and is therefore contained in it), it follows that $\overline{C} \subseteq C$.

Is this right? Something about it is fishy...

user193319
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    $\overline{C}$ is always connected when $C$ is: this is a standard result (you can even insert an intermediate subset). Could $\overline{C}$ be any different from $C$? – Randall Oct 23 '17 at 19:37
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    Recall the definition of closed: A is closed if A^c is open. – Sort of Damocles Oct 23 '17 at 19:39
  • @Randall To answer your question, no and that's what I thought I was showing. Isn't it true that a connected subspace intersects one, and only one, component; and does $C \subseteq \overline{C}$ imply that they intersect? From what I understand, this should immediately imply $\overline{C} \subseteq C$, since the components are equivalence classes. – user193319 Oct 23 '17 at 19:44
  • Yes, I think you are right. – Randall Oct 23 '17 at 19:49
  • Unless you belong to the church that declares $\varnothing$ to be disconnected, there is one connected subspace that doesn't intersect any component. From the connectedness of $\overline{C}$, the equality $C = \overline{C}$ follows since components are by definition the maximal connected subspaces. – Daniel Fischer Oct 23 '17 at 20:13
  • The empty set is not disconnected. It's just not connected. This post was made by the intuitionist gang. – Jackozee Hakkiuz Apr 25 '23 at 02:59

2 Answers2

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A component of $x$ is the largest connected set containing $x$.

Let $C$ be a component of $x$. Thus $x\in\overline{C}$ and $C \subseteq \overline{C}$.

However, $C$ is the largest connected set, therefore $\overline{C} \subseteq C$.
Hence $C = \overline{C}$, and $C$ is closed.

William Elliot
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Your proof is about right. I'd formulate it as follows: a connected component $C$ of $X$ is a maximally connected subset; this means 2 things:

  1. $C$ is connected.
  2. if $C \subseteq D$ and $D$ is connected, $C=D$.

Now use that $C$ connected implies $\overline{C}$ connected. Then applying 2. and noting that obviously $C \subseteq \overline{C}$, we conclude that $C = \overline{C}$, which is equivalent to $C$ being closed. QED.

Henno Brandsma
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