# Connected Components are Closed

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...

• $\overline{C}$ 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 $C$? Commented Oct 23, 2017 at 19:37
• Recall the definition of closed: A is closed if A^c is open. Commented Oct 23, 2017 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. Commented Oct 23, 2017 at 19:44
• Yes, I think you are right. Commented Oct 23, 2017 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. Commented Oct 23, 2017 at 20:13

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.

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.