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Let $X$ be a connected metric space, let $A \subseteq X$ be a connected set. Let $\mathcal{C}$ be a connected component of $X-A$. Show that $X-\mathcal{C}$ is connected.

Ok, so Im been dealing with this exercise for a while, and I am completely stucked.

I've managed the following (which I don't know if it is even true.)

Since $X-A \subseteq \mathcal{C}$ we have that $X-\mathcal{C} \subseteq A$. Now since $A$ is a connected set, and $\mathcal{C}$ is closed (since it's a connected component), one way to prove the statement is to see that $X-\mathcal{C}$ is closed in $A$ because in this case, $X-\mathcal{C}$ would be open and closed in $A$, connected which implies $A=X-\mathcal{C}$ and therefore $X-\mathcal{C}$ is connected.

However Im not being able to show that $X-\mathcal{C}$ is closed in $A$, (im not sure if in fact it is).

If not, If its possible, I would like a hint, not a complete solution!

Joaquin Liniado
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  • This doesn't sound right: "Since $X-A \subseteq \mathcal{C}$...". What if $X=\mathbb{R}^2$, and $A$ is the $x$-axis, and $\mathcal{C}$ is the upper half-plane. Then $X-A$ is not contained in $\mathcal{C}$. – Mankind Jun 21 '15 at 20:49
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    First, $X\setminus A$ is not a subset of $C$. In fact, the opposite is true. In these types of problems, it is often helpful to consider a separation of $X\setminus C$ and show that you can derive from there a separation of $A$, forming a contradiction. . – Plutoro Jun 21 '15 at 20:50

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