Connected and locally Connected Space. Show $\overline{U}\cap C \neq \emptyset$ for C close and U connected component in $X\setminus C$

Question

Let X be a connected and locally connected space and let C be a non-trivial closed subset. Let U be a non-empty connected component of $X\setminus C$.
1) Prove that $\overline{U}\cap C \neq \emptyset$
2) Assume that C is connected. Prove $X\setminus U$ is connected.

My attemp: I tried assuming that $\overline{U}\cap C = \emptyset$ and then got $(X\setminus \overline{U} )\cup (X\setminus C) = X$
Now I wanted to show that $$(X\setminus \overline{U} )\cap (X\setminus C)=\emptyset$$ for every point in the intersection there exists an open connected neighbourhood around x $$x\in B \subset (X\setminus \overline{U} )\cap (X\setminus C)$$

But I don't know how to continue or even if this is the right approuch.

Thank you,

Answer 1

Your idea does not work. In general $(X \setminus \overline U) \cap (X \setminus C) \ne \emptyset$. As an example take $X=\mathbb R,C=[0,1],U=(1,\infty)$.

We shall prove a slightly more general result.

Let $X$ be a connected space and let $C$ be a non-trivial closed subset such that $X \setminus C$ is locally connected. Let $U$ be a non-empty connected component of $X\setminus C$.

1. $\overline{U}\cap C \neq \emptyset$ .

2. Assume that C is connected. Then $C \cup U$ is connected and closed in $X$.

3. Assume that C is connected. Then $X\setminus U$ is connected and closed in $X$.

Proof of 1.

We need two facts.

• Each component of a space is closed.
• Each component of a locally connected space is open. For a proof see e.g. About locally path-connected spaces.

We conclude that $U$ is open in $X \setminus C$. Since $C$ is closed, $U$ is also open in $X$.

Assume that $\overline{U}\cap C = \emptyset$. Since $U$ is closed in $X \setminus C$, we have $\overline{U} \cap (X \setminus C) = U$ and therefore $\overline{U} = \left(\overline{U} \cap C \right) \cup \left(\overline{U} \cap (X \setminus C)\right) = U$. Hence $U$ is closed in $X$.

Therefore $U$ is a non-trivial clopen subset of $X$ which contradicts the fact that $X$ is connected.

Proof of 2.

This follows from 1.

Proof of 3.

Let $U_\alpha$, $\alpha \in J$, denote the components of $X \setminus C$ which are distinct from $U$. By 2.all $C \cup U_\alpha$ are connected, thus $X \setminus U = \bigcup_{\alpha \in J} C \cup U_\alpha$ is connected. Since $U$ is open in $X$, the complement $X\setminus U$ is closed in $X$.

Remark 1.

With no assumption on $X \setminus C$, 1. is true if $U$ is open in $X \setminus C$. Local connectivity of $X \setminus C$ is a just a general condition assuring this for all components. Note that if $X \setminus C$ only has finitely many components, then they are all open.

This transfers to 2.

Without local connectivity assumption on $X \setminus C$ 3. remains true if all components $U' \ne U$ are open; $U$ need not be open in this case.

Remark 2.

The assumptions on $X$ on $X \setminus C$ (or on $U$) cannot be dropped. See Union of a connected subset with a component is connected.