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

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,

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.

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.