Characteristic of a connected, locally compact, Hausdorff space X which is locally connected.

Q. Prove that a connected, locally compact, Hausdorff space X is locally connected if and only if for each compact subset K and each open set U containing K, all but a finite number of components of X-K lie in U.

I tried to solve this using the exercise "Let X be a Hausdorff, locally connected and locally compact space. Let U be a connected subset of X and let x,y∈U. Prove there exists a compact connected subset T of U such that T contains both x,y" solved by Henno Brandsma, but unable to conclude anything.

We shall prove the slightly more general

Theorem. For a locally compact Hausdorff space $$X$$ the following are equivalent:

(1) $$X$$ is locally connected and has only finitey many components.

(2) For each compact subset $$K \subset X$$ and each open set $$U \subset X$$ containing $$K$$, all but finitely many components of $$X \setminus K$$ are contained in $$U$$.

Remark: A locally compact Hausdorff space $$X$$ which is locally connected and connected may have compact subsets $$K$$ such that $$X \setminus K$$ has infinitely many components. An example is $$X = [0,1] \subset \mathbb R$$ which is compact Hausdorff, locally connected and connected. Now consider $$K = \{0\} \cup \{ 1/n \mid n \in \mathbb N \}$$.

Let us prepare the proof of the theorem by a definition and a lemma.

A splitting of a space $$Z$$ is a pair of nonempty open subsets $$W, W' \subset Z$$ such that $$W \cup W' = Z$$.

Lemma. For a space $$X$$ the following are equivalent:

(a) $$X$$ is locally connected, i.e. for each $$x \in X$$ and each open neighborhood $$U$$ of $$x$$ there exists a connected open neighborhood $$V$$ of $$x$$ such that $$V \subset U$$.

(b) For each $$x \in X$$ and each open neighborhood $$U$$ of $$x$$ there exists a connected (not necessarily open) neighborhood $$N$$ of $$x$$ such that $$N \subset U$$.

(c) The components of each open subset of $$X$$ are open in $$X$$.

This is a standard characterization of locally connected spaces. See About locally path-connected spaces and Definition of locally pathwise connected. (For the sake of completeness we give a proof at the end of the answer, since the referenced proofs deals with locally path-connected spaces.)

We now prove the theorem.

Let $$U$$ be an open neighborhood of the compact set $$K$$. Let $$\mathcal C = \{C_\alpha\}_{\alpha \in A}$$ denote the set of components of the open set $$X \setminus K$$. Since $$X$$ is locally connected, the $$C_\alpha$$ are open sets.

$$(1) \Rightarrow (2)$$:

Case 1. $$X$$ is connected.

Since $$X$$ is locally compact Hausdorff, there exists an open neighborhood $$V$$ of $$K$$ such that $$\overline{V}$$ is a compact subset of $$U$$. In the sequel we only consider the case that $$V \ne \emptyset$$. (If $$V =\emptyset$$, then $$K = \emptyset$$ and $$\mathcal C = \{ X \}$$ so that the assertion is trivial.)

We claim that $$C_\alpha \cap V \ne \emptyset$$ for all $$\alpha \in A$$. To see this, assume that $$C_{\alpha_0} \cap V = \emptyset$$ for some $$\alpha_0$$. Then $$(C_{\alpha_0}, V \cup \bigcup_{\alpha \ne \alpha_0} C_\alpha)$$ forms a splitting of $$X$$ which is impossible.

Let $$A'$$ be the set of all $$\alpha' \in A$$ such that $$C_{\alpha'}$$ is not contained in $$U$$. We shall show that $$A'$$ is finite.

$$B = \text{bd} V$$ is compact. For $$\alpha' \in A'$$ we have $$C_{\alpha'} \not\subset \overline{V}$$, i.e. $$C_{\alpha'} \setminus \overline{V} \ne \emptyset$$. We also know that $$C_{\alpha'} \cap V \ne \emptyset$$. If $$C_{\alpha'} \cap B = \emptyset$$, we get a splitting $$(C_{\alpha'} \setminus \overline{V}, C_{\alpha'} \cap V)$$ of $$C_{\alpha'}$$ which is impossible. Hence $$C_{\alpha'} \cap B \ne \emptyset$$ for all $$\alpha' \in A'$$.

Now let $$A''$$ denote the set of all $$\alpha'' \in$$ such that $$C_{\alpha''} \cap B \ne \emptyset$$. Then $$A' \subset A''$$. We show that $$A''$$ is finite which finishes the proof.

By construction $$\mathcal C'' = \{C_{\alpha''}\}_{\alpha'' \in A''}$$ is an open cover of $$B \subset X \setminus K$$. The $$C_{\alpha''}$$ are pairwise disjoint, hence no proper subset of $$\mathcal C''$$ covers $$B$$. Since $$B$$ is compact, $$\mathcal C''$$ has a finite subcover $$\mathcal C^* \subset \mathcal C''$$. This is possible only when $$\mathcal C^* = \mathcal C''$$.

Case 2. $$X$$ is arbitrary with finitely many components $$X_1,\dots,X_n$$.

The $$X_i$$ are open and closed subsets of $$X$$, hence locally compact Hausdorff, locally connected and connected. Case 1 applies to the subsets $$K_i = K \cap X_i$$ and $$U_i = U \cap X_i$$ of $$X_i$$. Thus, if $$\mathcal C_i$$ denotes the set of components of $$X_i \setminus K_i$$, then the subset $$\mathcal C'_i$$ of elements of $$\mathcal C_i$$ which are not contained in $$U_i$$ is finite. But the set of components of $$X \setminus K$$ is the union of the $$\mathcal C_i$$ and the subset $$\mathcal C'$$ of elements of $$\mathcal C$$ which are mot contained in $$U$$ is the union of the $$\mathcal C'_i$$, i.e. is finite.

$$(2) \Rightarrow (1)$$:

Taking $$K = U = \emptyset$$, we see that $$X$$ has only finitely many components.

Let $$x \in X$$ and $$U$$ be an open neighborhood of $$x$$. Since $$X$$ is locally compact Hausdorff, there exists an open neighborhood $$V$$ of $$x$$ such that $$\overline{V}$$ is a compact subset of $$U$$. Next choose any open neighborhood $$W$$ of $$x$$ such that $$\overline{W} \subset V$$. Then $$U' = U \setminus \overline{W}$$ is an open neighborhood of the compact set $$B = \text{bd} V$$.

Let $$\mathcal C$$ denotes the set of components of $$X \setminus B$$. We know that only finitely many $$C_1,\dots, C_n \in \mathcal C$$ are not contained in $$U'$$. Therefore $$W \subset \bigcup_{i=1}^n C_i$$. One of these components, say $$C_1$$, must contain $$x$$. Hence $$C_1 \cap V \ne \emptyset$$ and we conclude that $$C_1 \subset V \subset U$$ (otherwise we would get a splitting $$(C_1 \cap V, C_1 \cap (X \setminus \overline{V}))$$ of $$C_1$$). The $$C_i$$ are closed in $$X \setminus B$$, hence also $$C' = \bigcup_{i=2}^n C_i$$ is closed in $$X \setminus B$$. Thus $$C' \cap W$$ is closed in $$W \subset X \setminus B$$. Hence $$C_1 \cap W = W \setminus (C' \cap W)$$ is open in $$W$$ and therefore also open in $$X$$. This shows that $$C_1$$ is a connected neighborhood of $$x$$ which is contained in $$U$$.

Proof of the Lemma:

(a) $$\Rightarrow$$ (b) : Obvious.

(b) $$\Rightarrow$$ (c) : Let $$U \subset X$$ be open and $$C$$ be a component of $$X$$. Consider $$x \in C$$. There exist an open neighborhood $$V$$ of $$x$$ and a connected $$N \subset X$$ such that $$V \subset N \subset U$$. Since $$N \cap C \ne \emptyset$$, we see that $$N \cup C$$ is a connected subset of $$U$$ which contains $$C$$. By definition of $$C$$ we see that $$N \cup C = C$$, i.e. $$N \subset C$$. We conclude $$V \subset C$$.

(c) $$\Rightarrow$$ (a) : Let $$U$$ be a neighborhood of $$x \in X$$ and $$C$$ be the component of $$U$$ which contains $$x$$. This is an open neighborhood of $$x$$ contained in $$U$$.