A locally compact, connected, Hausdorff and locally connected space is not the countable disjoint union of nonempty closed subsets

In Chapter $$5$$, Section $$4$$ of the book Continuum theory by Sam B. Nadler, Jr., the auther defines $$\sigma$$-connectedness (Definition $$5.15$$) to be the property of not being a disjoint union of countably many (including finitely many) and more than one nonempty closed subsets. It is well known (for example, see here) that a continuum (i.e. compact, connected and Hausdorff space) is $$\sigma$$-connected. Continuum theory gives an example (Example $$5.14$$) that shows that compactness cannot be weakened to local compactness. This link gives an essentially same counterexample (see Figure $$1$$).

On the other hand, in the link above, the authors say that a locally compact, connected, Hausdorff and locally connected spaces is $$\sigma$$-connected by using one point compactification. However, they admit that they could not find a reference for the generalized result (the second footnote).

So my question is: Is the generalized result true, which is to say, a locally compact, connected, Hausdorff and locally connected space cannot be written as a countable disjoint union of nonempty closed subsets?

• If you add "metrizable" then the claim follows from the fact that such a space is arcwise connected. Commented Apr 27 at 14:54

I think you can simply adapt the proof as in the case for the continuum.

We may of course restrict ourselves to the case in which $$X$$ is compact by considering the one point compactification if needed (I can spell out the details if needed). For simplicity we say that a space is good if it is Hausdorff, connected, compact, and locally connected. Observe that for good spaces the Baire category theorem holds. In particular we have that if $$X$$ is a good space and it is union of a disjoint family of closed sets $$(C_i)_{i\in\omega}$$ (which we may assume are all non empty) then we have that the sets $$D_i=X\setminus (C_i\setminus int({C_i}))$$ are open dense. In particular this means $$\cap_{i\in\omega} D_i=\bigcup_{i\in\omega} int(C_i)$$ is a dense, disconnected, open proper subset of $$X$$.

Lemma: For any good space $$X$$ that is dijoint union of countably many closed sets $$(C_i)_{i\in\omega}$$. Without loss of generality we may assume that all of the $$C_i$$ are non empty. We show that there is a good subspace $$Y_1\subseteq X$$ that does not intersect $$C_1$$ but intersects infinitely many $$C_i$$.

Proof: To do so we need to find a $$y\in X$$ and a compact connected neighborhood of $$y$$ that intersects infinitely many $$C_i$$. We have that for any $$y$$ and any connected neighborhood $$U$$ of $$y$$ either intersects exactly one closed set $$C_i$$ or infinitely many. This means in particular that for any $$y\notin \bigcup_{i\in\omega}int(C_i)$$ then any neighborhood of $$y$$ intersects infinitely many $$C_i$$. So we set $$Y_1$$ be a closed compact connected neighborhood of a $$y\notin \bigcup_{i\in\omega}int(C_i)$$.

Now to prove the main result we just iterate the process to get a descending chain of good subspaces $$(Y_i)_{i\in\omega}$$ of $$X$$ such that $$Y_i\cap C_i=\emptyset$$. By Cantor's theorem we have that $$\bigcap_{i\in\omega} Y_i\neq\emptyset$$ since each $$Y_i$$ is closed and compact. But this is absurd since any $$z$$ in that intersection cannot be contained in any $$C_i$$.

Edit: It is easy to check that if $$X$$ is Hausdorff and locally compact then then $$X\cup\{\infty\}$$ is Hausdorff and compact. Furthermore you can check that $$X$$ is connected and Hausdorff and not compact then so will $$X\cup\{\infty\}$$ (This is because any compact set of $$X$$ will be a closed proper subset) and every neighborhood of $$\infty$$ is connected. Now if $$X=\bigcup_{i\in \omega} C_i$$ such that none of the $$C_i$$ are empty, then $$X\cup\{\infty\}=\bigcup_{i\in \omega} C_i \cup\{\infty\}$$. That is to say if $$X$$ is a Hausdorff connect locally compact and connected space and it is the dijoint union of countably many closed sets then either $$X$$ is compact or the one point compactification of $$X$$ is the disjoint union of countably many closed sets. This allows us to restrict our attention to the good spaces. However looking back at the proof you can easily avoid doing this step since you do not use the compactness of $$X$$ to construct the descending sequences of good spaces $$(Y_i)_{i\in\omega}$$.

• Thanks for the answer! (Of course, it would be very nice to spell out the details :) ) Commented Apr 27 at 20:54

Another approach is the following, based on a simple fact, which might be of its on interest:

If $$X$$ is connected and for each $$x \in X$$ there exists a (not necessarily open) neighborhood of $$x$$, which is $$\sigma$$-connected, then $$X$$ is $$\sigma$$-connected.

PROOF. Assume not. Then, since $$X$$ is connected, there is a pairwise disjoint family $$(X_n)_{n < \omega}$$ such that $$X = \bigcup_{n < \omega} X_n$$ and each $$X_n$$ is a closed, non-empty subset of $$X$$. (This is just to simplify notification. By considering possibly finite families, here we could avoid connectness.) But now, since $$X$$ is connected, $$X_0$$ is not open. Hence, there is a $$x \in X_0$$, which is not in the interior of $$X_0$$. Pick a $$\sigma$$-connected neighborhood $$U$$ of $$x$$. It is $$U = \bigcup_{n < \omega} (U \cap X_n)$$, $$x \in U \cap X_0 \neq \emptyset$$, hence $$U \cap X_1 = U \cap X_2 = \ldots = \emptyset$$, hence $$U \subset X_0$$. Since $$x$$ is in the interior of U, it is in the interior of $$X_0$$. Contradiction!

Now, if $$X$$ is locally compact, locally connected and Hausdorff, then each $$x \in X$$ has a compact, connected neighborhood, which is $$\sigma$$-connected by the above-mentioned, well-known result.

This proves that the answer to the question is affirmative.

• Thanks! It's just a pity that the other answer has been posted earlier :( Commented Apr 27 at 20:56
• @Jianing Song: Never mind ;-)
– Ulli
Commented Apr 28 at 5:40
• Very nice and short. By "open kernel" of a set, do you mean its interior? Commented Apr 28 at 18:20
• @PatrickR: yes! And certainly, "interior" is the more common name. Therefore, I changed notation in my answer. Thank you for notifying.
– Ulli
Commented Apr 28 at 19:34