$\sigma$-connected spaces and path-connected property

Question

A topological space $X$ is called $\sigma$-connected if it cannot be written as a union of at least two and at most countably infinitely many pairwise disjoint nonempty closed sets. In other words, $X$ is connected and cannot be covered by a sequence of pairwise disjoint nonempty closed sets.

A classical result (Is $[0,1]$ a countable disjoint union of closed sets?) is that the interval $[0,1]$ is $\sigma$-connected. More generally, so is any continuum (i.e. a compact connected Hausdorff space). See also (A locally compact, connected, Hausdorff and locally connected space is not the countable disjoint union of nonempty closed subsets) for a related result.

Question: What is the relation between the $\sigma$-connected and path connected properties? Proof of implications and interesting examples and counterexamples?

Answer 1

Ulli's answer gives a really simple and elegant proof of the result "path-connected $\Rightarrow$ $\sigma$-connected". Actually as we will see, $\sigma$-connectedness behave in many aspects just like connectedness, or other connectedness properties. Even if we have already know "path-connected $\Rightarrow$ $\sigma$-connected", I would still like to name some properties of $\sigma$-connectedness:

Fact 1. If $\{A_\alpha\}_\alpha$ is a family of $\sigma$-connected sets with nonempty intersection, then $\displaystyle\bigcup_\alpha A_\alpha$ is $\sigma$-connected.

Proof: Write $\displaystyle\bigcup_\alpha A_\alpha = \displaystyle\bigcup_{n\in\omega} C_n$, where $\{C_n\}_{n\in\omega}$ is a family of disjoint closed subsets with respect to the subspace topology of $\displaystyle\bigcup_\alpha A_\alpha$, then for each $\alpha$, $\{A_\alpha\cap C_n\}_{n\in\omega}$ is a family of closed subsets with respect to the subspace topology of $A_\alpha$. So each $A_\alpha$ must be included to the same $C_n$ since the intersection of all $A_\alpha$ is nonempty.

Fact 2. The continuous image of a $\sigma$-connected set is $\sigma$-connected.

Proof: Let $X$ be a $\sigma$-connected space, and $f:X\to Y$ be a continuous surjection (so I don't need to write "with respect to the subspace topology" all the time). Let $\{C_n\}_{n\in\omega}$ be a partition of $Y$ into closed subsets, then $$ X = f^{-1}\left(\bigcup_{n\in\omega}C_n\right) = \bigcup_{n\in\omega}f^{-1}(C_n), $$ which partitions $X$ into closed subsets, so one of $f^{-1}(C_n)$ is the whole $X$ and the others are empty. This shows that one of $C_n$ is the whole $Y$ the others are empty because $f$ is a surjection.

Corollary. A path-connected space is $\sigma$-connected.

Proof. Let $X$ be path-connected and fix $x_0\in X$. Let $S=\{f:[0,1]\to X\text{ continuous}:f(0)=x_0\}$, then $f([0,1])$ is $\sigma$-connected as the continuous image of the $\sigma$-connected space $[0,1]$, so the union of all $f([0,1])$ is also $\sigma$-connected since their intersection contains $x_0$. Since $X$ is path-connected, the union of all $f([0,1])$ is precisely $X$.

Now I will give an example of a space that is $\sigma$-connected but not path-connected.

Let $K=\left\{\dfrac{1}{n}:n\in\mathbb{N}^*\right\}$. Smirnov's deleted sequence topology $\mathbb{R}_K$, or what Munkres calls $K$-topology, consists of all sets of the form $U\setminus B$, where $U\subset\mathbb{R}$ is open in the standard topology and $B\subset K$. This is a topology strictly finer than the standard topology on $\mathbb{R}$. (I believe that this is among the two examples used by Munkres to teech the comparison of topologies; the other more well-known example is the Sorgenfrey line, or what Munkres calls the lower-limit topology).

It is standard that $\mathbb{R}_K$ is not path-connected. You will notice that the proof of its $\sigma$-connectedness is exactly the proof of its connectedness.

Fact 3. The closure of a $\sigma$-connected set is $\sigma$-connected.

Proof. Let $E$ be a $\sigma$-connected set of a space $X$, and write $\overline{E}=\displaystyle\bigcup_{n\in\omega} C_n$, where $\{C_n\}_{n\in\omega}$ is a family of disjoint closed subsets with respect to the subspace topology of $\overline{E}$ (so each $C_n$ is also closed since $\overline{E}$ is closed), then $\{E\cap C_n\}_{n\in\omega}$ is a family of closed subsets with respect to the subspace topology of $E$. By $\sigma$-connectedness we can suppose WLOG that $E\cap C_0 = E$ and $E\cap C_n=\emptyset$ for $n>0$, which means that $E\subset C_0$, then $\overline{E}\subset C_0$ since $C_0$ is closed.

Claim. $\mathbb{R}_K$ is $\sigma$-connected.

Proof. The restriction of $\mathbb{R}_K$ to the subspaces $(-\infty,0]$ and $(0,+\infty)$ coincides with their counterparts in standard topology, so both subspaces are $\sigma$-connected since they are path-connected, then we see that $[0,+\infty)=\overline{(0,+\infty)}$ (in $\mathbb{R}_K$) and $\mathbb{R}_K = (-\infty,0]\cup[0,+\infty)$ are also $\sigma$-connected.

Answer 2

For the one implication: Of course, there are tons of compact, connected, $T_2$, hence $\sigma$-connected, spaces, which are not path connected, for instance the lexicographic unit square, $\beta \mathbb R$, the closure of the topologist's sine curve, etc.

On the other hand, if $X$ is path-connected and $T_2$, it is $\sigma$-connected. ($T_2$ is not necessary, see below update)

This directly follows from the above-mentioned result and:

If for all $x, y \in X$, there is a $F \subset X$, such that $F$ is $\sigma$-connected and $x, y \in F$, then X is $\sigma$-connected.

PROOF. Assume not. Obviously, $X$ is connected. Hence, $X = \bigcup_{n \in \mathbb N} X_n$, where each $X_n$ is closed, non-empty and the $X_n$ are pairwise disjoint. Pick $x \in X_1, y \in X_2$ and let $F \subset X$ be $\sigma$-connected, $x, y \in F$. Then $F$ is the disjoint union of $F \cap X_n, n \in \mathbb N$, where each set is closed in $F$, and at least two are non-empty. Contradiction!

Update
There is an even stronger result:

If $X$ is path connected, then it is $\sigma$-connected.

PROOF. Assume not. Since $X$ is connected, $X = \bigcup_{n \in \mathbb N} X_n$, where each $X_n$ is closed, non-empty and the $X_n$ are pairwise disjoint. Pick $x \in X_1, y \in X_2$. Since $X$ is path connected, there is a continuous map $f: [0, 1] \rightarrow X$, such that $f(0) = x, f(1) = y$. Hence $[0, 1]$ is the disjoint union of $f^{-1} (X_n), n \in \mathbb N$, where each set is closed in $[0, 1]$, and at least two are non-empty, contradicting $\sigma$-connectness of $[0, 1]$.

Remark
I will keep the first, weaker result, since its lemma might be of its own interest.