This was mentioned in an article, but I have not been able to find a proof anywhere. Assume the original space has at least 2 points. The disjoint connected sets should be nontrivial (each has at least 2 points).

For instance $[0,2]=[0,1]\cup (1,2]$.

I tried to use the boundary bumping theorem.

  • $\begingroup$ Presumably you want the space to be Hausdorff as well? (Otherwise there are trivial counterexamples.) $\endgroup$ Dec 12 '15 at 21:19
  • $\begingroup$ Yes I assume Hausdorff, otherwise I guess the two point indiscrete space would be an easy counterexample. $\endgroup$ Dec 12 '15 at 21:25

I cannot imagine this theorem holds for all topological spaces. Nevertheless, if we restrict to spaces where the boundary-bumping theorem holds (usually compact, connected, Hausdorff spaces) then we can do the following.

Choose any point $x \in X$. Supposing if $(X-x)$ is connected we can write $X = \{x\} \cup (X-x)$ as required. Otherwise $(X-x)$ is partitioned into connected components. For each component $F$ of $(X-x)$ the closure $\overline F = F \cup \{x\}$ by boundary-bumping. Moreover the closure is a connected set. So split the set of connected components of $(X-x)$ into two nonempty parts $\mathcal F$ and $\mathcal G$ and consider the two sets

$A =\bigcup \{\overline F \colon F \in \mathcal F\}$

$B =\bigcup \{\overline G \colon G \in \mathcal G\}$

Every $\overline F$ includes the point $x$ and likewise for evry $\overline G$. So $A$ and $B$ are connected. By construction we have $X = A \cup B$ as required.

Edit: If you want the intersection of $A$ and $B$ to be nontrivial, then you can replace $\{x\}$ with any proper subcontinuum $K$ and repeat the proof. That proper subcontinua always exist is a well-known corollary to the boundary-bumping theorem. All you need to change is that, instead of $\overline F = F \cup \{x\}$ we have that $\overline F$ meets $K$ and hence $\overline F \cup K$ is connected (and closed). Then use the sets

$A =\bigcup \{\overline F \cup K \colon F \in \mathcal F\}$

$B =\bigcup \{\overline G \cup K \colon G \in \mathcal G\}$

  • $\begingroup$ But $A$ and $B$ intersect. Also, I want the connected sets to each be nontrivial - the union $\{x\}\cup (X-x)$ is too trivial. $\endgroup$ Dec 12 '15 at 21:32
  • $\begingroup$ nice try though. I ran into the same problem myself until I finally arrived at my answer below. Does it make sense? $\endgroup$ Dec 12 '15 at 21:34
  • $\begingroup$ Then you can replace the role of $\{x\}$ in the proof by a proper subcontinuum, obtained by employing a corollary to the boundary-bumping theorem. Your proof is more-or-less correct, if a bit messy. For example, why to you need to say $X= U \cup V$? And do you claim that $V$ is connected? $\endgroup$
    – Daron
    Dec 12 '15 at 21:35
  • $\begingroup$ I am saying $U$ and $V$ are disjoint open sets partitioning $X\setminus K$. It does not matter if $V$ is connected or not - by the lemma I mentioned $K\cup V$ is connected regardless. $\endgroup$ Dec 12 '15 at 21:37
  • $\begingroup$ The $A$ and $B$ in your edited version still intersect! I want the connected sets to be disjoint. $\endgroup$ Dec 12 '15 at 21:39

Here's my attempt. Please let me know if it looks ok.

Let $X$ be a compact connected space with at least two points. By the boundary bumping theorem there is a proper closed connected $K\subseteq X$ with at least 2 points. We may assume $X\setminus K$ is not connected, say $X\setminus K=U\cup V$. Let $p\in U$ and let $C$ be the component of $p$ in $U$. By the boundary bumping theorem $C$ is nontrivial. For each $q\in U\setminus C$ let $C_q$ be the component of $q$ in $U$. By the boundary bumping theorem each $C_q$ limits to $K$. Now we have $C$ and $K\cup V\cup \bigcup _{q\in U\setminus C} C_q$ are disjoint nontrivial connected sets unioning to $X$.

Note: I used the classic lemma: If $X$ is connected, $K\subseteq X$ is connected, and $X\setminus K=U\cup V$, then $K\cup U$ and $K\cup V$ are connected.

Simplification suggested by @Daron: Let $C$ be one component of $X\setminus K$; $C$ is nontrivial. Then $X\setminus C$ is also connected and nontrivial and as it is equal to $K$ with the potential addition of some components of $X\setminus K$ (these components limit to $K$).


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