I went back to studying Topology after quite a long time without doing so, and have been struggling with some concepts. I’ve been wanting to use the following, and I’m not quite sure if my proof is consistent or is even right.

Proposition: Let $X$ be a connected space. Then any open set $U\subseteq X$ can be decomposed into a union of disjoint connected open components $U_j$, with indices on some set $J$ such that $$U=\bigcup\limits_{j\in J}^{}U_j.$$

My attempt at a Proof:

Let $\mathcal B$ be a basis for the topology on $X$. We can then write an open set $U$ in terms of elements $B_k$ of $\mathcal B$ for some indices in $K$ as

$$U=\bigcup\limits_{k\in K}^{}B_k.$$

We now define $J$ as a partition $\{K_j\}$ of $K$ such that

$$U_j=\bigcup\limits_{k\in K_j}^{} B_k$$ is connected and $U_j\cap U_{j’}=\emptyset$ for $j\neq j’$.

By definition, each of the $U_j$ are open and since $J$ partitions $K$, the theorem holds. $\Box$

What I just can’t seem to prove is whether we can always find such partition $J$ for any connected space. Is there a systematic way to do this? Is there even a simple proof of existence?

  • $\begingroup$ There must be more to the problem than you've said, because this version is too trivial. Any set $U$ in any topological space is a disjoint union of connected sets: $U=\bigcup_{x\in U}\{x\}$. $\endgroup$ – David C. Ullrich May 23 '18 at 14:37
  • $\begingroup$ Oops, right. The $U_j$ have to be open $\endgroup$ – Gabriel Golfetti May 23 '18 at 14:40
  • $\begingroup$ @DavidC.Ullrich I believe there is a requirement that the sets be open. $\endgroup$ – Acccumulation May 23 '18 at 14:48
  • $\begingroup$ Ok, you should fix the question (note the "edit" link...). This would be trivial if $X$ was locally connected - the hypothesis is really that $X$ is connected? This seems to say that any connected space is locally connected, which really doesn't sound right. $\endgroup$ – David C. Ullrich May 23 '18 at 14:49
  • $\begingroup$ Are you given that the space is Hausdorff? $\endgroup$ – Acccumulation May 23 '18 at 15:19

If we were given that $X$ was locally connected this would be clear. In fact the standard example of a connected but not locally connected space, namely the "topologist's sine curve", gives a counterexample here. Let $U$ be a small neighborhood of the origin; then $U$ is not a disjoint union of connected open sets. (Because if it were then one of them would be a small connected neighborhood of the origin...)

  • $\begingroup$ Thanks, this makes sense $\endgroup$ – Gabriel Golfetti May 24 '18 at 10:20

First of all, that partition is not always true. I cannot give a counterexample now, but intuitively, it seems not true. And, $X$ should be a locally connected space. I’ll give you a reference. Munkress, Topology(2nd edition). On page 161, theorem 25.3 states that A space $X$ is locally connected if and only if for every open set $U$, each component of $U$ is open in $X$.


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