Prove that open sets in a connected space are the union of connected sets

Question

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?

Answer 1

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...)

Answer 2

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 $X$.