Given a Topological space $(X,\tau)$ and two disjoint open sets $A,B$, the following holds:
$$\partial(A\cup B) = \partial A\cup \partial B$$
I am interested in a modified version of this.
Construction: Let $\mathcal{R}$ be the collection of rays (one dimensional subspaces) of $\mathbb{R}^2$. We call a subset of $\mathbb{R}^2$ an open stick, if it is bounded, connected, contains $(0,0)$, is contained in some $r\in \mathcal{R}$, and is open in the subspace Topology of this $r$. Here is your typical stick (Desmos doesn't do open dots at the ends for some reason):
Now suppose we have some stick family $(s_r)$ indexed by $\mathcal{R}$. Also, for each stick $s_r$, denote by $\partial's_r$ its boundary in the subspace $r$.
Question: When is it possible to conclude $$ \partial \bigcup_{r\in \mathcal{R}} s_r = \bigcup_{r\in \mathcal{R}}\partial' s_r \hspace{ 0.5cm } ?$$ I conjecture that it is precisely when $ \bigcup_{r\in \mathcal{R}}\partial' s_r \simeq S^1$, the unit circle!
The intuition for this is rather simple, for this identity to hold, we need each stick to contribute only its endpoint to the boundary of the union, and to prevent a stick from contributing a point other than its endpoint, we need to ensure that there is no discontinuity in the length of the sticks, at that particular stick. But I am looking for help to turn this into a rigorous argument.