Talk:Union of Connected Sets with Common Point is Connected/Proof 1

I need help understanding this (and so by extension there may be others in the same situation).

a) Does it not first have to be assumed that there exist disjoint subsets $U, V$ such that $U \cup V = \displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$? Is it not also necessary to point out that if no such disjoint subsets exist, then Connectedness follows? (Yes, the proof is trivial because $A \cap B \ne \O \leadsto A^- \cap B \ne \O$ and so $A$ and $B$ are non-separated and the result follows by Definition 1 of Connected Sets.)


 * What is shown is that given any two open sets $U, V$ such that $U \cup V = \displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ and $U \cap V = \empty$ then either $U = \O$ or $V = \O$, and so the conditions for a separation can't hold. If no such sets existed then non-separation would hold trivially. But there are always two open disjoint sets whose union is the subspace, namely: $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ and $\O$ (admitteldy not very interesting)
 * Maybe a page showing the following would help:
 * $T$ is connected $\forall U, V \in \tau : U \cap V = \O \land U \cup V = S \implies U = \O \lor V = \O$
 * --Leigh.Samphier (talk) 04:37, 7 July 2019 (EDT)

b) It would be useful to state which definitions of which concepts are being invoked, say:


 * From Union is Smallest Superset:
 * $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha = U$


 * That is $V = \O$ (and invoke a specific result as to why)


 * Hence $U \mid V$ does not form a partition of $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$


 * Hence $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ admits no separation.


 * Thus the subspace $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ is not a connected topological space by definition 1.


 * Hence by $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ is not a connected set of $T$ by definition 3.

(Tidied up and all the links set up properly, of course.)

Does it make sense to do this, or is it overly detailed and pedantic? --prime mover (talk) 03:41, 7 July 2019 (EDT)


 * It would help if it was made clear which definitions were being reference --Leigh.Samphier (talk) 04:37, 7 July 2019 (EDT)