Talk:Union from Synthetic Basis is Topology/Proof 1

"AC is used". I presume one means that we have to choose a particular representation of any set in $\vartheta$? It sounds like it could be that AC is used here, indeed. But I suspect also that $\vartheta$ could be characterised in a more cumbersome way, thus maybe avoiding AC. This is worth some thought. --Lord_Farin (talk) 09:50, 24 September 2012 (UTC)

On this matter, could it be that one could write:


 * $U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

which apparently avoids the use of a choice function? Of course, the proof will have to be changed to use this construction, but I think that the parts $(2)$ and $(3)$ can be shown using this language - a bit cumbersome, but avoiding AC makes it worthwhile. --Lord_Farin (talk) 09:57, 24 September 2012 (UTC)


 * Yeah, that resolves the issue. Thanks. --abcxyz (talk) 20:05, 25 September 2012 (UTC)


 * I have applied above construction; it worked. I am quite sure I didn't use AC in the new proof, but please read it to ensure it is not essentially flawed. I have glossed over some applications of a theorem saying:


 * $\mathbb S \subseteq \mathbb T \implies \bigcup \mathbb S \subseteq \bigcup \mathbb T$


 * which I didn't manage to locate. --Lord_Farin (talk) 12:01, 24 September 2012 (UTC)

Axeman cometh
Bizarre. A carefully constructed proof is axed down to practically nothing. Get a grip, gentlemen! --prime mover (talk) 21:06, 25 September 2012 (UTC)