Definition talk:Set Union/Finite Union

"We need a proof that $S_1 \cup S_2 \cup \cdots \cup S_n$ equals the indexed union." Seriously? Off you go then.

"Also, finite unions are not always expressed in terms of 1-indexed sequences. They may be 0-indexed sequences, or unions of finite sets., etc. Source review might be a good idea." Oh stop trolling! It's just notation. Given that you have a finite set of sets, you can index them with $1$ to $n$, or alternatively you can index them with $0$ to $n-1$ or you can index them however you like. But the point is: given that you have indexed them from $1$ to $n$, the notation $\displaystyle \bigcup_{1 \mathop = 1} n$ means the same thing as $S_1 \cup S_1 \cup \cdots \cup S_n$.

So what's your solution? --prime mover (talk) 05:21, 10 May 2013 (UTC)


 * Not sure exactly. But $S_1 \cup S_2 \cup \cdots \cup S_n$ is a bunch of binary unions strung together, while $\bigcup \{S_1, S_2, \dots, S_n \} = \bigcup_{k=1}^n S_k$ is the union of the image of a finite sequence. They must be proven equivalent. The other point I was trying to make is that they might not even be indexed at all, in some usages. If $S$ is a finite set, $\bigcup S$ is a finite union, no? --Dfeuer (talk) 05:35, 10 May 2013 (UTC)


 * The latter is covered in Definition:Set Union/General Definition. Srsly, I can not see your problem. --prime mover (talk) 05:38, 10 May 2013 (UTC)


 * The way it is currently is two definitions mashed together. That is a problem all by itself, which should be fairly easy to rectify. The second problem is that although the general definition does in fact cover the case I'm talking about, it doesn't expressly indicate that such a thing is considered a finite union. --Dfeuer (talk) 19:44, 10 May 2013 (UTC)