Determine $\langle\bigcup_{\alpha \in \mathbb{I}} B_{\alpha}\rangle$

Question

I am having problems with the next assigned homework:

Let $\{B_{\alpha} \}_{\alpha\in \mathbb{I}}$ a family of abelian subgroups. Determine $\langle\bigcup_{\alpha \in \mathbb{I}} B_{\alpha}\rangle$

I think I can proove that: $$ \left\langle\bigcup_{\alpha = 1}^{n} B_{\alpha}\right\rangle = \sum_{\alpha=1}^{n}(B_i)$$

where $ B_1+ B_2 + ... + B_n = \{c = \sum_{i=1}^{n}b_i:b_i \in B_i\}$

That is the case when $\mathbb{I} = \{1,2,...n\}$. But in general I do not know what is. I would try to $$\sum_{\alpha\in \mathbb{I}}B_i $$ but how we can determine c? in the case when $\mathbb{I} = \mathbb{N}$ for example. I supposed that for a group the sum is closed but only when is finite. So, any idea?

I take it your group is abelian? Otherwise the use of additive notation would be discourage; and if it is the case, you should mention it somewhere. — Arturo Magidin, Apr 06 '19 at 19:03
What is your definition of “subgroup generated by”? And have you established both the top-down and the bottom-up description? — Arturo Magidin, Apr 06 '19 at 19:03
@ArturoMagidin my definition is stated here: https://math.stackexchange.com/questions/3175448/let-s-subseteq-a-a-is-a-group-then-langle-s-rangle-is-the-smallest-su?noredirect=1#comment6537729_3175448 — José Marín, Apr 06 '19 at 23:25
The top-down definition. And have you proven the equivalence to the bottom-up description? — Arturo Magidin, Apr 07 '19 at 02:54
No, I just did the characterization that is the smallest element that contains the set. — José Marín, Apr 07 '19 at 04:10

score 1 · Answer 1 · answered Apr 06 '19 at 17:28

A generally useful strategy for proving two sets $S$ and $T$ are equal is to prove two implications: if $s\in S$ then $s\in T$, and if $t\in T$ then $t\in S$. This lets you work with specific objects $s$ and $t$, which (by virtue of the hypotheses) have certain forms that you can manipulate.

In this particular case, remember also that elements of spans and sums are, by definition, always finite sums of other elements, even if those other elements can potentially be drawn from infinitely many sets.

So, in that case I think I have the proof because it will be similar to $ \left\langle\bigcup_{\alpha = 1}^{n} B_{\alpha}\right\rangle = \sum_{\alpha=1}^{n}(B_i)$ — José Marín, Apr 06 '19 at 17:46

Determine $\langle\bigcup_{\alpha \in \mathbb{I}} B_{\alpha}\rangle$

1 Answers1