Does $\bigcup_{i=1}^\infty$ represent the arbitrary union ? Does the union here reach up to $i=\infty$?

Question

I think the answer to this question is yes. Since up to this union it will cover all the natural as well as transfinite numbers. But since the term $arbitrary$ has wide area what about others (like $empty\ union$ ) ? However sometimes we have seen this symbol for countable union, while some authors use this for arbitrary union (in definition of topology).

Please clarify. What I'm missing ?

Answer 1

No, this represents a countable union. In this context you have a set $A_i$ for every $i \in {\mathbb N}$ and you form the union $$\bigcup_{i=0}^\infty A_i = \bigcup \{ A_i \mid i \in {\mathbb N}\} = \{ x \mid \exists i \in {\mathbb N}: x \in A_i\}.$$ In particular, there is no $A_\infty$ included in the union.

For an arbitrary union, you'd have an (arbitrary large) index set $I$ and a set $A_i$ for every $i \in I$ and you form the union $$\bigcup_{i \in I} A_i = \bigcup \{ A_i \mid i \in I\} = \{ x \mid \exists i \in I: x \in A_i\}.$$ Or, as suggested in the comments by Mark S., you'd have a set ${\cal A}$ whose members are sets and you form the union $$\bigcup{\cal A} = \{x \mid \exists A \in {\cal A} : x \in A \}.$$

If you take a union up to and including some (possibly transfinite) ordinal $\beta$, you'd probably be understood if you wrote $\bigcup_{\alpha=0}^\beta A_\alpha$, but I think it is more common to write $\bigcup_{\alpha \leq \beta} A_\alpha$ in that case, especially because you can then also write $\bigcup_{\alpha < \beta} A_\alpha$ to signify that $A_\beta$ is not included in the union.

If you want to take the union over all ordinals, you could write $\bigcup_{\alpha \in \text{Ord}} A_\alpha$, ignoring for a moment that this gets you in the domain of proper classes instead of sets.

Answer 2

If you want the union to reach an infinite index then you need a number system that includes infinite numbers, such as the hyperreals. Over the hyperreals you can indeed have infinite indices $H$ and hyperfinite unions of the type $\bigcup_{i=1}^{H}$.

In other words, what you are missing is that infinite unions "up to a cardinal" don't make sense (at least not without further effort).