Topology - The arbitrary union axiom

So, the common answer to why we need the concept of topology is that we need it to talk about things like limits of infinite sequences and continuity. But, when we define the axioms of topology, we have an axiom which says that an arbitrary(countable and uncountable) union of open sets exists(or is this not true?) and is open. But doesn't the notion of arbitrary union in itself hide an infinite sequence, namely the the sequence of partial unions of the open set? Isn't the union of infinitely many open sets equal to the "limit" of the partial unions? Is there an alternate interpretation of arbitrary unions which doesn't requires us to define a topology on the power set of the set of interest first?

• The union of a family of sets is the set of all those things that are a member of at least one of the aforementioned sets. No limits needed (and the family of sets does not need to be ordered in any way). – anon Jun 25 '12 at 1:46

No, no sequence or limit process in involved. This is a purely set-theoretic concept. If you have a collection of sets $\mathcal{F}$ in a universe of discourse $\Omega$, you define $$\bigcup \mathcal{F} = \{x\in\Omega| \exists F\in \mathcal{F}\; {\rm with}\; x\in F\}.$$ It's just an existential quantifier. No order or structure is involved.
• Well, this is a kind of limit (you can think of it as the pointwise limit of the indicator function of the sets, so the topology is the topology of pointwise convergence on the set of functions $\Omega \to \{ 0, 1 \}$), but fortunately it's a kind of limit that always exists (in a kind of topology that one canonically always has available). – Qiaochu Yuan Jun 25 '12 at 4:19
Let $X$ be a set and $\alpha$ be an ordinal, a $\alpha$-sequence is a function $f : \alpha \rightarrow X$. There is nothing about topology needed here to define a sequence. You may be confusing this with the notion of a convergence of $\omega$-sequence which do require topology.
Also arbitrary union does not require the use of sequence. Let $\mathcal{F}$ be a collection of subsets of $X$. You can define $\bigcup \mathcal{F} = \{x \in X : (\exists F \in \mathcal{F})(x \in F)\}$.