Are indexing sets always well-ordered? Since if we have an operation on a collection indexed by $\Gamma$, such as a sum $\sum_{\gamma\in\Gamma}a_\gamma$ or Tychonoff product $\prod\{X_\gamma:\gamma\in\Gamma\}$, where the order of the operands is thought of as important, is an implicit ordering assumed?
My guess is that $\sum_{\gamma\in\Gamma}a_\gamma$ is not a well-defined object for infinite $\Gamma$, since even if a well-ordered indexing set is assumed, the value in the case of a conditionally convergent sequence is unclear, and for the product $\prod\{X_\gamma:\gamma\in\Gamma\}$, I suppose order is not important per se, since we can access the "$\gamma$th coordinate" using projections $\pi_\gamma\colon\prod\{X_\gamma:\gamma\in\Gamma\}\to X_\gamma$.
But is there no sense of "the first entry" with infinite products, only the "$\gamma$th" one? (Assuming the indexes are not themselves ordinals.)
Some remarks: the Cartesian product $\prod_{\gamma \in \Gamma} X_\gamma$ needs no order on the index set; it's just the set of functions $f: \Gamma \to \bigcup_{\gamma \in \Gamma} X_\gamma$ such that $f(\gamma) \in X_\gamma$ for every $\gamma$. It's not a "sequence". The $\gamma$'th coordinate is just the value of the function at $\gamma$. Projections are point evaluations.
$\sum_{\gamma \in \Gamma}a_\gamma$ is possible to define for any set $A$ in which all $a_\gamma$ lie and on which we have a commutative group operation $+$ and also a topology: let $G$ be the set of all finite subsets of $\Gamma$ with partial order $\subseteq$; this is a standard directed poset. Define a net $s: G \to A$ by $s(G)= \sum_{\gamma \in G} a_\gamma$ (which is a well-defined finite sum, for an Abelian group) and the limit in $A$ of this net, if it exists, (see a good book on general topology) is denoted $\sum_{\gamma \in \Gamma} a_\gamma$. So also no order is needed. Theorem, if $A= \mathbb{R}$, standard $+$ and topology, $\sum_{\gamma \in \Gamma}a_\gamma$ exists iff $\{\gamma \in \Gamma: a_\gamma \neq 0\}$ is at most countable and the countable sum for this converges absolutely in any order we enumerate it. So it's quite a strong notion of infinite sum, in that light.