Are there any interesting instances of "super-semigroups", where we can use infinite products like:

  • $b = (aaa\dots)$
  • $c = (ababab\dots)$
  • $d = (abcaabbccaaaabbbbccccaaaaaaaabbbbbbbbcccccccc\dots)$

and have these arbitrary infinite products make sense?

I've seen the concept of an utralimit, but are there any structures that don't involve a metric space or even a topology?

  • $\begingroup$ did you hear of formal power series rings? they might be such an object. also they are really nicely powerful! $\endgroup$ – Enkidu Dec 10 '18 at 10:15
  • $\begingroup$ I think arbitrary infinite sums are defined in an ordinal algebra but I'm not sure. I'm even less sure about arbitrary infinite products. $\endgroup$ – bof Dec 10 '18 at 10:27
  • $\begingroup$ Formally it would be complicated to make sense of arbitrary infinite products. If you say products indexed by sets of cardinality less than $\kappa$ there's no issue, but arbitrary products you would run into (at least) formal problems, if not foundational problems $\endgroup$ – Max Dec 10 '18 at 11:05

Yes, plenty. For example, a suplattice is like a monoid but where the monoid operation can have arbitrarily large arity. A simple example is the powerset $P(X)$ of subsets of a set $X$, with the operation given by union. This is the free suplattice on $X$, and one can take arbitrarily large unions.

More generally, a cocomplete category has in particular all coproducts, and coproduct again is a monoid operation which can have arbitrarily large arity. A simple example here is the category $\text{Set}$ of set, where the coproduct is disjoint union. Note that by Freyd's theorem every cocomplete small category is a poset, and hence a suplattice. So most examples are large.

Note also that monoid operations that allow infinite arities and which are also associative in the appropriate sense are subject to the Eilenberg swindle, which refers to the following: if we consider the product $a^{\infty}$ of infinitely many copies of $a$, then

$$a^{\infty} \cdot a = a^{\infty}.$$

It follows that if the operation is cancellative, then it collapses: if we could cancel $a^{\infty}$ from both sides, then $a$ must be equal to the identity $1$.

There are also profinite groups, which allow some (but not all) infinite multiplications, and which unlike the examples I've given so far aren't commutative. You can think of these in terms of topologies but you also don't have to.


One of the simplest objects supporting such operations is a topological semigroup, i.e. a semigroup $S$ equipped with a topology $(S,\tau)$ for which the multiplication map $m\colon S\times S\to S$ is continuous. Then for a sequence $(s_n)_{n\in\mathbb N}$ of elements of $S$ the infinite product is defined whenever the set $$ \bigl\{\prod_{i=1}^ns_i\bigr\}_{n\in\mathbb N} $$ has a unique limit point.


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