Topology as an Algebraic Structure 2

In this post, we continue the exploration started in Topology as an Algebraic Structure.

The previous discussion is enough for a desired negative answer to our first question. But what if we asked for a non-varying $(\Sigma, A)$ that applied just to finite sets? In other words, replace $c$ by a finite number $n$ , and find $(\Sigma, A)$ such that $T(n) = U(n,A)$ .

We can probably use already proven results from the literature here and a similar approach to the infinite case. We’d use a similar exponential lower bound for $T(n)$ , and then given that $U(n,A)$ must basically be at-most polynomial, we can easily see that we can find $n$ that would lead to our desired contradiction.

According to Finite topological space – Wikipedia, the number of topologies on $S_{n}$ is the number of preorders on $S_{n}$ . A preorder on $S_{n}$ is a binary relation and hence a subset of $S_{n}^{2}$ . Look at the definition of preorders: Preorder – Wikipedia. Intuitively, we might think that relations are quite related (ha!) to operations, therefore we might think this equivalence might actually give us a way to express the axioms of a topological space for finite sets equivalently with a non-varying $(\Sigma, A)$ .

Let’s pursue that route. How can we rewrite the axioms of a preorder as an algebraic structure? (Bonus points if we can do an equational class!)

A binary relation on $S$ is a subset of $S^{2}$ , while a binary operation is a subset of $S^{3}$ . But actually, we can try to add a unary operation a la equational-class-axiomatic-rewriting-of-groups and yield our desired algebraic structure.

So how can we turn the relation into an operation? Well, it can just be an indicator operation! Specifically, given a preorder $\leq$ , we define an operation $f:S^{2} \rightarrow S$ that outputs one of two constants in $S$ depending on whether the inputs are in the relation or not. In other words, our $(\Sigma, A)$ can consist of:

Two unary operations, $y$ and $n$
Constancy guarantee for the unary operations (motivationally, this turns them into constants): $y(a) = y(b)$ for all $a,b$ , and $n(a) = n(b)$ for all $a,b$
The “preorder” binary operation $f$ , defined such that $f(a,b) = y(a)$ if $a \leq b$ and $f(a,b) = n(a)$ otherwise
We do need $y \neq n$ with this setup, which unfortunately destroys the candidacy for an equational class … maybe there is a better way to rewrite this.

For now, though, this could be our algebraic structure!

Let’s write this out and see if we can prove our desired equivalency, basically to check our work on this. We’ll introduce the term “algebraic preorder” now, but we’re not trying to conflict with any possible usage of “algebraic preorder” already out there in the literature; this is just a “temporary” term to establish a rewriting of the preorder axioms, after which we won’t use the term anymore.

Definition. An algebraic preorder on a set $S$ is a choice of a unary operation $y$ , a unary operation $n$ , and a binary operation $f$ , satisfying the following axioms:

For all $a,b$ , we have $y(a) = y(b)$
For all $a,b$ , we have $n(a) = n(b)$
There exists $a$ such that $y(a) \neq n(a)$
For all $a$ , we have $f(a,a) = y(a)$
For all $a,b,c$ , we have: if $f(a,b) = y(a)$ and $f(b,c) = y(b)$ , then $f(a,c) = y(a)$

Actually, on second thought, we can get rid of the unary operations and simply think of $a \leq b$ as equivalent to $f(a,b) = a$ . (On further inspection, this is actually somewhat reminiscent of the duality of the algebraic and geometric approaches to lattices.) Thus, we can rewrite this more simply:

Definition. An algebraic preorder on a set $S$ is a binary operation $f$ satisfying the following axioms:

For all $a$ , we have $f(a,a) = a$
For all $a,b,c$ , we have: if $f(a,b) = a$ and $f(b,c) = b$ , then $f(a,c) = a$

Now, we basically want to conclude that “algebraic” preorders (as we’ve defined them) are equivalent to preorders. Then, we can conclude that for finite sets, we can rewrite topology equivalently as an algebraic structure!

So given a preorder $\leq$ , we can define an algebraic preorder by $f(a,b) = a$ if $a \leq b$ , and $f(a,b) = b$ otherwise. Conversely, given an algebraic preorder, define a binary relation $\leq$ as follows: $a \leq b$ if and only if $f(a,b) = a$ . The last two axioms guarantee that $\leq$ is a preorder.

But these constructions aren’t actually necessarily inverses of each other: if we start with an algebraic preorder $f$ , we can construct a preorder $\leq$ , and from that we can construct an algebraic preorder $f^{'}$ , but it is not necessarily the case that $f = f^{'}$ . This can happen if $f(a,b)$ is something other than $a$ or $b$ , if $\neg(a \leq b)$ . In other words, the mapping of algebraic preorders to preorders is many-to-one; multiple algebraic preorders will lead to the construction of one preorder. This isn’t sufficient for equivalency.

We can guarantee equivalency by adding one more axiom:

For all $a$ , we have $f(a,a) = a$
For all $a,b,c$ , we have: if $f(a,b) = a$ and $f(b,c) = b$ , then $f(a,c) = a$
For all $a,b$ , we have: $f(a,b) = a$ or $f(a,b) = b$

Thus, for finite sets, topology can be expressed equivalently as an algebraic structure with a non-varying $(\Sigma, A)$ !

A next interesting question would be whether we can write topology for finite spaces as an equational class. We can try to determine the viability of this using Birkhoff’s theorem.

It remains to investigate this.

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We’ve already seen that a non-varying $(\Sigma, A)$ doesn’t exist for topologies on infinite sets. Regardless, as we’ve already mentioned, we can consider the theory that results from allowing $A$ or even $\Sigma$ to vary; to this end, we can define the symbols $\Sigma(c)$ and $A(c)$ for an infinite cardinal $c$ . We can consider characterizing sufficient values for $\Sigma(c)$ , and if we can even make the corresponding $A(c)$ all identities then many results from universal algebra could be applied (since equational classes have a very rich theory.)

One way we could do this is the following. We know that $U(c,A) \leq 2^{cI}$ (using the definition of $I$ from earlier), regardless of whether or not $I$ depends on $c$ . To make this easier to read, we can say that $U\left( c,A(c) \right) \leq 2^{cI(c)}$ . We can say that we will find $\Sigma(c)$ to make this an equality, and then we just need to find $I(c)$ such that

$\displaystyle 2^{2^{c}} = 2^{cI(c)}.$

This is equivalent to $2^{c} = cI(c)$ , which we can show is equivalent to $I(c) = 2^{c}$ . We use the fact that if $a,b$ are infinite cardinals with $a \leq b$ then $ab = b$ . If $I(c) < 2^{c}$ , then we can have either $I(c) \leq c$ , in which case $cI(c) = c < 2^{c}$ , or $I(c) > c$ , in which case $cI(c) = I(c) < 2^{c}$ ; if $I(c) > 2^{c}$ , then $cI(c) = I(c) > 2^{c}$ . If $I(c) = 2^{c}$ , then $cI(c) = I(c) = 2^{c}$ . Thus, the only solution is $I(c) = 2^{c}$ . But here, it doesn’t matter what the arities of the operations are, nor what axioms they satisfy. It should then be pretty doable to find values of $\Sigma(c)$ that yield equational classes.

So with the discussion so far, any value of $\Sigma(c)$ with an indexing set that has cardinality $2^{c}$ would work (for example, a module over a ring with cardinality $2^{c}$ .) But this seems a little too “loose” to me. A pertinent question here would be whether bijectivity is enough to imply further equivalencies that would be “natural.” For instance, would homomorphisms for our signature be guaranteed to be equivalent to topological homomorphisms, or does this result in extra conditions that constrain what $\Sigma(c)$ we should pick? And actually, is there a more generic way to express the full scope of “natural” equivalencies we’d want, rather than slapping on ad-hoc expectations (like equivalence of homomorphisms)?

To see what equivalencies we’d want, let’s summarize how we’d want to apply universal algebra to topology. For any (infinite) cardinal $c$ and any set $S$ of cardinality $c$ , this discussion shows we can produce a $\left(\Sigma(c), A(c)\right)$ and a bijection $f$ from topologies on $S$ to collections of operations satisfying $A(c)$ on $S$ . Now, say we are given an (infinite) topological space $(T,\tau)$ . From this, we can produce suitable $(\Sigma, A)$ and $f$ , and we can calculate $f(\tau)$ , which would be a particular set of operations on $T$ satisfying $A$ . (Actually, one point to note here is that we have been dealing with existence of a bijection based on cardinality; we haven’t actually discussed how we’d construct one yet in a way such that we can include it in calculations.)

So we started with a topological space $(T,\tau)$ , and now we have $T$ plus a set of specific operations on it. We know that these operations satisfy $A$ , and if we can produce a calculable form of $f$ and not just assert its existence in our theory, then we can calculate $f(\tau)$ and yield exactly what the operations are. Then, we can do algebra on $T$ with these operations. But say we have a logical statement in this algebra on $T$ — basically, say we have a logical statement that consists of a syntactically correct string of symbols from the set of primitive logical symbols and the specific operation symbols of $f(\tau)$ . How can we go back from this statement to saying something about the topology $\tau$ ?

This is the generic form we’d want that would express all the “natural” expectations we have for our theory. Basically, we want to be able to translate statements about the algebra back to statements about the topology. Whatever that yields would be our “natural” condition on $(\Sigma, A)$ and $f$ . We expect that this condition can be formally expressed and studied via mathematical logic.

It remains to figure this out. A similar setup would be pertinent to the finite cardinality case too.

Changed to more standard notation in February 2023.

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We continue this discussion in the next post.

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