Topology as an Algebraic Structure

Can we express the axioms of topology equivalently as the axioms of an algebraic structure?

Phrased rigorously, the question is:

Question. Find a universal algebra signature $\Sigma$ and a set of $\Sigma$ -axioms $A$ such that, for any set $S$ , we can associate every topology on $S$ equivalently with every set of operations on $S$ of type $\Sigma$ satisfying $A$ . In other words, find $\Sigma$ and $A$ such that, for every set $S$ , the set of topologies on $S$ is bijective with the set of collections of operations on $S$ satisfying $A$ .

Guess. No such $\Sigma$ and $A$ exist. However, we think it can be possible to have a $(\Sigma, A)$ that varies with the set $S$ . We could even restrict the kind of $S$ we study to allow this to work. Then, we could apply universal algebraic methods to topology! (For such a scenario, we wouldn’t know the signature or axioms before knowing the set, so for a general discussion we’d need universal algebra, not just abstract algebra.) We can call that “universal algebraic topology.”

We investigate these ideas in this exploration.

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Before we begin, a terminology note on “universal algebraic topology”: I did a Google-search of this term, in quotes (which allows searching the phrase as a unit without looking at individual words), on December 8, 2022, to see whether this term already had a standard meaning in the literature. Looking at the few links that came up, there was only one real instance of the phrase “universal algebraic topology” (there were two links, but one pointed to the other.) This phrase was used only once in that linked page, in reference to a cited source that itself only used the term “universal algebraic geometry,” not topology. The subject of universal algebraic geometry is already established, so it seems safe to say that there won’t be any confusion or conflict in using the term “universal algebraic topology.”

Now, let’s try to establish a negative answer to the initial question (of a $(\Sigma, A)$ that does not vary with the set.)

Follow-Up. It would be even nicer if the bijection could be an isomorphism that preserves some kind of structure, say around mapping open sets / continuity to the algebraic structure. Phrased rigorously, can we find a bijection $f:\left\{ \text{topologies on }S \right\} \rightarrow \{ \text{collections of operations on }S\text{ satisfying }A\}$ such that … actually, I’m not sure if there is a “standard” or “natural” operation between different topologies whose structure can be preserved. So maybe just a bijection would be the most “natural” question.

We could reduce this to discussing cardinalities. Given a set $S$ , what is the cardinality of the set of topologies on $S$ ? Given a $(\Sigma, A)$ , what is the cardinality of the set of collections of operations that satisfy $A$ ?

(In what follows, for a set $S$ we will use $P(S)$ to denote the power set of $S$ and $T(S)$ to denote the set of topologies on $S$ .)

First, we note that, if two sets $S$ and $S'$ are in bijection with each other, then the topologies on $S$ and $S'$ are in bijection with each other too. The set of collections of operations on $S$ satisfying $A$ is also in bijection with the corresponding set of collections of operations on $S'$ satisfying $A$ . (The bijection essentially allows “substituting” elements “within the logic.” This is clear to see, but for my own learning I do step-by-step formal proofs in an appendix.)

This means that we can just consider “representative” sets for each cardinality; the actual nature of the set doesn’t matter for us.

(In what follows, given two sets $S$ and $S'$ , let $S \equiv S'$ denote that the sets are in bijection with each other.)

Thus, let $c$ be a cardinal; the notion of $T(c)$ , or the (common) cardinality of the set of topologies on a set of cardinality $c$ , is well-defined. Similarly, the notion of the cardinality of the set of collections of operations satisfying a set of axioms $A$ on a set of cardinality $c$ , is well-defined, which we call $U(c,A)$ . The question reduces to finding (or disproving existence of) $(\Sigma, A)$ such that $T(c) = U(c,A)$ for any cardinality $c$ . Counting $T(c)$ will lead to the resolution of this question.

Bijection tends to get wonky and different for infinite versus finite sets, so we tackle each separately.

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First, let’s look at infinite sets. If $c$ is infinite, then what is $T(c)$ ?

Well, actually, look at the universal algebra angle first. Any collection of operations satisfying a given $A$ must be a collection of operations of different arities, indexed by a common set. So fix a given indexing set $I$ , then the collection must look like $\left\{ o_{i}:i \in I \right\}$ , where $o_{i}$ is an operation of arity, say, $n_{i}$ . (We don’t consider infinite arities for now.) Regardless of the axioms in $A$ , maximally $U(c, A)$ can contain all collections of this form. How many such collections are there?

Well, for an operation of arity $n$ , which is basically a function $S^{n} \rightarrow S$ , and thus a subset of $S^{n + 1}$ , the number of choices is at most the number of subsets of $S^{n + 1}$ , $P\left( c^{n + 1} \right)$ . Since $c^{n + 1} \equiv c$ for infinite $c$ , this is at most $2^{c}$ . Thus, each operation has $2^{c}$ choices, and the operations are indexed by $I$ ; the indexing is basically a function that has domain $I$ and outputs an arity along with an operation of that arity. In other words, this is a function $I \rightarrow \mathbb{Z}_{+} \times 2^{c}$ . We know that $\mathbb{Z}_{+}$ is the smallest infinite cardinal, so $2^{c} > \mathbb{Z}_{+}$ and the range has at most the cardinality of $2^{c}$ . Hence, the number of choices for this collection of operations is at most the number of choices of functions $I \rightarrow 2^{c}$ , or $\left( 2^{c} \right)^{I}$ . Importantly, $I$ is not dependent on $c$ .

On the other hand, $T(c)$ is a subset of $P\left( P(c) \right)$ , or $2^{2^{c}}$ . Thus, $T(c) \leq 2^{2^{c}}$ . If we can conclude for infinite $c$ that in fact $T(c) = 2^{2^{c}}$ , then this will yield a contradiction to the existence of a suitable $(\Sigma, A)$ for infinite sets, as desired. This is because we can always choose $c$ such that $2^{2^{c}} > \left( 2^{c} \right)^{I}$ ; in fact, just pick $c = I$ . (Based on Cardinal number – Wikipedia, we have $\left( 2^{a} \right)^{b} = 2^{ab}$ for infinite cardinals $a,b$ , thus $\left( 2^{c} \right)^{I} = 2^{cI} = 2^{I} < 2^{2^{I}}$ .)

But we have that $T(c)=2^{2^{c}}$ from this Math StackExchange answer (which I turn found through this Quora answer.) Thus, we conclude our desired result: a $(\Sigma, A)$ (that is non-varying) cannot exist for infinite sets, and thus cannot exist for general topological spaces.

(Before finding this answer, I thought to use the information cited in Topological field – Encyclopedia of Mathematics, namely the point starting with “on any field of infinite cardinality there exist X many distinct topologies …” The argument would have depended on showing that there exists a field of any given infinite cardinality, or equivalently, for any set of any infinite cardinality there is a field structure on it. That is an interesting question in its own right, which helped inspire a separate exploration.)

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We include formal proofs of bijective equivalencies here, for my own learning. (Now, $A$ doesn’t stand for the set of axioms as before, but rather just any set.)

Lemma. If $A \equiv B$ , then $P(A) \equiv P(B)$ , with the “natural” bijection.

Proof. Given the bijection $f:A \rightarrow B$ , define the function $f^{'}:P(A) \rightarrow P(B)$ by, for $A^{'} \subseteq A$ , $f^{'}\left( A^{'} \right) = f\left( A^{'} \right)$ . (This is the “natural” construction.) We show that $f^{'}$ is a desired bijection.

First, we show that $f^{'}$ is injective. Let $A^{'},A^{''} \subseteq A$ be such that $f^{'}\left( A^{'} \right) = f^{'}\left( A^{''} \right) \subseteq B$ . By definition of $f^{'}$ , this means $f\left( A^{'} \right) = f\left( A^{''} \right)$ . Let $a^{'} \in A^{'}$ ; then, $f\left( a^{'} \right) \in f\left( A^{'} \right) = f\left( A^{''} \right)$ . Thus, there exists $a^{''} \in A^{'}$ such that $f\left( a^{'} \right) = f\left( a^{''} \right)$ , but since $f$ is bijective we have $a^{'} = a^{''}$ . Hence, $a^{'} \in A^{''}$ . But the same exact logic works to show that $a^{''} \in A^{''} \rightarrow a^{''} \in A^{'}$ . (The situation is symmetric.) Thus, $A^{'} = A^{''}$ , and $f^{'}$ is injective.

Next, we show that $f^{'}$ is surjective. Let $B^{'} \subseteq B$ , and define $A^{'} = \left\{ f^{- 1}(b):b \in B^{'} \right\} \subseteq A$ . We show that $f^{'}\left( A^{'} \right) = B^{'}$ ; this implies that $f^{'}$ is surjective. By definition of $f^{'}$ , this is the same as saying $f\left( A^{'} \right) = B^{'}$ . Let $b^{'} \in B^{'}$ . Then, $f^{- 1}\left( b^{'} \right) \in A^{'}$ by definition of $A^{'}$ , thus $b^{'} = f\left( f^{- 1}\left( b^{'} \right) \right) \in f\left( A^{'} \right)$ . Next, let $b^{'} \in f\left( A^{'} \right)$ , so $b^{'} = f\left( a^{'} \right)$ for some $a^{'} \in A^{'}$ . Then, $f^{- 1}\left( b^{'} \right) = a^{'}$ , and by definition of $A^{'}$ there exists $b \in B^{'}$ such that $a^{'} = f^{- 1}(b)$ . But then $f^{- 1}\left( b^{'} \right) = f^{- 1}(b)$ , so $b^{'} = b$ , and $b^{'} \in B^{'}$ . Thus, $B^{'} = f\left( A^{'} \right)$ , yielding the desired surjectivity.

Together, these prove the claim. $\blacksquare$

Lemma. If $A \equiv B$ , then $T(A) \equiv T(B)$ , with the “natural” bijection.

Proof. First, we see that $P(A)$ is in bijection with $P(B)$ , as above. Thus, let $f:P(A) \rightarrow P(B)$ be a bijection. Actually, we can in the same way yield a bijection $f:P\left( P(A) \right) \rightarrow P\left( P(B) \right)$ . Then, we can just use this bijection for the topologies as well, since $T(A) \subseteq P\left( P(A) \right)$ . Specifically, we can consider the restriction of $f$ to $T(A)$ , $f\left. \right|_{T(A)}:T(A) \rightarrow P\left( P(B) \right)$ . We know this function is injective on $T(A)$ since $f$ on all of $P\left( P(A) \right)$ is. To conclude that $f$ is in fact a bijection from $T(A)$ to $T(B)$ , we just need to show $f\left( T(A) \right) = T(B)$ .

This step is not actually obvious without using information about what defines a topology; it is possible that a particular bijection may not preserve some property of its elements, in this case that property being that it’s still a valid topology. Nevertheless, since in a fundamental sense the topology axioms deal with set-theoretic operations like unions/intersections/etc., and since these set-theoretic operations are preserved by bijections, which intuitively define “equivalence for sets,” our result is reasonable.

Thus, we first show formally that these kinds of set-theoretic operations are in fact preserved by bijections; this will then be useful for topologies.

Lemma. Let $f:A \rightarrow B$ be a bijection, and let $\left\{ U_{i} \right\}_{i \in I}$ be a collection of subsets of $A$ . We have:

$f\left( \cup_{i \in I}U_{i} \right) = \cup_{i \in I}f\left( U_{i} \right)$
$f\left( \cap_{i \in I}U_{i} \right) = \cap_{i \in I}f\left( U_{i} \right)$
The $U_{i}$ are pairwise disjoint if and only if the $f\left( U_{i} \right)$ for $i \in I$ are pairwise disjoint.

Proof. We show each in turn.

First, we show statement 1. Let $x \in f\left( \cup_{i \in I}U_{i} \right)$ , so that $x = f(y)$ where $y \in \cup_{i \in I}U_{i}$ . Thus, $y \in U_{i}$ for some $i \in I$ , hence $x = f(y) \in f\left( U_{i} \right) \subseteq \cup_{i \in I}f\left( U_{i} \right)$ . Conversely, let $x \in \cup_{i \in I}f\left( U_{i} \right)$ , so $x \in f\left( U_{i} \right)$ for some $i \in I$ . Then, $x = f(y)$ where $y \in U_{i}$ . This implies $y \in \cup_{i \in I}U_{i}$ , thus $x = f(y) \in f\left( \cup_{i \in I}U_{i} \right)$ . This shows statement 1.

Statement 2 is similar and exactly analogous to statement 1.

Now, we show statement 3, which we suspect will be important for our topology result. Assume the $U_{i}$ are pairwise disjoint, and consider arbitrary unequal $i,j \in I$ ; we show $f\left( U_{i} \right) \cap f\left( U_{j} \right) = \varnothing$ . Assume $x \in f\left( U_{i} \right) \cap f\left( U_{j} \right)$ , so that $x \in f\left( U_{i} \right) \rightarrow x = f(y)$ where $y \in U_{i}$ , and similarly $x \in f\left( U_{j} \right) \rightarrow x = f(z)$ where $z \in U_{j}$ . We have $f(y) = x = f(z) \rightarrow y = z$ since $f$ is a bijection, so $y \in U_{i}$ and $y \in U_{j}$ . But since the $U_{i}$ are pairwise disjoint, this is impossible. Conversely, assume the $f\left( U_{i} \right)$ are pairwise disjoint, and consider arbitrary unequal $i,j \in I$ ; we show $U_{i} \cap U_{j} = \varnothing$ . Assume $x \in U_{i} \cap U_{j}$ , so that $x \in U_{i} \rightarrow f(x) \in f\left( U_{i} \right)$ and $x \in U_{j} \rightarrow f(x) \in f\left( U_{j} \right)$ . But since the $f\left( U_{i} \right)$ are pairwise disjoint, this is impossible. This concludes the proof of statement 3. $\blacksquare$

Remark: it seems that for statement 3 it is sufficient that $f$ be injective, although I would need to carefully trace the proof to make sure that surjectivity isn’t implicitly used anywhere.

Now, we return to our lemma on topologies. We show $f\left( T(A) \right) = T(B)$ .

First, we show that $f\left( T(A) \right) \subseteq T(B)$ , or that every image of a topology on $A$ is a topology on $B$ . Let $\tau_{A}$ be a topology on $A$ , and consider $f\left( \tau_{A} \right) = \left\{ f(S) \middle| S \in \tau_{A} \right\}$ . We show that $f\left( \tau_{A} \right)$ is a topology on $B$ :

We show that $\varnothing \in f\left( \tau_{A} \right)$ . Since $\varnothing \in \tau_{A}$ , we can conclude this if $f(\varnothing) = \varnothing$ . But this is clear: if $x \in f(\varnothing)$ then there must exist $y$ such that $x = f(y)$ and $y \in \varnothing$ , contradiction.
We show that $B \in f\left( \tau_{A} \right)$ . Since $A \in \tau_{A}$ and $f(A) = B$ , this is immediate.
We show that arbitrary unions are closed in $f\left( \tau_{A} \right)$ . If $\left\{ U_{i} \right\}_{i \in I}$ is a collection of elements of $f\left( \tau_{A} \right)$ , then for each $i \in I$ there must exist $V_{i} \in \tau_{A}$ such that $f\left( V_{i} \right) = U_{i}$ . Then, we have $\cup_{i \in I}U_{i} = \cup_{i \in I}f\left( V_{i} \right) = f( \cup_{i \in I}V_{i})$ . But $\cup_{i \in I}V_{i} \in \tau_{A}$ since $\tau_{A}$ is closed under arbitrary unions, thus $\cup_{i \in I}U_{i} = f\left( \cup_{i \in I}V_{i} \right) \in f\left( \tau_{A} \right)$ , as desired.
We show that finite intersections are closed in $f\left( \tau_{A} \right)$ . If $\left\{ U_{i} \right\}_{i = I}$ is a finite collection of elements of $f\left( \tau_{A} \right)$ , then for each $i \in I$ there must exist $V_{i} \in \tau_{A}$ such that $f\left( V_{i} \right) = U_{i}$ . Then, we have $\cap_{i \in I}U_{i} = \cap_{i \in I}f\left( V_{i} \right) = f( \cap_{i \in I}V_{i})$ . But $\cap_{i \in I}V_{i} \in \tau_{A}$ since $\tau_{A}$ is closed under finite intersections, thus $\cap_{i \in I}U_{i} = f\left( \cap_{i \in I}V_{i} \right) \in f\left( \tau_{A} \right)$ , as desired.

Next, we show that $T(B) \subseteq f\left( T(A) \right)$ . Since $f$ is a bijection this is equivalent to saying $f^{- 1}\left( T(B) \right) \subseteq T(A)$ , but this is just the same result we just proved, but using the bijection $f^{- 1}:B \rightarrow A$ . Thus, this part is complete as well, and we can conclude $f\left( T(A) \right) = T(B)$ . This proves our desired claim. $\blacksquare$

The next lemma would be equivalency for bijective sets of collections of operations satisfying an arbitrary signature. This seems pretty clear to me, but I will revisit this and write a formal proof here for my learning.

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We continue this exploration in a follow-up post.

Started on November 2018, and revisited/revised on September 2020 and December 2022. Changed to more standard notation in January 2023. Revised with newly found information in February 2023.

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