What is known about links with a countably-infinite number of tame components?

Question

I want to be clear about my phrasing, but I'm not a topologist, so when I say "knot" or "link" I mean the equivalence class under ambient isotopy of an embedding of the circle into $\mathbb{R}^3$.

For a while I have been looking for references to what I've taken to calling the "Fraïssé link" - this thing would have (an instance of) every tame link occurring as a subset, and you would build it the same way that the model-theoretic Fraïssé construction builds the rational numbers from amalgamations of finite linear orders, or the random graph from finite graphs. However, when applying Fraïssé's construction, just ignore the model-theoretic question of what language you're using, or what a structure on a knot is. Just use "is a sublink" to replace the idea of a (logical) embedding, isotoping components where you need to to make the amalgamation property work.

I believe the chain-construction part of Fraïssé's work still succeeds as a plain-old union of sets this way. I also think the construction should commute with passing from a knot-instance to a knot, so realizations of this big snarly thing in $\mathbb{R}^3$ should all be ambient-isotopic to each other, too, but I'm not certain about that. If so, this would be an example of the kind of object I'm asking about in my question: links with a non-finite number of components, but the lowest-level components are ordinary knots.

This does not look like the usual sense of "wild knot" that people talk about, but if I'm wrong please let me know. Searches for "universal homogeneous link" or things like that have not led me to anything that looks similar to this object. Same with casual discussions with some model theorists. It's natural enough that I suspect I'm just missing the right terminology.

Ultimately, I'd like to find the right language to do some model theory with this object - in particular, a language that can express the Reidemeister moves as axioms, that works with the Fraïssé construction. But my specific request is whether knot theorists have noted any results about this particular object, or about any other links that have a countably-infinite number of tame components?

Update: Model Theory Background Let me say a bit more about the model theory background to address Ryan's comment: take an at-most countably-infinite collection of finitely-generated structures (such as all finite graphs, all f.g. torsion-free Abelian groups, or all finite sets that are linearly ordered) up to the appropriate idea of isomorphism for that structure. Given three properties on that collection - "hereditarity", "joint embedding", and "amalgamation" - Fraïssé showed that there is a unique (up to isomorphism) countable structure (the Fraïssé limit) that

1) admits every member of the collection as a substructure and has no other non-isomorphic finitely-generated-substructures (the technical term is the original collection is the age of the Fraïssé limit.)

2) in which any isomorphism between two of its f.g. substructures extends to an automorphism of the entire structure

Hereditarity: the original collection is closed under taking finitely-generated substructures of its members.

Joint Embedding: for any $A, B$ in the collection, there is a common $C$ in the collection into which they both embed.

The property that avoids Ryan's trivial limit is the third:

Amalgamation: given $A_1, A_2$ in the collection, with a common substructure $B$ and embeddings $f_1:B \rightarrow A_1$ and $f_2:B \rightarrow A_2$, there is a $C$ in the collection and embeddings $g_1:A_1 \rightarrow C$, $g_2:A_2 \rightarrow C$ where $g_1 \circ f_1 = g_2 \circ f_2$.

So, for example, if $A_1$ is the Hopf link, and $A_2$ is the unknot linked in some way with a right trefoil, putting them side-by-side would be a joint embedding, but an amalgamation would have to treat one of the unknots in $A_1$ as the same unknot in $A_2$. So Joel's comment is right about what I want - this process builds a structure that has every finite link relating to every other finite link in all possible ways.

"The amalgamation property" is also the model-theorist's response to "why is the limit of finite linear orders $\mathbb{Q}$ rather than $\mathbb{Z}$?"

In the context of model theory, the definition of embedding and substructure depends critically on the choice of logical language for the structure, but the construction of a chain of embeddings using amalgamation leading to the final limit object does not depend on the underlying language. My thought was to drop the logical part (for now), and proceed analogously replacing "finitely generated substructure" with "sublink with finitely many components".

Answer 1

I don't think that knot theorists are going to be very interested in such infinite links, but they do occur sometimes in the wider area of geometric topology, for instance in the proof of theorem 1.1 here. Still I doubt that they have been studied per se.

My understanding is that you are thinking about proper tame links with infinitely many components, i.e. such that every compact subset of $\Bbb R^3$ meets only finitely many components of the link (in general, a continuous map is called proper if the preimage of every compact set is compact). If so, then they reduce to "tame" embeddings of $X=(S^1\times\Bbb N)^+$ in $\Bbb R^3$, where $\Bbb N$ denotes the countable discrete space and $+$ stands for the one-point compactification. To see this, consider the one-point compactification of $\Bbb R^3$, which is $S^3$, and remove some distant point $pt$ (which won't make any difference since in moving the $1$-dimensional space $X$ in $S^3$ one can easily arrange to avoid $pt$).

In general, there is not so much literature about knotting of spaces more general than polyhedra; most of what is known is summarized in the recent book by Daverman and Venema. In this light the space $X$ and its "tame" embeddings don't look very attractive, to be honest.

so realizations of this big snarly thing in $\Bbb R^3$ should all be ambient-isotopic to each other, too, but I'm not certain about that.

Are you saying that if $L$ and $L'$ are proper tame links of countably many circles in $\Bbb R^3$ with the property that every finite link is a sublink of both $L$ and $L'$, then $L$ is ambient isotopic to $L'$?

This is not the case: take $L$ to be the union of links $L_n$, where each $L_n$ is a finite link in a ball $B_n$ of radius $1/3$ centered at $(0,0,n)$, and the ambient isotopy types of the $L_n$ are precisely those of all non-split finite links (each occurring once). On the other hand pick some infinite link $L''$ that is `non-split at infinity'; for instance such that its $i$th component has a nonzero linking number with the $(i+1)$st component for each $i$. As long as $L$ and $L''$ are disjoint, their union $L'$ is a again link that is non-split at infinity; by this I more specifically mean that it is not isotopic to any link whose $i$th component lies in the ball $B_i$ for each $i$. In particular, $L$ is not isotopic to $L'$ (even non-ambiently).

EDIT: One has of course to specify exactly what is the language and the first-order theory, but it seems that with a sufficiently reasonable setup, the model theoretic conditions given in Scott's edit should amount to the following.

Rather than unions of sequences $\mathcal{L}$ of finite tame links $L_1\subset L_2\subset\dots$ up to ambient isotopy, we consider the sequences themselves, up to a weaker equivalence relation of pro-isotopy (defined below).

A "Fraisse link" is then the pro-isotopy class of a sequence $L_1\subset L_2\subset\dots$ satisfying the property in my comment above: whenever some $L_n$ occurs as a sublink of a finite link $L$, then this $L$ must be equivalent to a sublink of some $L_m$ relative to $L_n$ (that is, the identity on $L_n$ must extend to a self-homeomorphism of $\Bbb R^3$ sending $L$ onto a sublink of $L_m$).

Two sequences $\mathcal{L}$ and $\mathcal{L}'$ as above are pro-isotopic if each $L_i$ is equivalent to a sublink of some $L_j'$, $j=j(i)$, via a self-homeomorhpism $h_i$ of $\Bbb R^3$; and each $L_j'$ is equivalent to a sublink of some $L_k$, $k=k(j)$, via a self-homeomorphism $h_j'$ of $\Bbb R^3$, so that for each $i$, the composition $h_{j(i)}'h_i$ sends $L_i$ onto itself; and for each $j$, the composition $h_{k(j)}h_j'$ sends $L_j$ onto itself.

Pro-isotopy is indeed very similar to pro-homotopy, which brings to attention an equivalent definition of pro-isotopy (a la Pontryagin's original definition of pro-isomorphism and Siebenmann's definition of shape): pro-isotopy is the equivalence relation generated by the following two relations: 1) the relation of being a subsequence, 2) sequences $\mathcal{L}$ and $\mathcal{L}'$ are related if there exists a sequence of self-homeomorphisms $H_i$ of $\Bbb R^3$ such that $H_i(L_i)=L_i'$ and $H_{i+1}(L_i)=H_i(L_i)$.

[The proof that the two definitions of pro-isotopy are equivalent is by a standard argument: given $\mathcal{L}$ and $\mathcal{L}'$ that are pro-isotopic in the original sense, the sequence $h_1(L_1)\subset L_{j(1)}'\subset h_{k(j(1))}(L_{k(j(1))})\subset\dots$ has a subsequence that is also a subsequence of $\mathcal{L}'$; and on the other hand is related in the sense of (2) to the sequence $L_1\subset h_{j(1)}'(L_{j(1)}')\subset L_{k(j(1))}\subset\dots$, which in turn has a common subsequence with $\mathcal{L}$.]

The point of these definitions is that a Fraisse link (as defined above) is, obviously, unique.

Now that a Fraisse link is defined, I understand the original question as follows: can a Fraisse link be identified with an (ambient isotopy class of) an infinite link? The answer is no. It is clear that sequences $L_1\subset L_2\subset\dots$ and $L_1'\subset L_2'\subset\dots$ are pro-isotopic if their unions are ambient isotopic. But the converse does not hold, by the reasons that Bill Thurston gave in the end of his answer. I'm not sure that the above definitions of pro-isotopy and a Fraisse link are exactly what he had in mind, but anyhow with these definitions his argument makes sense.

Answer 2

One of the best ways to understand knots and links geometrically is to create orbifolds with the knot or link as singular locus. When they are order 2 singular sets, there's a lot of good theory involving the sphere/torus decomposition (or JSJ decomposition). But if you make them into say order 7 orbifold loci, there are fewer decomposing surfaces, and less to think about. The geometrization theorem for orbifolds says that each of these in the ordinary sense decomposes into pieces modeled on one of the eight kinds of locally homogeneous geometry. I'm not familiar with the Fraïssé theory, but we're used to passing to geometric limits of these objects as they become more and more complicated, and they can be quite interesting even in relatively simple situations (for instance, the limits of $k$-bridge knots or links. A great deal is understood.

One can similarly pass to limits with more and more components, which I guess probably includes the kind of limit you're interested in. One could consider these orbifolds, together with erasing maps that erase some of the components, mapping more complicated ones back to simpler ones; such maps always decrease the Gromov norm, which is the sum of volumes of the hyperbolic pieces. I think these limits of geometric structures are probably much more understandable than a particular embedding in space. The geometric structures could be arranged so the links tend not to crowd together, in this system of geometric structures.

Assuming from Joel David Hamkins' comment that it's not the trivial kind of example Ryan Budney mentioned (with disjoint copies in disjoint balls), the link components would need to have a limit set, and the way the link components converge to the limit set could take many possible different inequivalent forms, without changing the topology and relationships of all the link components. There's probably no one best limit set, so I would be surprised if such an object could be made canonical up to isotopy of $\mathbb R^3$.