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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.

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.

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: the "Fraisse link" is an equivalence class of sequences $\mathcal{L}$ of finite tame links $L_1\subset L_2\subset\dots$ (and not just of their union!) 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 $L_m$).

The equivalence relation is as follows. I will call it pro-isotopy. 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 every $n$ there exists an $i>n$ such that the composition $h_{j(i)}'h_i$ is the identity on $L_n$; and a $j>n$ such that the composition $h_{k(j)}h_j'$ is the identity on $L_n$.

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 $H_{i+1}|_{L_i}=H_i$. 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$\mathcal{L}$.

Now 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; the question becomes, does the converse hold? The answer is no, by the reasons that Bill Thurston gave in the end of his answer. (I'm not sure that my definition of pro-isotopy is exactly what he had in mind, but anyhow his argument applies.)

However, pro-isotopy of sequences obviously implies non-ambient isotopy of their unions. But here, of course, the converse implication doesn't hold.

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.

EDIT: One has of course to specify exactly what is the language and the first-order theory, but it looks like with sufficiently reasonable choices, the model theoretic conditions given in Scott's edit should amount to the following: the "Fraisse link" is an equivalence class of a sequence $L$ of finite tame links $L_1\subset L_2\subset\dots$ (and not just of their union!) 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 $L_m$).

The equivalence relation is as follows. I will call it pro-isotopy. Two sequences $L$ and $L'$ 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 every $n$ there exists an $i>n$ such that the composition $h_{j(i)}'h_i$ is the identity on $L_n$; and a $j>n$ such that the composition $h_{k(j)}h_j'$ is the identity on $L_n$.

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 $L$ and $L'$ 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 $H_{i+1}|_{L_i}=H_i$. The proof that the two definitions of pro-isotopy are equivalent is by a standard argument: given $L$ and $L'$ 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 $L'$; 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$L$.

Now 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; the question becomes, does the converse hold? The answer is no, by the reasons that Bill Thurston gave in the end of his answer. (I don't know if my definition of pro-isotopy is what he had in mind, but anyhow his argument applies.)

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: the "Fraisse link" is an equivalence class of sequences $\mathcal{L}$ of finite tame links $L_1\subset L_2\subset\dots$ (and not just of their union!) 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 $L_m$).

The equivalence relation is as follows. I will call it pro-isotopy. 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 every $n$ there exists an $i>n$ such that the composition $h_{j(i)}'h_i$ is the identity on $L_n$; and a $j>n$ such that the composition $h_{k(j)}h_j'$ is the identity on $L_n$.

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 $H_{i+1}|_{L_i}=H_i$. 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$\mathcal{L}$.

Now 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; the question becomes, does the converse hold? The answer is no, by the reasons that Bill Thurston gave in the end of his answer. (I'm not sure that my definition of pro-isotopy is exactly what he had in mind, but anyhow his argument applies.)