Return to Answer

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, consider the space of smooth maps $f : N \to X$, $Map(N,X)$ has as an open and dense subspace maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, the space of smooth maps $f : N \to X$ ( $Map(N,X)$) has as an open and dense subspace maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

edit, continued from the comments above: The notion of co-dimension used by Vassiliev is very much in the spirit of the above comments. When you have a map $D^n \to X$ such that its restriction to $S^{n-1}$ can be perturbed to be disjoint from $A$, yet no small perturbation of the original map can be made to be disjoint from $A$, that's when you are in the co-dimension $n$ setting. A map $D^n \to X$ can't be a neighbourhood of a point in its image since $X$ isn't finite-dimensional, but you can consider such maps to be a "slice" of a neighbourhood of a point in the image. By that I mean you have a map $f : V \to X$ whose image is open in $X$, and you decompose $V$ into an $n$-dimensional subspace and a complementary subspace. The the image of the $n$-dimensional subspace under $f$ will be your "resolutions" of your knot-with-multiple (double/triple, etc) points. The complementary subspace will consist of perturbations of the knot complementary to the resolution bump functions.

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, consider the space of smooth maps $f : N \to X$, $Map(N,X)$ has as an open and dense subspace the maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, consider the space of smooth maps $f : N \to X$, $Map(N,X)$ has as an open and dense subspace maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$ and smooth maps $f : N \to X$, there exists a neighbourhood $N_f$ of $f$ in the mapping space $Map(N,X)$ such that the set of $g \in N_f$ such that $g(N) \cap A = \emptyset$ is open and dense in $N_f$. The co-dimension of $A$ in $X$ is $1$ larger than the sup of all $n$ such that the above is true.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".

If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, consider the space of smooth maps $f : N \to X$, $Map(N,X)$ has as an open and dense subspace the maps which are disjoint from $A$.

Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".