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Edit: My misunderstanding was in the definition of a space filling curve. $f([0,1])$ need only be the unit square in $\mathbb{R}^2$, not that entire Euclidean plane.

First, I wasn't exactly sure which question I should ask, for the main title, but hopefully this explanation helps.

It appears to be well known that the complement of a knot in $\mathbb{R}^3$ is always path connected. But it seems to me that a space-filling knot could exist, which would make the complement not path connected.

More specifically, my understanding is that a space filling curve is a continuous function $f:[0,1]\to\mathbb{R}^2$ such that $f([0,1])=\mathbb{R}^2$. Similarly, a knot is a continuous function $K:[0,1]\to\mathbb{R}^3$ such that $K(0)=K(1)$ (and the function is injective otherwise). Given a space filling curve $f$, I see no reason why one couldn't slightly modify it to turn it into a closed loop by traveling along the original $f$ on the domain $[0,1/2]$ and then moving from $f(1)$ to $f(0)$ on the domain $(1/2, 1]$ The only difficulty is that going from $f(1)$ to $f(0)$ would require moving in the third dimension to maintain injectivity of the knot, but the construction appears to still work to me. It's still continuous, and the image is still all of $\mathbb{R}^2$, but now it satisfies the definition of a knot.

Is my mistake in that construction above? The only other place I could see myself be mistaken is assuming that $\mathbb{R}^3-\mathbb{R}^2$ is not path connected, but that seems obvious.

Calvin Godfrey
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  • @MichaelMorrow that's not one of the constraints, the knot $K$ just has to satisfy $K(0)=K(1)$ and, in this specific case, $\mathbb{R}^2\subseteq K([0,1])$. – Calvin Godfrey May 29 '21 at 20:14
  • Ah ok. I think I misunderstood. So your knot doesn't have to fill $\mathbb{R}^3$, just $\mathbb{R}^2$? – morrowmh May 29 '21 at 20:17
  • Yes, that is right. – Calvin Godfrey May 29 '21 at 20:18
  • Well then I think compactness is still an issue. If $\mathbb{R}^2\subseteq K([0,1])$ then $K([0,1])$ is not bounded, so it can't be compact. – morrowmh May 29 '21 at 20:19
  • Why is compactness of $K([0,1])$ relevant? – Calvin Godfrey May 29 '21 at 20:20
  • Because the continuous image of a compact set must be compact. So, since $[0,1]$ is compact, whatever $K([0,1])$ is, it needs to be compact. Which, since $\mathbb{R}^3$ is a metric space, means it needs to be bounded, hence cannot fill space. – morrowmh May 29 '21 at 20:21
  • Let us continue this discussion in chat. – Calvin Godfrey May 29 '21 at 20:23
  • I should have also clarified: compact implies closed and bounded if $\mathbb{R^3}$ is given the Euclidean metric (which is what you're working with), not for general metrics. – morrowmh May 29 '21 at 20:30
  • Your construction is self-contradictory. You cannot simultaneously require that "the image is still all of $\mathbb R^2$" and that the knot moves "in the third dimension to maintain injectivity". – Lee Mosher May 29 '21 at 22:09
  • In fact even if you loosen the first requirement to say only that the image is contained in $\mathbb R^2$, you cannot simultaneously require that the knot moves "in the third dimension to maintain injectivity". – Lee Mosher May 29 '21 at 22:39

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The key is in this part of the definition of a knot:

(and the function is injective otherwise)

This cannot be satisfied by a space-filling curve. The modification of the space-filling $f$ which you construct will indeed be periodic, continuous, and its image will contain the unit square, but it won't be a knot because there will be many points in $[0,1/2]$ that get mapped to the same point of $\mathbb{R}^3$. Your "moving in the third dimension" might make it injective on $[1/2,1]$, but it won't do anything about the non-injectivity of the original $f$.

It's a standard, and easy, fact from general topology that an injective continuous map with compact domain will be a homeomorphism onto its image. So if your modified function were a knot, its image would be homeomorphic to a circle, which intuitively cannot be true of any set containing a unit square. For instance, you can show that if you remove any two points from the image of a knot, the resulting set must be disconnected.

See for instance Is it true that a space-filling curve cannot be injective everywhere? and In what way is the Peano curve not one-to-one with $[0,1]^2$?.

Nate Eldredge
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