Is it true that a space-filling curve cannot be injective everywhere?

Question

Here, it is said that a space-filling curve cannot be injective because "that will make the curve a homeomorphism from the unit interval onto the unit square", since every continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

That does mean there can be no injective mapping from $[0,1]$ onto $[0,1]\!\times\![0,1]$, but it doesn't really tell us anything about mappings from noncompact intervals onto the unit square. In the case of the ordinary Peano-Hilbert curve, it does not help much: the construction has infinitely ^{(only countably, though)} many inherent points of overlap (discussed here: In what way is the Peano curve not one-to-one with $[0,1]^2$? ), but I can't really see any reason why there could not be a space-filling curve without such points. Of course, the curve's inverse could not be continuous, but when mapping from a noncompact interval it wouldn't need to be! After all, there are other curves mapping noncompact spaces bijectively to compact ones in a continuous way, but with noncontinuous inverse, like the obvious $[0,2\pi[\to S^1$.

Answer 1

There are no continuous bijections from $\mathbb{R}$ to $\mathbb{R}^2$, or to $[0,1]^2$.

Suppose $f$ is a continuous injection from $\mathbb{R}$ into $\mathbb{R}^2$. Then for each $n \in \mathbb{N}$, $f|_{[-n,n]}$ is a continuous injection from a compact space to a Hausdorff space, and hence a homeomorphism onto its image. Thus the image $f([-n,n])$ is nowhere dense: it's compact, hence closed, hence if it were somewhere dense it would contain a closed ball. But then there would be infinitely many points that could be deleted from it without disconnecting it, contradicting it being homeomorphic to $[-n,n]$.

Thus the image of $f$ is a countable union of nowhere-dense sets. So by the Baire category theorem it can't be all of $\mathbb{R}^2$, or indeed any set with nonempty interior.

Answer 2

A proof of "No continuous bijection for $[0, 1]$ and $[0, 1]^2$":

Let $f$ be a continuous bijection from $[0, 1]$ to $[0, 1]^2$, $x \in [0, 1]$

$A = [0, 1] - x$, $B = [0, 1]^2 - f(x)$

A is not path-connected but B is, which leads to a contradiction.

Answer 3

One can actually extend the above Baire Category theorem proof by user Chris Eagle with the invariance of domain theorem to prove that there is no continuous injection $f$ from any closed "1-dimensional set" $S \subsetneq \mathbb R^2$ (namely, an $S$ with empty interior --- for example $S:=$ the $x$-axis, or $[0,1]$ in the $x$-axis, all horizontal and vertical gridlines for the lattice $\mathbb Z^2$, etc.) to any set $V\subseteq \mathbb R^2 $ with non-empty interior. Slogan: continuous injections can only map "1-dimensional sets" to "1-dimensional sets".

(This also proves that any "1-dimensional", i.e. empty interior, set $S\subseteq \mathbb R^2$ contained in a closed "1-dimensional" set can not be mapped via continuous injection to a subset of $\mathbb R^2$ with non-empty interior.)

Proof of 1st paragraph claim, by contrapositive: as in Chris Eagle's proof, we see that $S_n:=\overline{B(0,n)} \cap S$ is a compact set, so $f$ restricted to it becomes a homeomorphism from $S_n$ to $f(S_n)$, which is compact and hence closed, and form an increasing union to $V:= f(S)$. If $S$ has nonempty interior, the Baire Category Theorem forces some $f(S_n)$ to have nonempty interior $U_n$. But $f^{-1}|_{f(S_n)}: f(S_n) \to S_n$ is a homeomorphism, and remains a homeomorphism (to its image) after restricting further to $U_n \subseteq f(S_n)$. But the invariance of domain theorem says that its image $f^{-1}(U_n) \subseteq S_n$ must be open in $\mathbb R^2$, so $S_n$ must have nonempty interior.

---

P.S. if we do not ask that $S$ be closed, it is possible to have a continuous injection to a "large" set: Does there exist a bijective, continuous map from the irrationals onto the reals?. Using the continuous bijection $b$ from the irrationals $\mathbb I \to \mathbb R$, we get $(x,y)\mapsto (b(x),b(y)): \mathbb I \times \mathbb I \to \mathbb R^2$ is a continuous bijection from $\mathbb I^2$ (which has empty interior) to $\mathbb R^2$.