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

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 $[0,2\pi[\to S^1$.

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.

• Where exactly does "if it were somewhere dense it would contain a closed ball" come from? Commented Jun 3, 2011 at 23:15
• @leftaroundabout: if it were somewhere dense its closure would contain a closed ball, and since it's already closed, it contains a closed ball. Commented Jun 3, 2011 at 23:20
• About deleting points, maybe you should emphasize that the homeomorphism $f^{-1}$'s restriction to the closed ball is also a homeomorphism. So the closed ball is homeomorphic to a closed connected set in $[-n,n]$. Then if we delete points.... Commented Dec 15, 2021 at 4:27

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.

• Your proof seems to rely on the argument that if $A$ is not path-connected but $B$ is, there can be no continuous bijection $A\to B$. But consider $A = \mathbb{R}$ with the discrete topology and $B = \mathbb{R}$ with the usual topology. The identity map $A\to B$ is a continuous bijection, $AA$ is not path-connected, and $B$ is path-connected. Commented Jun 27, 2020 at 17:51
• Your argument works to show that there is no continuous bijection $[0,1]^2\to [0,1]$, but not the other way around. Commented Jun 27, 2020 at 17:51

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 $$\mathbb R^2$$.