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It is not hard to see that there exist infinitely differentiable curves in the plane that are injective and dense.*

There are also real-analytic curves that are dense in the plane.**

Question: Does there exist a real-analytic curve in the plane that is both injective and dense?

I would most prefer a real-analytic curve whose derivative never vanishes.

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* For example, see an answer to the MSE question https://math.stackexchange.com/questions/321646/is-there-an-injective-c1-curve-dense-in-the-plane .

** See preprint of "Dense analytic curves generated by iteration of complex periodic functions", Maksim Vaskouski, Nikolai Prochorov, Nikolay Sheshko, 2018. (This says it was discovered by H. Bohr and R. Courant in 1914 that for fixed x in the interval (1/2, 1], the image {zeta(x + it)} by the Riemann zeta function of the affine line {x + it}, t ∈ ℝ, is dense in ℂ.)

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    $\begingroup$ Yes. A standard theorem (you can find it in most textbooks on approximation theory) is that for every positive continuous function $\delta$ on $\mathbb R$ and any continuous $f$ there, one can find a real analytic (even entire) $g$ such that $|g-f|\le\delta$ on the line. This doesn't yet give you what you want if used just as a black box, but if you look at the proof, you'll be able to modify it pretty easily to answer your question. To get 1-1, it helps to approximate the derivative together with $f$ itself. If you still have trouble with it, let me know and I'll post more details :-) $\endgroup$ Commented Jan 3, 2025 at 2:00
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    $\begingroup$ This is certainly so if the curve is parametrized by a non-compact interval. But if the curve is parametrized by $[a,b]$ the answer is negative. $\endgroup$ Commented Jan 3, 2025 at 15:38
  • $\begingroup$ @fedja Can you say a bit more about how controlling/approximating the derivative allows you to prevent self-intersections between distant input points? $\endgroup$ Commented Jan 3, 2025 at 22:10
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    $\begingroup$ @KevinCasto Distant points are controlled by the small error of the approximation to a continuous curve you draw adding piece by piece already. The derivative is needed only to avoid self-intersections in a very near vicinity (if you have a non-self intersecting continuous $f$, its uniform approximation $g$ certainly won't glue faraway parameters, but can glue very close ones as much as it wants unless you control something like the derivative as well). $\endgroup$ Commented Jan 3, 2025 at 23:23

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Let $\wp$ be the Weierstrass p-function associated to the lattice ${\bf Z}+ i{\bf Z}$ and $\theta$ an irrational number. The analytic curve $$t \mapsto \wp(e^{t+i2\pi \theta}) \in {\bf C}$$ should do the trick. The Weierstrass function sends the torus ${\bf C}/({\bf Z}+i{\bf Z})$ to ${\bf C}\cup \{\infty\}$. This is a 2-sheeted ramified covering, two points having the same image if and only if they are opposite (up to the lattice) if I remember well. We parametrize an half-line with irrational slope on the torus. It is dense on the torus, so its projection is also dense and it is injective because we took only half of the line.

Edit: Jacobi elliptic functions are implemented in several CAS softwares, e.g. maxima. So if we take the Jacobi sn function instead of the Weierstrass function, we can actually plot the curve. Here is the graph with the function $sn^2$ which is even, and parameter $1/\sqrt{2}$ for a square fundamental domain.

/* maxima code */
f(x) := jacobi_sn(x*exp(%i*sqrt(2)), 1/sqrt(2))^2;
plot2d([parametric, realpart(f(t)), imagpart(f(t)), [t,0,200]], [x,-1,2], [y,-1.5,1.5], [yx_ratio, 1]);

enter image description here

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    $\begingroup$ Right, this is also a horizontal trajectory of the push-forward via $\mathcal{P}$ of a constant multiple of the quadratic differential $dz^2$. $\endgroup$ Commented Jan 5, 2025 at 0:33
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Here is one way to prove this. Let $X$ be a Riemann surface of finite type, i.e. one obtained from a compact Riemann surface by removing finitely many points. Consider the space $Q(X)$ consisting of (holomorphic) quadratic differentials $\omega$ on $X$, which have finite norm. Each (horizontal) trajectory of $\omega$ either contains one of the zeroes of $\omega$ (and, hence, is not smooth) or is closed or is an injectively immersed biinfinite real-analytic curve in $X$. It is known that (unless $Q(X)=0$) there is a dense subset in $Q(X)$ consisting of quadratic differentials $\omega$ which have smooth trajectories each of which is dense in $X$. See for instance Theorem 25.1 in

Strebel, Kurt, Quadratic differentials, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge, Band 5. Berlin etc.: Springer-Verlag. XII, 184 p. (1984). ZBL0547.30001.

Now, take $X$ which is $S^2$ with at least four punctures (to ensure that $Q(X)\ne 0$), one of which you declare to be the point at infinity. Then a dense trajectory of a quadratic differential on $X$ will be dense in $\mathbb C$. It is also real-analytic and injectively immersed.

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