I found this marvellous thing while playing around in Desmos!

clearly a square!

It seems to me that the curve $\vec{r}(t) = \left[ \cos ( t^2 ), \sin t \right] \left( t \in [0, +\infty) \right)$ is filling a square. But, as @BarryCipra kindly pointed out, it crosses the X axis only when $t$ is an integer multiple of $\pi$, so it crosses the X axis countably many times. And we know that $[-1, 1]$ is uncountable, so it seems the curve doesn't actually fill the square. A question that would be fair to ask: Does the curve get arbitrarly close to every point in the square? Or: Is $[-1, 1]^2$ the infinite closure of the curve?

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    $\begingroup$ It's not literally filling the square. For example, the curve $(x(t),y(t))=(\cos(t^2),\sin(t))$ crosses the $x$-axis only when $t$ is an integer multiple of $\pi$, so there are only countably many such points. What the curve does seem to do is get arbitrarily close to every point in the unit square. $\endgroup$ Mar 10 '21 at 20:34
  • $\begingroup$ @BarryCipra Damn, that's kind of a bummer. Thanks for the clarification. Please post this comment as an answer, I'd like to accept it and close the question. $\endgroup$ Mar 10 '21 at 21:09
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    $\begingroup$ Since there is as yet no answer, it would be fine with me if you want to edit your question to ask if the curve does indeed get arbitrarily close to every point (i.e., the infinite curve's closure is the unit square). That strikes me as worth asking, in part because I'm not sure how to go about proving it. (If you decide to do this, you might want to temporarily delete the current question while you edit, to prevent an answer to the original question slipping in.) $\endgroup$ Mar 10 '21 at 21:14
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    $\begingroup$ It would be sufficient to show that $\left\{\left(\{\frac t{2\pi}\},\{\frac{t^2}{2\pi}\}\right):t\in\Bbb R\right\}$ is dense in $[0,1]^2$, where $\{\cdot\}$ is the fractional part. $\endgroup$
    – Karl
    Mar 10 '21 at 22:19

Intuitively, as $t$ gets larger and larger, we see that $\cos(t^2)$ oscillates faster and faster relative to $\sin(t)$, so we can choose very large $t$ that gives the right $y$-value and then just perturb it a bit to get the right $x$-value while only changing $y$ by a very small amount.

Let's make this more precise. Let $(x_0, y_0) \in [-1, 1]^2$ and $\varepsilon > 0$ be arbitrary. Let $t_0 \geq \frac{\pi}{\varepsilon}$ such that $\sin(t_0) = y_0$. We have $\cos(t_0^2 + c) = x_0$ for some $c \in [0, 2\pi]$. Let $t = \sqrt{t_0^2 + c}$. Then $\cos(t^2) = x_0$. Moreover, $$\lvert \sin(t) - y_0 \rvert = \lvert \sin(t) - \sin(t_0) \rvert \leq \lvert t - t_0 \rvert = \sqrt{t_0^2 + c} - t_0$$ $$ \mathrel{\leq} \sqrt{t_0^2 + 2\pi} - t_0 = \frac{2\pi}{\sqrt{t_0^2 + 2\pi} + t_0} < \frac{\pi}{t_0} \leq \varepsilon.$$ Thus, the distance between $(\cos(t^2), \sin(t))$ and $(x_0, y_0)$ is less than $\varepsilon$.

In particular, the curve gets within distance $\varepsilon$ of every point in $[-1, 1]^2$ as $t$ ranges over the interval $[0, \frac{\pi}{\varepsilon} + 2\pi]$. (I'm not claiming this bound is necessarily optimal, to be clear, but I'd guess it's at least pretty close.)


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