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Can you draw a curve such that it intersects every x at y=0 without ever crossing over itself (that is to say, without hitting the same value twice)?

Wouldn't accomplishing this feat mean infinitely approaching every point on the numberline? Can that be done without a straight horizontal line at y=0?

If so what would stop you from creating a function that does the same, not to just a numberline but a 2 dimensional plane?

If such a line was attempted, isn't there an infinite about of space between all points that a line should never need to cross itself to get out of a spiral it's created?

In order to do this would you need an infinitely complex function? If so is it possible to prove that the function to do this could in principle exist, but we able to prove that since it must be infinitely complex the function can't be known?

In short, is it possible to intersect every infinite point on a plane once and only once with a continuous line?

Am I missing something fundamental about lines and infinities?

Has this kind of question been explored before?

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    $\begingroup$ The answer to the question in your first paragraph, "Can you draw a curve such that it intersects every x at y=0 without ever crossing over itself (that is to say, without hitting the same value twice)?", is just the line along the $x$-axis. For the rest of your question, you may enjoy reading about space-filling curves. $\endgroup$ – Rahul Aug 31 '18 at 5:14
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    $\begingroup$ It is not clear to me that the $x$-axis is the only such curve even if we ask for something as strong as continuity. $\endgroup$ – Fimpellizieri Aug 31 '18 at 6:05
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    $\begingroup$ I can imagine the function $f(x) = sin(x/C)$, where $C$ is a constant. Now I take the limit of $C$ to $+0$. You end up with an "infinitely fast" oscillating function that intersects the x-axis "everywhere". Is that the kind of answer you were looking for ? $\endgroup$ – M. Wind Aug 31 '18 at 6:06
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    $\begingroup$ Do I understand correctly that you are looking for a continuous bijection $\mathbb{R} \to \mathbb{R}^2$? $\endgroup$ – Paul Frost Aug 31 '18 at 7:43
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    $\begingroup$ A continuous space filling curve that is also injective would be a continuous bijection $f:\mathbb R\longrightarrow \mathbb R^2$. This is impossible; see this answer. $\endgroup$ – Fimpellizieri Sep 1 '18 at 6:49

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