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As explained in this answer, it is possible to create a bijection from $[0,1]\rightarrow[0,1]^2$. However, the example provided is clearly not continuous. It seems either very complicated or impossible to create a continuous bijection between the unit interval and the unit square. Does this mapping exist? If so, what does it look like? If not, how does one prove that there does not exist a continuous bijective mapping between $[0,1]\rightarrow[0,1]^2$?

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  • $\begingroup$ you can google space filling curve $\endgroup$ – Ben Nov 26 '16 at 17:43
  • $\begingroup$ See the Peano curve. $\endgroup$ – user378947 Nov 26 '16 at 17:43
  • $\begingroup$ The OP asks for a bijection, which the Peano curve is not (it is not injective). $\endgroup$ – Dániel G. Nov 26 '16 at 17:45
  • $\begingroup$ Why isn't the Peano curve injective? $\endgroup$ – Romain S Nov 26 '16 at 17:58
  • $\begingroup$ @relep I read that but posted the link anyway because a lot of books that explain the Peano curve also explain that we can't ask it to be a bijection. I would have posted your answer instead but I just got up. +1 btw! :) $\endgroup$ – user378947 Nov 26 '16 at 17:59
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No such bijection exists: it is a standard result that any continuous bijection from a compact space (such as $[0,1]$) onto a Hausdorff space is a homeomorphism, and there's no homeomorphism from the unit interval to the unit square, since removing an interior point will make the first one disconnected, while leaving the latter connected.

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(1)... [0,1] is not homeomorphic to $[0,1]^2.$

Proof:

(1). Every $t\in [0,1]$ has a local base $B_t$ (of intervals) such that $ \partial b$ has at most $2$ members for each $b\in B_t.$

(2). Let $p=(x,y)\in [0,1]^2$ and let $B_p$ be a local base at $p.$ For any $r>0$ let $$B(p,r)=[0,1]^2\cap \{(u,v):(u-x)^2+(v-y)^2<r^2\}$$ Choose $b_r\in B_p$ such that $b_r\subset B(p,r).$

$$\text {For } \theta \in [0,2\pi) \text { let } V(r,\theta)=\sup \{z\geq 0: 0\leq z'\leq z\implies p+z'(\cos \theta,\sin \theta) \in b_r\}.$$ Observe that $b_r\supset B(p,r')$ for some $r'\in (0,r]$.

It follows that for all sufficiently small positive $r$, the set $$\{\theta: V(r,\theta)>0\}$$ is infinite. And that for all sufficiently small $r$, the set $$\{(p+V(r,\theta)(\cos \theta,\sin \theta):V(r,\theta)>0\}\}$$ is an infinite subset of $\partial b_r.$

Remark. There are continuous surjections from $[0,1]$ to $[0,1]^2$ (see "space-filling curves" or "Peano curve") but they are not injective.

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Another way to see that this is not possible is to try to draw a continuous curve starting from origin that covers all the area of square without overlapping the previous path. The best you could do is to travel around the perimeter and once you approach origin after a round, no matter how close you go to origin and start the same process, there will be a few points between starting point of second round and the origin.

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    $\begingroup$ This isn't even close to a formal argument. Why would such a path have to start at the corner? Even if it did, why would such a curve have to travel around the perimeter? $\endgroup$ – Wojowu Nov 26 '16 at 19:18
  • $\begingroup$ I know it is not formal. I just took origin as a starting point. One could start from anywhere, and I tried to cover maximum length before you get to a point, where if a turn is taken, then a few points are lost. I know it's not a formal answer with rigour and theorems but it was just a geometrical intuition. $\endgroup$ – jnyan Nov 26 '16 at 19:24

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