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Setting: Let $F':[0,1] \rightarrow [0,1]$ be the Cantor Function.

Goal: Show that there exists a well-defined, surjective, continuous function from $[0,1]$ to $[0,1]^2$ (i.e., a space-filling curve).

Attempt:

EDIT: It turns out my function is non-sensical in that the domain isn't even $[0,1]$! I'm leaving it here to show an attempt was made at answering the question, but I've yet to come up with a suitable candidate function from $[0,1]$ to $[0,1]^2$.

  1. Let $G: [0,1] \rightarrow [0,1] \times [0,1]$ s.t.

    $$ G(x,y) = (F'(x),F'(y)) $$

  2. I claim that $G$ is a well-defined, surjective, continuous map.

  3. First $G$ is clearly well-defined (there are no issues with representatives or equivalence classes).

  4. To see that $G$ is surjective, let $(a,b) \in [0,1]^2$.

  5. Consider $F'$ is surjective (I've already shown this on my own).

  6. Hence we have that $\exists x,y \in [0,1]$ s.t. $F'(x) = a$ and $F'(y) = b$.

  7. Then consider that $G(x,y) = (F'(x), F'(y)) = (a,b)$ so that $G$ is surjective as desired.

Question: Why is $G$ continuous? I've shown elsewhere that $C \cong [0,1]$, and I've been given the hint that $C \cong C \times C$ would be a useful fact to prove. Why would this fact be useful?

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$G(x,y)=(F'(x),F'(y))$ defines a map $[0,1]\times[0,1]\to[0,1]\times[0,1]$, not $[0,1]\to[0,1]\times[0,1]$. –  Christoph Feb 24 at 13:24
    
Did you really show $\mathcal C\cong [0,1]$? Because one of them is totally disconnected! –  Christoph Feb 24 at 13:51
    
I showed $C \cong [0,1]$ in the cardinality sense. Perhaps this was the wrong notation to use. –  user1770201 Feb 24 at 14:48
1  
When talking about topological spaces, we usually reserve the symbol $\cong$ to mean homeomorphic rather than the weaker relation of them being equinumerous which is what it appears you're using the symbol for. It is the case that $\mathcal{C}\times\mathcal{C}$ is homeomorphic to $\mathcal{C}$. That is, $\mathcal{C} \times \mathcal{C} \cong \mathcal{C}$. –  Daniel Rust Feb 24 at 15:02

1 Answer 1

The Cantor function $F':[0,1]\to[0,1]$ gives you a surjective continous map $F'|_{\mathcal C}:\mathcal C\to[0,1]$ from the Cantor set to the unit interval. Thus you can construct a surjective continuous map $$ \mathcal C \overset\cong\longrightarrow \mathcal C\times\mathcal C \overset{F'|_{\mathcal C}\times F'|_{\mathcal C}}\longrightarrow [0,1]\times [0,1] $$ and extend that to $[0,1]$. The extension can be done by linear interpolation explicitly or by applying the Tietze extension theorem.

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