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How do I establish a bijection from $[1,0]$ to $$[1,0]\times [1,0]$$ that is continuous?

I have not been able to succeed.

Edit Sorry, it's $[0,1]$ in all cases.

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What is meant by $[1, 0]^2$? –  Amateur Math Guy Feb 3 '13 at 16:38
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Presumably both should be $[0,1]$, and $[0,1]^2=[0,1]\times[0,1]$. –  Clayton Feb 3 '13 at 16:39
    
@MaisamHedyelloo: THat will only get the diagonal, and it won't hit points like $(1,0)$ or $(0,1)$, hence it isn't surjective. –  Clayton Feb 3 '13 at 16:42
    
Thank You all. Never thought it was a trick question. –  user45099 Feb 3 '13 at 17:03

3 Answers 3

up vote 2 down vote accepted

Since $[0,1]$ is compact, such a function $f$ would yield a homeomorphism of $[0,1]$ onto $[0,1]^2$.

Now remove $1/2$ from $[0,1]$. You would get that $[0,1/2)\cup(1/2,1]$ is homeomorphic to $[0,1]^2\setminus\{f(1/2)\}$.

Note that the latter is connected, while $[0,1/2)\cup(1/2,1]$ is not.

Such a function can't exist.

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If by $[1,0]$ you mean $[0,1]$ then there is no such function.

Indeed if there was, then the reciprocal function $g=f^{-1}$ would be continuous since $[0,1]^2$ is compact, so that it would be a homeomorphism. To see that this isn't possible, remark that if you remove an interior point of the square $[0,1]^2$, it remains connected ("in one piece"), whereas if you remove an interior point of the line segment $[0,1]$, it becomes disconnected.

(more formally, the contradiction comes from the fact the continuous image by $g$ of the connected set $[0,1]^2 - \{a\}$ should be connected).

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A continuous bijection from a compact set to an Haussdof set is an homeomorphism.

But, there exists a continuous surjective function $[0,1] \to [0,1]^2$ : Peano's curve.

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