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The problem is about proving that $[0,1]^2$ is uncountable. I have proven that $[0,1]$ is uncountable, and so I would conclude that $|[0,1]^2|=|[0,1]|$ by finding a bijection $[0,1]^2\to [0,1]$. I think about letting $f:[0,1]^2 \to [0,1]$ to be defined as $f(x,y)=(x+y)/2$. This is surjective, but I do not know how to prove that it is injective. If $f(x,y)=f(x',y')$ is assumed, I have no idea how to make $x+y=x'+y'$ to imply $x=x'$ and $y=y'$.

PS: If this can not be injective, I'll edit the post to add another approach. Any hints are appreciated.

UnknownW
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1 Answers1

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There is an injective map sending $x\to (x,0)$. In particular the cardinality of the unit square is at least as large as of the unit interval.

Sri-Amirthan Theivendran
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  • I used that bijection $[0,1]\to [0,1]\times {0}$ and using the fact that $[0,1]\times {0}\subseteq [0,1]^2$. Thanks! – UnknownW Oct 04 '17 at 21:35