I wanted to prove that there exist no continuous one to one and onto function form $[0,1] \to [0,1]\times[0,1]$.
My attempt : image of f on $[0,1]$ , a compact set is again compact .$[0,1]\times[0,1]$ is also compact.
On contrary suppose there exist continuous one one onto function between $[0,1]\to [0,1]\times[0,1]$ then its inverse function is also continuous one one onto.
Upto this I can write from given information
No idea how to proceed further .Any Help will be appreciated .
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José Carlos Santos
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Curious student
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Everything you want to know is contained in here. – Jan Bohr May 05 '18 at 08:53
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@JanBohr it is not injective – Lorenzo Q May 05 '18 at 08:54
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I did not claim that. But the article discusses in detail why it cannot be injective under 'Properties'. – Jan Bohr May 05 '18 at 08:54
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There is no such function because, since the domain is compact, it would be a homeomorphism. But $[0,1]$ and $[0,1]\times[0,1]$ are not homeomorphic: if you remove $\frac12$ from $[0,1]$, it becomes disconnected. But there is no such point in $[0,1]\times[0,1]$.
José Carlos Santos
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Efficient one. However there must be some 1st year analysis way of doing this, I suspect... – mol3574710n0fN074710n May 05 '18 at 08:55
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