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 .


marked as duplicate by Martin R, José Carlos Santos real-analysis May 5 '18 at 9:25

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  • $\begingroup$ Everything you want to know is contained in here. $\endgroup$ – Jan Bohr May 5 '18 at 8:53
  • $\begingroup$ @JanBohr it is not injective $\endgroup$ – Lorenzo Quarisa May 5 '18 at 8:54
  • $\begingroup$ I did not claim that. But the article discusses in detail why it cannot be injective under 'Properties'. $\endgroup$ – Jan Bohr May 5 '18 at 8:54

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 $[0,1]\times[0,1]$.

  • $\begingroup$ Efficient one. However there must be some 1st year analysis way of doing this, I suspect... $\endgroup$ – mol3574710n0fN074710n May 5 '18 at 8:55
  • $\begingroup$ @mol3574710n0fN074710n I suspect otherwise… $\endgroup$ – José Carlos Santos May 5 '18 at 8:56

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