How to show

There is no continuous bijection between $[0,1]$ and $[0,1] \times [0,1]$ ?

My Try:

I think, between $[0,1]$ and $[0,1] \times [0,1]$, continuous onto function exist. But the one to one continuous map does not exist.

Is my guess correct?


marked as duplicate by Martin R, Chris Culter, Community Jun 26 '17 at 5:44

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Take a point out of both $[0,1]$ and $[0,1]\times[0,1]$. Then one is connected, while the other is not.

Hope this helps.

  • $\begingroup$ Certainly this is the way to go; however, there are two points you can take out of $[0,1]$ to preserve connectedness. If we're picking points at random though, it's true that $[0,1]^2$ is guaranteed connectedness whereas $[0,1]$ is not. $\endgroup$ – Kaj Hansen Jun 26 '17 at 5:29
  • $\begingroup$ This actually only proves there is no continuous bijection with a continuous inverse between the two sets (i.e., it shows the two sets are not homeomorphic. $\endgroup$ – 5xum Jun 26 '17 at 5:47
  • 1
    $\begingroup$ @5xum Yes, we need the compactness of $[0,1]$ and the Hausdorffness of $[0,1]\times[0,1]$ to show that every continuous bijection is a homeomorphism. $\endgroup$ – awllower Jun 26 '17 at 6:01
  • $\begingroup$ @KajHansen Indeed there are two non-cut points of $[0,1]$. Thanks for pointing it out. :) $\endgroup$ – awllower Jun 26 '17 at 6:01