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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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint:

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.

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  • $\begingroup$ 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
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    $\begingroup$ @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