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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?

Community
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  • See https://math.stackexchange.com/questions/510573/is-there-a-continuous-bijection-between-an-interval-and-a-square-0-1-mapsto? – Robert Z Jun 26 '17 at 05:26

1 Answers1

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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.

awllower
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  • 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. – Kaj Hansen Jun 26 '17 at 05:29
  • 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. – 5xum Jun 26 '17 at 05:47
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    @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. – awllower Jun 26 '17 at 06:01
  • @KajHansen Indeed there are two non-cut points of $[0,1]$. Thanks for pointing it out. :) – awllower Jun 26 '17 at 06:01