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

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

• – Robert Z Jun 26 '17 at 5:26

Take a point out of both $[0,1]$ and $[0,1]\times[0,1]$. Then one is connected, while the other is not.
• 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 5:29
• @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 6:01
• @KajHansen Indeed there are two non-cut points of $[0,1]$. Thanks for pointing it out. :) – awllower Jun 26 '17 at 6:01