TITLE: Show that $[0, 1)\times[0, 1)$ is homeomorphic to $[0, 1]\times[0, 1)$ but not to $[0, 1]\times[0, 1]$.
QUESTION [6 upvotes]: Show that $[0, 1)\times[0, 1)$ is homeomorphic to $[0, 1]\times[0, 1)$ but not to $[0, 1]\times[0, 1]$. When I sketch these spaces it this statement makes sense intuitively because $[0, 1]\times[0, 1]$ is closed and $[0, 1)\times[0, 1)$ and $[0, 1]\times[0, 1)$ are each "missing" part of their boundary which would make them closed. I also know, $[0, 1]\times[0, 1]$ is compact and compact spaces are not homeomorphic to non-compact spaces. 
I am having trouble saying this explicitly, as is pretty obvious in the clumsiness of the wording above.

REPLY [7 votes]: Let's make a homeomorphism $[0,1)\times[0,1)\to [0,1)\times[0,1]$:
   (0,1)    (1,1)
   B-------C
   |\     /|
   |  \ /  |
   A---M---D
   |  / \  |
   |/     \|
   O-------E (1,0)

The domain includes the boundaries OAB and OE. The codomain includes the same boundaries, plus the top line BC. In both cases, the point on the boundary that separates "included" from "not included" is itself not included.
Our strategy will be to stretch OAB to cover BC too, so let the homeomorphism map

triangle OMA to OMB
triangle AMB to BMC
triangle BMC to CMD
triangle CME to DME
triangle EMO to itself

Each of these mappings is a simple affine stretch and (in two cases) turn of the triangle in question, and the edges and corners line up correctly, so we do indeed get a homeomorphism.