TITLE: Homeomorphisms in $\mathbb{R}^2$.
QUESTION [5 upvotes]: I'm getting confused about homeomorphisms.  I believe that $[0,1]\times [0,1)$ and $[0,1)\times [0,1)$ are homeomorphic but $[0,1]\times (0,1)$ and $[0,1)\times (0,1)$ are not.  Please can you try and explain why this is the case?

REPLY [4 votes]: Let
$$
A = [0,1]\times [0,1)\\
B = [0,1)\times [0,1)\\
C = [0,1]\times (0,1) \\
D = [0,1)\times (0,1)
$$
In your question, "explain why this is the case", the meaning of "this" is not very clear. It appears to be that $C$ and $D$ are not homoemorphic. If that's it, then here's a decent reason:
The boundary of $C$ has two connected components, while those of $A$, $B$, and $D$ all have one. 
As for why $A$ and $B$ are homeomorphic: 
The boundary of $A$ is sort of a squared-off letter $U$ (with the tips of the $U$ being open), while that of $B$ is an open interval. It's not too tough to see how you might "retract" the boundary of $U$ to that of $B$, dragging along the rest of the contents of the set in a continuous way. That's informal mumbo-jumbo, but it's also a hint about constructing an actual homeomorphism. One way to do it is to look at a complex variables book, and find an explicit transformation that sends three sides of a square to one side, in the chapter on conformal maps. Another is to actually just try to manufacture the thing with polynomials and inventiveness -- it's not too difficult.