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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?

Daniel Ward
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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.

John Hughes
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    However, in order to make the first part (that $C$ and $D$ are not homeomorphic) precise, one has to give an internal characterization (inside the spaces $A$, $B$, etc., not relative to an overspace) of what you call "boundary". Proposal: "boundary" $={x\in X\mid X\setminus{x}\text{ is simply connected}}$. – user 59363 Mar 27 '15 at 14:39
  • Good point (and nice characterization)! – John Hughes Mar 27 '15 at 14:59