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.

  • 5
    $\begingroup$ draw a picture and look at the boundary $\endgroup$ – Mirko Nov 27 '14 at 20:29

Your compactness argument is fine for half of the problem.

Proving that $[0,1)\times[0,1)$ is homeomorphic to $[0,1]\times[0,1)$ takes a bit more work if you’re to do it at all rigorously. One way is to use radial expansion from the centre of each square to map $[0,1]\times[0,1]$ homeomorphically onto the closed disk of radius $\frac12\sqrt2$ centred at $\left\langle\frac12,\frac12\right\rangle$. This map takes $[0,1)\times[0,1)$ onto all of that disk except the closed arc of the boundary lying on or above the diameter of slope $-1$, and it takes $[0,1]\times[0,1)$ onto all of that disk except the closed arc on the boundary lying between that diameter and the diameter of slope $1$ and above the centre of the disk. Then you can do an angular deformation to show that these two disks missing closed segments of their boundaries are homeomorphic. That last step will be more easily done if you first translate the disks to put their centres at the origin.

  • $\begingroup$ I'm wondering if it would be possible to come up with a more general proof that would cover convex sets with the right sorts of boundaries. $\endgroup$ – dfeuer Oct 10 '13 at 5:30

Let's make a homeomorphism $[0,1)\times[0,1)\to [0,1)\times[0,1]$:

   (0,1)    (1,1)
   |\     /|
   |  \ /  |
   |  / \  |
   |/     \|
   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.


As for the odd man out, you've already said it all: only one of the three spaces is compact, and compactness is a topological property, so it's not homeomorphic to the others.

As for showing that the other two are homeomorphic, I'd ask the instructor whether he really expects you to write out the equations, or he would accept something hand-wavy like "it's intuitively obvious that each of those wretched things is homeomorphic to a closed disc minus a closed arc on the circumference", or he has in mind some clever way of proving it rigorously without writing a mess of equations.


Yes, they are. One approach is to show that each is homeomorphic to the set of points whose polar coordinates $\langle r,\theta\rangle$ satisfy $0\le\theta\le\frac{\pi}2$ and $0\le r<1$. You could also use the set of points whose polar coordinates satisfy $0\le\theta\le\pi$ and $0\le r<1$. (Even if you don’t adopt exactly one of these approachs, they should let you see what’s going on.)


You're absolutely right that the proof that $[0,1) \times [0,1)$ is not homeomorphic to $[0,1]^2$ is most easily done by a compactness argument as you described. It's not clear what problem you have with that. To prove that it is homeomorphic to $[0,1]\times[0,1)$ might most easily be done by considering how you might write a homeomorphism using polar coordinates (basically, you want to squeeze $\theta$ and then shake $r$ out to match).


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