# Prove that $[0,1) \times [0,1)$ is homeomorphic to $[0,1] \times [0,1)$

I need to prove that:

$$[0,1) \times [0,1)$$ is homeomorphic to $$[0,1] \times [0,1)$$

I am not confident in writing homeomorphisms, but I picture the situation as:

I noticed that the each side of the half open (on the left) and U-shaped (on the right) squares is homeomorphic to another, then the interior of these squares are the same so one can use the identity map in there.

However, I could not manage to write a specific homeomorphism $$f$$ for the mentioned sides.

• Here's a strategy. Show that both of them are homeomorphic to the union of the open disc and only part of the circle. i.e. $\{z:|z|< 1\}\cup\{e^{i\theta}:0\leq \theta \leq \pi\}$ And then use transitive property of homeomorphisms to say that both spaces are homeomorphic Dec 1, 2023 at 14:53
• Can you write down the function on the boundary? It's allowed to define it piece-wise. Then, extend the formula. Dec 1, 2023 at 14:54
• I think the prevalent opinion among topologists I know is that having the right "deformation" picture in your mind is really sufficient understanding of this problem and writing down an explicit map is an unproductive exercise. If you really want to, I think one nice way to start is to write down an explicit homeomorphism between $[0, 1]^2$ and a closed disc, and then look what happens to these two spaces under that map (elaborating a little bit on Mr.Gandalf Sauron's suggestion). Dec 1, 2023 at 16:46
• I agree with "writing down an explicit map is an unproductive exercise" , @IzaakvanDongen , that was one reason why I did not write it out in my intuitive explanation. Visualizing the map is enough.
– Prem
Dec 1, 2023 at 17:21
• @Prem, indeed. I like your answer and I upvoted it! OP, I just spotted your sentence "the interior of these squares are the same so one can use the identity map in there". This isn't right! There is no way to extend the identity map on $(0, 1)^2$ to a homeomorphism of your spaces. The only map it could possibly extend to is the identity map, by density of $(0, 1)^2$ in $[0, 1]^2$, but that's not even a well-defined map. So inevitably such a homeomorphism will have to "shuffle about" the interior a bit. Dec 1, 2023 at 17:31

My approach would be to prove two lemmas:

• Every pair of triangles in the plane are homeomorphic (use a linear transformation).

• If two spaces may be covered by finitely-many homeomorphic closed subspaces, such that the homeomorphisms coincide where they overlap, then the union of these homeomorphisms is itself a homeomorphism.

Then I'd draw a picture like this:

Then we have a homeomorphism that sends the blue path to the blue path; restrict to the rest of the square to obtain the desired map.

• That's a very nice answer. For the second ingredient, don't we need the subspaces to be closed? Dec 1, 2023 at 20:32
• I think so; I mentioned overlapping boundaries but I should have clarified they were defined on those boundaries. Dec 1, 2023 at 21:47

Here is my Intuitive way to see the "mapping" , where I am giving the high-level operations which are required.

You should be able to make the necessary functional relationship with that. Let me know if you encounter issue with that.

Convert every Blue Solid line $$y=x+a$$ ( $$-1.0 ) to half-way to axis along the line : We will get the Blue Dotted line.
We have got the triangle.

Convert every Grey line $$y=x+a$$ ($$-0.5) to half-way to diagonal. The other $$a$$ value do not change.
We have got the shaded area.

Convert every Diagonal Purple line $$y=-x+a$$ ($$-0.5) to half-way towards the Diagonal $$y=x$$ to get the shaded area.

Rotate by $$45^\circ$$ to get this :

That can easily map to the required area.

Combine all the invertibile maps to get the required homeomorphism.
Do let me know if you encounter Issues with that.

Over-all view :
We are shrinking various lines either towards axis or towards middle. We will not shrink certain lines which are already where we want them.
We will then rotate to align with required area.
That gives easy way to map two rectangles with the top line missing.
Putting all these maps together , we get the single source-to-destination invertiable map.