# Are $(0,1) \times [0,1)$ and $[0,1) \times [0,1)$ homeomorphic?

I do not know how to approach. I tried removing points and see if the remaining spaces are connected or not, but i couldn't conclude anything without doubt.

• The discussion in wikipedia might be a direction to consider: "The one-dimensional intervals [ 0 , 1 ] and ( 0 , 1 ) are not homeomorphic because no continuous bijection could be made.[4]" See under "non-examples" en.wikipedia.org/wiki/Homeomorphism. I hope this helps. – ad2004 Dec 26 '19 at 22:13
• Seems like it should be. Imagine a pizza crust on a square tray with two adjacent closed edges. You can twist and stretch that around to move the closed edges to just being on one side. I don't know if it would be pleasant to describe it mathematically, but there's your homeomorphism. – Matthew Daly Dec 26 '19 at 22:14
• They are both homeomorphic to half of the open unit disc, with diameter included. – WhatsUp Dec 26 '19 at 22:14
• Almost the same: math.stackexchange.com/questions/3229483/… – Asaf Karagila Dec 27 '19 at 1:03
• Does this answer your question? Are the spaces $\mathbb{H}^n$ and $\overline{\mathbb{R}}^n_+$ homeomorphic? – Yanior Weg Dec 30 '19 at 18:36

Here's anther recipe for finding an explicit homeomorphism $$\ f:[0,1)\times[0,1)\rightarrow (0,1)\times[0,1)\$$. On the blue region of the first figure below, on and above the line $$\ x+y=1\$$, choose $$\ f\$$ to be the identity. In the region below that line choose $$\ f\$$ to map the green lines (of slope $$1$$) in the first diagram onto the vertical green lines in the second.

• I believe that's $[0,1)\times (0,1)$ on the second picture, but otherwise it looks good. – mechanodroid Dec 27 '19 at 9:26
• Yes, of course you're right. I've now updated the diagram. – lonza leggiera Dec 27 '19 at 12:41

We can try rotating and stretching $$[0,1)^2$$ to $$(-1,1)\times [0,1)$$ in the sense that points on the $$x$$-axis remain in place, points on the $$y = x$$ line rotate to points on the $$y$$-axis, and points on the $$y$$-axis rotate to points on the negative $$x$$-axis. More formally, a point with polar angle $$\phi \in \left[0, \frac\pi2\right]$$ will rotate to a point with polar angle $$2\phi$$.

We also have to scale the distances of points from the origin. Introduce the following continuous function $$L : [0,2\pi] \to [1,\sqrt{2}]$$ as $$L(\phi) := \begin{cases} \frac1{\left|\cos\phi\right|}, &\text{if }\phi \in \left[0, \frac\pi4\right] \cup \left[\frac{3\pi}4,2\pi\right]\\ \frac1{\sin\phi}, &\text{if }\phi \in \left[\frac\pi4, \frac{3\pi}4\right] \end{cases}$$ which measures the width of $$(-1,1)\times [0,1)$$ along the line with polar angle $$\phi$$.

Now check that $$f : [0,1)^2 \to (-1,1)\times [0,1)$$ given in polar coordinates by $$f(0,0) = (0,0)$$ and $$r(\cos\phi, \sin\phi) \mapsto r(\cos2\phi, \sin2\phi)\frac{L(2\phi)}{L(\phi)}\quad \text{ for } r > 0, \phi\in\left[0,\frac\pi2\right]$$ is a homeomorphism with inverse $$(0,0) \mapsto (0,0)$$ $$r(\cos\phi, \sin\phi) \mapsto r\left(\cos\frac12\phi, \sin\frac12\phi\right)\frac{L\left(\frac12\phi\right)}{L(\phi)}\quad \text{ for } r > 0, \phi\in[0,2\pi]$$

Finally, notice that $$g : (-1,1) \times [0,1) \to (0,1) \times [0,1)$$ given by $$g(x,y) = \left(\frac{1+x}2, y\right)$$ is a homeomorphism.

Consider the open unit disk together with an arc of finite length from its bounding unit circle.

I think it's clear that any two such sets are homeomorphic, independent of the length of the arc.

For the sets in question, which are squares, project (stretch) each homeomorphically from its center to the circumscribing disk. The result is two sets as above, one with a fourth and one with a half the circumference.