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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.