Product of Closed and Half-Open Unit Intervals is Homeomorphic to Product of Half-Open Unit Intervals

Theorem
Let $\closedint 0 1$ denote the closed unit interval $\set {x \in \R: 0 \le x \le 1}$.

Let $\hointr 0 1$ denote the half-open unit interval $\set {x \in \R: 0 \le x < 1}$.

Let both $\closedint 0 1$ and $\hointr 0 1$ have the Euclidean topology.

Then the product space:
 * $\closedint 0 1 \times \hointr 0 1$

is homeomorphic to:
 * $\hointr 0 1 \times \hointr 0 1$

Proof
First we take the square $\Box ABCD$ embedded in the Cartesian plane such that $AD$ corresponds to $\closedint 0 1$ and $AB$ corresponds to $\hointr 0 1$:


 * Closed-0-1-by-halfopen-0-1.png

This corresponds to the set $\closedint 0 1 \times \hointr 0 1$.

It is noted that the line segment $BC$ which corresponds to $\closedint 0 1 \times \set 1$ is not in the set $\closedint 0 1 \times \hointr 0 1$.

Then we apply a homeomorphism which maps the perimeter of $\closedint 0 1 \times \hointr 0 1$ to the circle whose center is $\tuple {\dfrac 1 2, \dfrac 1 2}$ and whose radius is $\dfrac {\sqrt 2} 2$.

The points $A$, $B$, $C$ and $D$ are fixed by this homeomorphism.


 * Closed-0-1-by-halfopen-0-1--2.png

Note how the point $E$ is mapped to the point $E'$.

Then we apply a homeomorphism to the circle $\bigcirc ABE'CD$ which maps:
 * the arc $BADC$ to the arc $BAD$
 * the arc $BE'C$ to the arc $BEC$

which in the process maps:
 * $E'$ to $E''$ which is the same as $C$.
 * $C$ to $C''$ which is the same point as $D$.


 * Closed-0-1-by-halfopen-0-1--3.png

Then we apply a homeomorphism to the circle $\bigcirc ABEC$ back to the square $\Box ABEC$.

It is seen that the line segment $E''C$ is now identified with the $\set 1 \times \hointr 0 1$.

Hence $\closedint 0 1 \times \hointr 0 1$ has been transformed via $3$ homeomorphisms to $\hointr 0 1 \times \hointr 0 1$.