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 $\hointr 0 1$:
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.
Note how the point $E$ is mapped to the point $E'$.
Then we apply a homeomorphism to the circle $\bigcirc ABE'CD$ which maps:
which in the process maps:
- $E'$ to $E$ which is the same as $C$.
- $C$ to $C$ which is the same point as $D$.
Then we apply a homeomorphism to the circle $\bigcirc ABEC$ back to the square $\Box ABEC$.
It is seen that the line segment $EC$ 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 $\hointr 0 1 \times \hointr 0 1$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 21$