Topological Products of Topological Spaces

Finite Topological Products of Topological Spaces

If $X$ and $Y$ are topological spaces then we can define a special topology on the Cartesian product $X \times Y$ to obtain a new topological space.

Definition: Let $X$ and $Y$ be topological spaces and let $X \times Y$ be the Cartesian product of these sets. The Topological Product of these two spaces is the set $X \times Y$ with the topology $\tau$ with basis $\displaystyle{\mathcal B = \{ U \times V : U \: \mathrm{is \: open \: in \:} X, V \: \mathrm{is \: open \: in \:} Y \}}$. If $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then the topological product is the set $\displaystyle{\prod_{i=1}^{n} X_i = X_1 \times X_2 \times ... \times X_n}$ with the topology $\tau$ with basis $\displaystyle{\mathcal B = \left \{ \prod_{i=1}^{n} U_i : U_i \: \mathrm{is \: open \: in \:} X_i, \: \forall i \in \{ 1, 2, ..., n \} \right \}}$.

In other words, the topological product of two topological spaces is the new topological space $(X \times Y, \tau)$ where $\tau$ is topology with the basis of Cartesian sets $U \times V$ which we call open if $U$ is open in $X$ and $V$ is open in $Y$.

Alternatively, if $X$ and $Y$ are topological spaces then the topology on $X \times Y$ described above will be induced as the initial topology from the projection maps $p_1 : X \times Y \to X$ and $p_2 : X \times Y \to Y$. (Recall that the initial topology induced by a collection of maps is the coarsest topology which makes each of the maps continuous). We prove this on the subsequence Projection Mappings of Finite Topological Products page for the more general finite topological products.

When it comes to two topological spaces $X$ and $Y$, we can look at two products, $X \times Y$ and $Y \times X$. Fortunately these two spaces are homeomorphic as we prove in the following theorem.

Theorem 1: Let $X$ and $Y$ be topological spaces. Then the topological products $X \times Y$ and $Y \times X$ are homeomorphic and an explicit homeomorphism is given by $f : X \times Y \to Y \times X$ defined by $f(x, y) = (y, x)$.

• Proof: To show that $f$ is a homeomorphism we must show that $f$ is a bijective, continuous, and open map.

• It should not be too hard to see that $f$ is bijective. Let $(x, y), (z, w) \in X$ and suppose that $f(x, y) = f(z, w)$. Then $(y, x) = (w, z)$ which implies that $y = w$ and $x = z$, so $(x, y) = (z, w)$ and $f$ is injective. Now let $(z, w) \in Y$. Then $f(w, z) = (z, w)$ which shows that $f$ is surjective. Hence $f$ is bijective.

• We now show that $f$ is continuous. Let $V \times U \subseteq Y \times X$ where $V$ is open in $Y$ and $U$ is open in $X$. Then:

(1)

\begin{align} \quad f^{-1}(V \times U) = U \times V \end{align}

• But $U$ is open in $X$ and $V$ is open in $Y$, so $f^{-1}(V \times U)$ is open in $U \times V$ so $f$ is continuous.

• We lastly show that $f$ is open by showing that $f^{-1}$ is continuous. Let $U \times V \subseteq X \times Y$ where $U$ is open in $X$ and $V$ is open in $Y$. Then:

(2)

\begin{align} \quad f(U \times V) = V \times U \end{align}

• But $V$ is open in $Y$ and $U$ is open in $X$, so $f^{-1}$ is continuous.

• Therefore $f$ is a bijective, continuous, and open map, so $f$ is a homeomorphism between $X \times Y$ and $Y \times X$ so these spaces are homeomorphic. $\blacksquare$