General Topology: Topologies
The Box Topology
Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces. We can construct a new topological space out of the cartesian product of these two spaces. The box topology on $X \times Y$ is defined by the basis $\mathcal{T}_X \times \mathcal{T}_Y$, and the resulting topological space is called the box space of $X \times Y$.
Box topologies can be constructed from more than two topological spaces. Given a collection of topological spaces $\{(X_\alpha, \mathcal{T}_{X_\alpha})\}_{\alpha \in I}$, the box topology on $\prod\limits_{\alpha \in I} X_{\alpha}$ is defined by the basis
$$\prod\limits_{\alpha \in I} \mathcal{T}_{X_\alpha}$$
and the resulting topological space is again called the box space.
When the cartesian product is taken of a finite number of topologies, the resulting topology is called the product topology, and the resulting space is called the product space. However, the term "product topology" technically refers to a more particular definition than the one given here for the box topology. The topologies that result from that definition are identical to the topologies that result from the box topology, but only when the cartesian product is taken over a finite number of topologies. When an infinite product is taken, the product topology and the box topology differ in many important ways. The term "box topology" is typically used in the context of infinite products of topological spaces to distinguish it from the more particular product topology. For this section, we'll use the term "product topology" unless discussing the infinite case where the box topology is meaningfully different.
Problems

Let $X = \{1, 2, 3\}$, $\mathcal{T}_{X} = \{\varnothing, \{1\}, \{2\}, \{1, 2\}, X \}$, $Y = \{a, b, c, d\}$, and $T_\mathcal{Y} = \{\varnothing, \{a\}, \{a, b\}, \{a, b, c\}, Y \}$. Write out the basis for the product topology on $X \times Y$.
 $\varnothing$
 $\{(1, a)\}$
 $\{(1, a), (1, b)\}$
 $\{(1, a), (1, b), (1, c)\}$
 $\{(1, a), (1, b), (1, c), (1, d)\}$
 $\{(2, a)\}$
 $\{(2, a), (2, b)\}$
 $\{(2, a), (2, b), (2, c)\}$
 $\{(2, a), (2, b), (2, c), (2, d)\}$
 $\{(1, a), (2, a)\}$
 $\{(1, a), (1, b), (2, a), (2, b)\}$
 $\{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}$
 $\{(1, a), (1, b), (1, c), (1, d), (2, a), (2, b), (2, c), (2, d)\}$
 $\{(1, a), (2, a), (3, a)\}$
 $\{(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)\}$
 $\{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c)\}$
 $\{(1, a), (1, b), (1, c), (1, d), (2, a), (2, b), (2, c), (2, d), (3, a), (3, b), (3, c), (3, d)\}$

Let $X$ and $Y$ be topological spaces, and let $A \subseteq X$ and let $B \subseteq Y$. Show that the product topology on $A \times B$ is the same as the subspace topology $A \times B$ inherits from $X \times Y$.
We show that the basis for the subspace topology on the product is the same as the basis for the product topology of the subspaces.
Let $C$ be a basis element in the product topology on $A \times B$. Then $C = (A \cap T_x, B \cap T_y)$, where $T_x$ is open in $X$ and $T_Y$ is open in $Y$. Note that $(A \cap T_x, B \cap T_y) = (A, B) \cap (T_x, T_y)$, which is an element of the subspace topology on $A \times B$ inherited from $X \times Y$. Because the bases for the two topologies are equal, the topologies themselves are therefore equal.