Munkres topology section 17 question 6

Question

Show that the countable collection {$(a, b) × (c, d) $| $a < b$ and$ c < d$, and $a, b, c, d$ are rational} is a basis for $R^2.$

I was thinking let us take open set $X \times Y $ for which $x\in (a,b) \in X$ and $y \in (c,d) \in Y$...now by definition of the product topology there is a basis element of $ x \times y \in U \times V \subset X \times Y$,,,,,,,

now i can not able to proceed further,,,pliz help me....

Answer 1

In the reals, every open non-empty open interval contains a rational number.

So if $O$ is open in $R^2$ in the product topology and $(x,y) \in O$, pick $U$ and $V$ open in $R$ such that $x \in U, y \in V$ and $U \times V \subseteq O$. (This can be done by the definition of the product topology).

As $U$ is open in $R$ pick an open interval $(l_x, r_x) \subseteq R$ such that $x \in (l_x,r_x) \subseteq U$. (definition of (order) topology on $R$).

Pick rationals $q_x$ in the open interval $(l_x, x)$ and a rational $q'_x \in (x, r_x)$.

So we have $$x \in (q_x, q'_x) \subseteq (l_x, r_x) \subseteq U$$

Do the same for $y$ and $V$ to get rationals $q_y, q'_y$ such that

$$y \in (q_y, q'_y) \subseteq (l_y, r_y) \subseteq V$$

And note that $$(x,y) \in (q_x, q'_x) \times (q_y, q'_y) \subseteq U \times V \subseteq O$$

and as we can find a member of the countable collection between any open set $O$ of the product topology and a member of $O$. This is exactly what it means to be a base for the product topology.

Answer 2

Having a grasp of the following two facts is all that is required to answer the OP's question:

(1) The open intervals of the form $(a,b)$, with both $a$ and $b$ rational, forms a basis for $\mathbb R$.

(2) Let $X$ and $Y$ be two topological spaces each with a chosen basis. The set of all ${\displaystyle U \times V}$, where ${\displaystyle U}$ is an element of the basis of ${\displaystyle X}$ and ${\displaystyle V}$ is an element of the basis of ${\displaystyle Y}$, is a basis for the product topology of ${\displaystyle X \times Y}$.

To be able to subscribe to (2), one must know this set theoretic identity:

$\tag 1 \cup \, U_i \times \cup \, V_j = \bigcup_{(i,j)}\; U_i \times V_j$