Solution: At first we will show that the given collection is a basis. We recall that for a topological space $X$, a collection of subsets of $X$, say $\mathcal{B} $, is called a basis if the following things hold.

• Given any $x \in X$, there exists $B \in \mathcal{B} $ such that $x \in B$.
• If $x \in B_1 \cap B_2$ for $B_1, B_2 \in \mathcal{B} $, then there exists $B_3\in \mathcal{B} $ such that \[ x \in B_3 \subseteq B_1 \cap B_2. \]

We will denote an element of $\mathbb{R} ^2 $ as $x \times y$. If $x \times y\in \mathbb{R} ^2$, then $x,y \in \mathbb{R} $. As $\mathbb{Q} $ is dense in $\mathbb{R} $, we can find $a,b,c,d \in \mathbb{Q} $ such that \[ a < x < b \text{ and } c < y < d \implies x \times y \in (a,b) \times (c,d) \in \mathcal{B} . \] For the second property, if $x \times y \in \left[ (a_1, b_1) \times (c_1, d_1) \right] \cap [(a_2, b_2) \times (c_2, d_2)]$. Choose rational numbers $a,b,c$ and $d$ such that \begin{align*} & \max \{ a_1, a_2 \} < a < x < b < \min \{ b_1, b_2 \} \\ & \max \{ c_1, c_2 \} < c < y < d < \min \{ b_1, b_2 \}. \end{align*}

Now it remains to show that the topology generated by $\mathcal{B} $, say $\mathcal{T}_{\mathcal{B} } $, is the same as the standard topology $\mathcal{T} _E$. It is clear that $\mathcal{T} _\mathcal{B} \subseteq \mathcal{T} _E $. We just need to show that the other inequality. We know that $\mathcal{B} _E = \{ a \times b: a,b\in \mathbb{R} \} $ is a basis for the standard topology on $\mathbb{R} ^2$. We recall that $\mathcal{T} _E \subseteq \mathcal{T} _\mathcal{B} $ is equivalent to show that given any $x \times y \in \mathbb{R} ^2$ and for each $B \in \mathcal{B}_E $ containing $x \times y$, there exists $B' \in \mathcal{B} $ such that $x \times y \in B' \in \mathcal{B} _E$. Look at the \autoref{fig:prob-13-06-24_02}, the choice of $B'$ is similar to previous explanation.