Topological basis on $\mathbb R^2$ Let $\mathcal B$ be the collection of all partial opened rectangles $[a,b)\times [c,d)$, where $a \lt b, c \lt d$. 
Show that $\mathcal B$ is a topological basis on $\mathbb R^2$. 
Is the topology generated by $\mathcal B$ finer than the usual topology ?
 A: Hints 
Part 1: Showing $\mathcal{B}$ is a topological basis on $\mathbb{R}^2$.
First, given $(x, y) \in \mathbb{R}^2$, can you construct a partial opened rectangle that contains that point? Giving an explicit example is relatively straightforward.
Now, suppose $U_1, U_2 \in \mathcal{B}$ and $(x', y') \in U_1 \cap U_2$. Is there $V \in \mathcal{B}$ such that $(x', y') \in V$ and $V \subseteq U_1 \cap U_2$? Again, if you draw a picture and give it a little thought, you can construct such a $V$.
Part 2: Showing the topology generated by $\mathcal{B}$ is finer than the usual topology.
Let $\mathcal{T}$ be the usual topology and $\mathcal{T}_\mathcal{B}$ be the topology generated by $\mathcal{B}$. Suppose $(x, y) \in (a, b) \times (c, d) \in \mathcal{T}$. Then $(x, y) \in [x, b) \times [y, d) \subseteq (a, b) \times (c, d)$, and $[x, b) \times [y, d) \in \mathcal{T}_\mathcal{B}$. This means that $\mathcal{T}_\mathcal{B}$ is at least as fine (or just finer depending on your language) as $\mathcal{T}$. Can we go further and say that $\mathcal{T}_\mathcal{B}$ is strictly finer than $\mathcal{T}$? Consider $(x, y) \in [x, b) \times [y, d) \in \mathcal{T}_\mathcal{B}$. Is there any $U \in \mathcal{T}$ such that $(x, y) \in U \subseteq [x, b) \times [y, d)$? If not, why?

Edit: Further hints
Consider $(x, y) \in \mathbb{R}^2$. Then $[x - 1, x + 1) \times [y - 1, y + 1)$ does the trick. Draw a picture to see why.
As for your other other question, we already found that $\mathcal{T}_\mathcal{B}$ is at least as fine as $\mathcal{T}$. However, we cannot say that $\mathcal{T}$ is at least as fine as $\mathcal{T}_\mathcal{B}$, meaning that $\mathcal{T}_\mathcal{B}$ is strictly finer than $\mathcal{T}$. Consider that partial opened rectangle I gave $[x, b) \times [y, d)$ and draw a picture of it, including the point $(x, y)$. Is it possible to fit a completely opened rectangle in this partially opened rectangle that still contains $(x, y)$? Note that if $(x, y) \in (a, b) \times (c, d)$, then $x > a$. But this is a problem because then $(a, b) \times (c, d) \not\subseteq [x, b) \times [y, d)$.
