Understanding illustrations of topologies in Munkres The description of the following illustration in Munkres reads: 
One can now see that the collection $\mathscr{B}$ of all circular regions in the plane generates the same topology as the collection $\mathscr{B}'$ of all rectangular regions.

Can someone please elaborate on the meaning of this illustration? I may just be oblivious, but I'm not understanding the relation between the caption and the image. Also, how does this relate to bases?
 A: The figure is referring to the preceding lemma, which states that the topology generated by basis $\mathcal{B}'$ is finer than (larger than) the topology generated by basis $\mathcal{B}$, if and only if for each $x \in X$ and $B \in \mathcal{B}$ containing $x$ there exists $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$.
The image on the right shows that for any circle $B$ containing $x$, you can fit a rectangle $B'$ inside $B$ that contains $x$. The image on the left shows that the same holds with the roles of the bases reversed. Thus each topology is finer than the other, so they are the same.
A: $\mathscr B$ is the base for a topology $T$ and $\mathscr B'$ is the base for a topology $T'.$ 
Each  $B\in \mathscr B$  is equal to $\cup F$  for some  $F\subset \mathscr B',$ so each $B\in T'.$ So $\mathscr B\subset T'.$ 
Each $t\in T$ is equal to $\cup G$ for some   $G\subset \mathscr B,$ and since $\mathscr B\subset T',$ we have  $G\subset T'$ so $t=\cup G\in T'.$ Therefore $T\subset T'.$
Interchanging the primed and un-primed superscripts we also get $T'\subset T$.
One of the pictures is showing that whenever $p\in B\in \mathscr B$ there exists $B'\in \mathscr B$ such that $p\in B'\subset B.$ So for each $p\in B$ we may  take $B'_p\in \mathscr B'$ such that $p\in B'_p\subset B.$ Then $$B=\cup \{\{p\}:p\in B\}\subset \cup \{B'_p: p\in B\} \subset B$$ so $B=\cup F$ where $F=\{B'_p:p\in B\}\in T'.$
