Equivalent bases on $I$ Consider $a\in I=[0,1]$, $m\in(b,1]$ with $0<b<1$. Then $\mathcal{B}=\{\beta_{m-b}(a)\cap I\}$ is a base for a topology on $I$ which coarser than the usual topology on $I$ (i.e., the subspace topology on $I$ induced by the real line $\mathbb R$), where $\beta_r(a)$ is the open ball of radius $r > 0$ centered at a point $a\in I$.
Is the converse of this statement correct? That is, the subspace topology on $I$ is coarser than the topology generated by $\mathcal B$.
Clarifying the Question: Fix a set of the form $F=(b,1]$ with $0<b<1$. The open ball of radius $m-b$ for $m\in F$ centered at $a\in I$ is denoted by $\beta_{m-b}(a)$. Then $\beta=\{\beta_{m-b}(a)\cap I: a\in I, m\in F\}$ is a base for a topology on $I$. Main goal is to show the topology generated by $\beta$ is coincides with the usual topology on $I$.
 A: Let $\mathscr T$ be the topology induced by $\mathcal B$, and let $\mathscr T_{\mathrm{std}}$ be the usual topology on $I$. To show that $\mathscr T_{\mathrm{std}}$ is coarser than $\mathscr T$ is by definition to show that $\mathscr T_{\mathrm{std}}\subseteq \mathscr T$.
Let $V$ be an open set in $\mathscr T_{\mathrm{std}}$, which by definition has the form $V=U\cap I$ for an open set $U\subseteq\Bbb R$. If $a\in V$, then since $U$ is open we can find a neighborhood $\beta_r(a)\subseteq U$ for some $r>0$. Now, if $r<1-b$, then we can let $m=r+b$, so that $m\in(b,1]$, and we have 
$$a\in\beta_{m-b}(a)\cap I=\beta_r(a)\cap I\subseteq U\cap I=V.$$
The other possibility is that $r>1-b$, in which case we can take $r'=\frac{1-b}{2}<1-b<r$. Then we can do the same argument instead with $r'$ to get
$$a\in\beta_{m-b}(a)\cap I=\beta_{r'}(a)\cap I\subseteq\beta_{r}(a)\cap I\subseteq U\cap I=V,$$
and we see either way we've found an element $B\in\mathcal B$ with $a\in B\subseteq V$, showing that $V$ is open in $\mathscr T$.
Basically, the point is if you're allowed to take $m\in(b,1]$ then you're getting balls of radius $r\in(0,1-b]$, so while you have an upper bound on the radius you can still make them arbitrarily small, which is all that matters for the topologies to be the same.
