Topologyy induction by metric If $d=d_1+d_2$ then, show that $\it T_d$ is finer than both $\it T_{d_1}$, and $\it T_{d_2}$. Where $d_1$, and $d_2$ is a metric space, $\it T_d$ is topology induced on d, $\it T_{d_1}$ is topology induced on $d_1$, and $\it T_{d_2}$ is topology induced on $d_2$.
I don't know what to do with the question asked. Please provide some hints. Efforts will be appreciated, thanks.
 A: If $d$ and $d’$ are two metrics on a set $X$ that induce the topologies $T$ and $T'$, then $T'$ is finer than T if for any x in X and any $\epsilon > 0$ , there is some $\delta > 0$ such that the $\delta-neighborhood$ of $x$ in the $d'$ metric is contained in the $\epsilon-neighborhood$ of the $d$ metric.
In other words, we only need to check that for any two points $x$ and $y$ in $X$, $d(x,y)\geq
d_1(x,y)$ and $d(x,y)\geq d_2(x,y)$.
A: Let $d$, $d_1$, and $d_2$ be metric spaces on $X$ inducing topologies $\it T_d$, $\it T_{d_1}$, and $\it T_{d_2}$.
I want to show that $\it T_d$ is finer than both $\it T_{d_1}$, and $\it T_{d_2}$.
$\it T_d$ is finer than $\it T_{d_1}$ if and only if for all $x\epsilon X$ and $\epsilon>0$, there is a $\delta>0$ such that
\begin{equation*}
B_d(x;\delta)\subset B_{d_1}(x;\epsilon).
\end{equation*}
Suppose that $\it T_{d_1}\subset \it T_d$. Let $x\epsilon X$ and $\epsilon>0$,
 $B_{d_1}(x;\epsilon)$ is open in $\it T_{d_1}$ so, it is open in $\it T_d$. Since the open $d$-ball form a basis for $\it T_d$, there is an open ball $B_d(x;\delta)$ such that
\begin{equation*}
x\epsilon B_d(x;\delta)\subset B_{d_1}(x;\epsilon).
\end{equation*}
Conversely,
Suppose that for all $x\epsilon X$ and $\epsilon >0$ there is a $\delta>0$ such that $B_d(x;\delta)\subset B_{d_1}(x;\epsilon)$. I will show that $\it T_{d_1}\subset \it T_d$.
Let $O$ be an open set in $\it T_{d_1}$. I will show that it is open in $\it T_d$. Let $x\epsilon O$ since, open $d_1$-ball form a basis for $\it T_{d_1}$ there exists an $\epsilon>0$ such that 
\begin{equation*} x\epsilon B_{d_1}(x;\epsilon)\subset O \end{equation*}
By assumption, there is a $\delta>0$ such that 
\begin{equation*}
x\epsilon B_d(x;\delta)\subset B_{d_1}(x;\epsilon)
\end{equation*}
Therefore, 
\begin{equation*} x\epsilon B_d(x;\delta)\subset O.\end{equation*}
Now $B_d(x;\delta)$ is a $\it T_d$-open set containing $x$ and is contained in $O$. Since $x\epsilon O$ was arbitrary therefore, $\it T_{d_1}\subset \it T_d$.
Similarly, it can be shown that $\it T_{d_2}\subset \it T_d$. Which shows that $\it T_d$ is finer than both $\it T_{d_1}$, and $\it T_{d_2}$.
Did I come up with the correct solution?
