I have to show the following:
Let $X$ be a set with metrics $d_1$ and $d_2$ inducing metric topologies $\tau_1$ and $\tau_2$.
Define a new metric on $X$ where $d(x,y) = d_1(x,y) + d_2(x,y)$ for all $x,y$ in $X$.
a) Show that the topology $\tau_d$ induced by $d$ is finer than $\tau_1$ and finer than $\tau_2$.
b) Show that if $\tau_1 = \tau_2$, then $\tau_d = \tau_1$.
Part a:
Since the set of all metric open balls is a basis for the metric topology, I'll show that $\tau_d$ is finer than $\tau_1$ by showing that any $d_1$ metric open ball contains a $d$ metric open ball.
Let $B_x^{d_1}(\delta)$ be a metric open ball of radius $\delta$ centered at an arbitrary point $x$ in $X$ for metric $d_1$.
Let $\epsilon$ = $\delta$ + 0 = $\delta$. Thus, the metric open ball of radius $\epsilon$ centered at x for metric $d = d_1 + d_2$ is equal to the $d_1$ metric open ball, ie $B_x^{d1+d2}(\epsilon)$ = $B_x^{d_1}(\delta)$. Thus, $B_x^{d1+d2}(\epsilon) \subseteq B_x^{d_1}(\delta)$. Thus, the topology induced by metric $d$ is finer than the topology induced by metric $d_1$.
A similar argument shows that the topology induced by metric $d$ is finer than the topology induced by metric $d_2$.
I'm stuck on part b. I know that to show $\tau_1$ = $\tau_d$ we just need to show that $\tau_1$ is finer than $\tau_d$, since in part a we showed that $\tau_d$ is finer than $\tau_1$, but I'm not sure how to do that.
At first I wanted to say that $\tau_1 = \tau_2$ means that metric $d=d_1+d_2 = 2d_1$ has multiple kd(x,y) bounded above by $d_1(x,y)$ for all $x,y$ in $X$ when k $\le$ $\frac 12$, but then I realized that $\tau_1 = \tau_2$ only means that they have the same open sets and does not mean that $d1$ and $d2$ are the same metric. So, I'm not sure that there's an argument by means of bounding a constant multiple of $d(x,y)$ by $d_1(x,y)$.
I'm not sure how to approach an argument that every $d$ metric open ball must contain a $d_1$ metric open ball. It's seems kind of obvious that if you extend a $d_1$ metric open ball of radius $\delta_1$ by a non-negative distance $\delta_2$ given by metric $d_2$, a $d_1$ metric open ball of radius no more than $\delta_1$ must be contained in it. But it doesn't seem that we need $\tau_1 = \tau_2$ to argue that, so I think I'm missing something here.