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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.

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To show that $\tau_d\supseteq\tau_1$ it’s not enough to show that each $d_1$-open ball contains a $d$-open ball: you must show that it contains a $d$-open ball with the same centre. You seem to have tried to do this, but it needs to be said as well, and your argument isn’t quite right, though you have found a $d$-open ball that works. All you have to show is that $B_x^d(\delta)\subseteq B_x^{d_1}(\delta)$. To do this, suppose that $y\in B_x^d(\delta)$; then $d_1(x,y)=d(x,y)-d_2(x,y)\le d(x,y)<\delta$, since $d_2(x,y)\ge 0$, so $y\in B_x^{d_1}(\delta)$. As you say, a similar argument yields the conclusion that $\tau_d\supseteq\tau_2$.

For (b) suppose that $U\in\tau_d$; we need to show that $U\in\tau_1$. Let $x\in U$; by hypothesis there is an $\epsilon>0$ such that $B_x^d(\epsilon)\subseteq U$, and we want a $\delta>0$ such that $B_x^{d_1}(\delta)\subseteq U$. We know that $B_x^{d_2}(\epsilon/2)\in\tau_1$, so there is a $\delta_1>0$ such that $B_x^{d_1}(\delta_1)\subseteq B_x^{d_2}(\epsilon/2)$. Let $\delta=\min\{\delta_1,\epsilon/2\}$; if $y\in B_x^{d_1}(\delta)$, what can you say about $d(x,y)$?

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  • $\begingroup$ Yay! You're back. Cool. $\endgroup$ – Rudy the Reindeer Oct 11 '14 at 4:24
  • $\begingroup$ @Matt: Yep, though I may not be as active as before. $\endgroup$ – Brian M. Scott Oct 11 '14 at 4:24
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    $\begingroup$ @user92638: Okay. I’d do it a little differently, but that works. I’d simply say that if $y\in B_x^{d_1}(\delta)$, then $d_1(x,y)<\delta_1$, so $d_2(x,y)<\epsilon/2$. Moreover, $d_1(x,y)<\epsilon/2$, so $d(x,y)=d_1(x,y)+d_2(x,y)<\epsilon$. (I think that it’s a little easier to read this way.) $\endgroup$ – Brian M. Scott Oct 11 '14 at 8:27
  • $\begingroup$ How does this sound? $$If y \in B_x^{d_1}(\delta), then$$ $$d(x,y) = d_1(x,y) + d_2(x,y) = min\{\delta_1, \epsilon/2\} + d_2(x,y) \tag*{}$$ $$< min\{\delta_1, \epsilon/2\} + \epsilon/2 (Since B_x^{d_1}(\delta) \subseteq B_x^{d_2}(\epsilon/2)) \le \epsilon,$$ so $$y \in B_x^{d_1}(\delta).$$ $\endgroup$ – user92638 Oct 11 '14 at 8:46
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    $\begingroup$ @user92638: No, you can’t say that $d_1(x,y)=\min\{\delta_1,\epsilon/2\}$: what you know is that $d_1(x,y)$ is actually less than that minimum. Otherwise it’s okay. $\endgroup$ – Brian M. Scott Oct 11 '14 at 8:51
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It suffices to show that if $\{x_i\}_{i\in I}$ is a net converging to $x$ in $\tau_1$ then $\{x_i\}_{i\in I}$ converges to $x$ in $\tau_d$ (because then any set closed in $\tau_1$ will be closed in $\tau_d$). But if $x_i\rightarrow x$ in $\tau_1$ then by assumption $x_i\rightarrow x$ in $\tau_2$ also, i.e. both $d_1(x_i,x)$ and $d_2(x_i,x)$ tend to $0$. Hence their sum does, so $x_i\rightarrow x$ in $d$.

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