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I am trying to show that if $d$ is a metric, and $\delta(x,y) = \min(d(x,y), 1)$ is also a metric, that $d$ and $\delta$ generate the same topology.

I am stuck on the part of showing that $\mathbf{B_{\delta}(x,r)\subseteq B_{d}(x,r)}$, where $B_{\delta}$ is an open ball in the $\delta$-metric and $B_{d}$ is an open mall in the $d$-metric.

So far, this is what I have done:

Let $y \in B_{\delta}(x,r)$. Then, $\delta (x,y) < r$. Since $\delta (x,y) = \min(d(x,y),1)$, if $\delta(x,y) = d(x,y)$, then $\delta(x,y) = d(x,y) < r$, which implies that $y \in B_{d}(x,r) \implies B_{\delta}(x,r) \subseteq B_{d}(x,r)$ in this case.

On the other hand, if $\delta(x,y) = \min(d(x,y),1) = 1$, then $\delta(x,y) = 1 < d(x,y)$, but this is exactly the opposite of what I want here. In order to get $B_{\delta}(x,r) \subseteq B_{d}(x,r)$, I kind of need $d(x,y) < \delta(x,y)$, don't I? But I'm not getting that here.

Will somebody please help me? And also, be willing to answer follow-up questions that I have? Because if you just give me a hint, I have no doubt that I will need to ask you lots of follow-ups! Thank you in advance.

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    It's not true (in general) that $B_\delta(x,r)\subseteq B_d(x,r)$. But that's not what you need to prove in order to prove that $d$ and $\delta$ generate the same topology. – 2017-02-14
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    @EricWofsey better to show that $d$ and $\delta $ are (strongly) equivalent metrics? – 2017-02-14
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    @EricWofsey guess what? I'm trying to show now that $\exists a,b > 0$ such that $a \delta(x,y) \leq d(x,y) \leq b \delta(x,y)$. I figured out that $a$ must be $1$ (trivial), but I have no idea how to figure out what $b$ is. Can you help me? – 2017-02-14
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    If $r=2$ then $B_{\delta}(x,r)$ is the whole space but $B_d(x,r)$ may not be. – 2017-02-14

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You've got the wrong condition for 'generates the same topology'. You should have encountered a lemma that gives an iff condition: Two metrics $d$ and $\delta$ induce the same topology iff for every $x$, every $d$-ball around $x$ contains a $\delta$-ball around $x$, and conversely. The radii of these balls don't have to be the same.

So for your example, what remains to show is that for every $x$ and $r>0$, the ball $B_d(x,r)$ contains $B_\delta(x,r')$ for some $r'>0$. The choice $r':=\min(r,1)$ should do. To see this, suppose $z\in B_\delta(x,r')$. There are two cases: (1) $r>1$ (2) $r\le1$. In case (1), we have $$ \min(d(x,z),1)=:\delta(x,z) 1$. Conclude $B_\delta(x,r')\subset B_d(x,r)$, qed. The second case is argued similarly. (Or, you can dispense with the second case by arguing "wlog, $r<1$".)

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    Okay, so the radii don't have to be the same. Showing that every $\delta$-ball contains a $d$-ball around $x$ is easy, because it follows directly from the fact that $\exists a > 0$ such that $a \delta (x,y) \leq d(x,y)$. It's showing the containment in the other direction that I'm having a problem with. So, let's say that for arbitrary $x$, we have the $d$-ball, $B_{d}(x,r)$. Then, how do I show that a $\delta$-ball is contained inside here? – 2017-02-14
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    @JessyunBourne The radius $r':=\min(r, 1)$ should work. – 2017-02-14
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    so I phrase that as, suppose $\exists z \in B_{d}(x,r)$. Then, because by definition, $\delta(x,z) = \min(d(x,z), 1) \leq \min(r, 1) = r^{\prime}$, $B_{\delta}(z, r^{\prime}) \subset B_{d}(x,r)$? I'm a little sleep-deprived so writing English sentences that sound good is proving difficult... – 2017-02-14
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    @JessyunBourne Haha, need to be a bit more careful. See edit – 2017-02-14
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    thank you, Big Cat! (Or is it Great Cat?) – 2017-02-14
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    actually, I think the two cases are $r \geq 1$ and $r<1$, aren't they? If $r = 1$, then $\min(r,1) = \min(1,1) = 1$, which reverts back to the $r>1$ case. – 2017-02-14
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    @JessyunBourne I think $r=1$ could fit in either case, since $\min(r,1)$ is both $r$ and $1$ when $r=1$. (You are welcome!) – 2017-02-14