Show that $d'(x,y)=\min\{1,d(x,y)\}$ induces the same topology as $d$
Hint: Show that:
- For any ball defined with the $d$ metric, you can find a ball defined with the $d'$ metric contained in the $d$-ball.
- For any $d'$-ball, show you can find a $d$-ball contained in it. You only need to work with balls since they form a basis for the metric topology.
I'll help you do the second part, and I'll leave the first part to you. Suppose we have $B_{d'}(x,r)$. I claim $B_{d}(x,r) \subseteq B_{d'}(x,r)$. Why? Let $y \in B_{d}(x,r)$. Then $d(x,y) < r$. But since $d'$ is the minimum of $d$ and $1$, then $d' \leq d$. So we have $d'(x,y) \leq d(x,y) < r$, which implies $d'(x,y) < r$, so $y \in B_{d'}(x,r)$. That shows us $B_{d}(x,r) \subseteq B_{d'}(x,r)$.
By the way: The purpose of this assignment, and the take home lesson you should take from it once you're done proving it, is that in the metric topology, only the small open balls matter. We can discard all larger open balls. So if you threw away all open balls of radius $3$ or larger, you would still have the same topology. If you threw away all balls of radius $.0000002$ or larger, you'd still have the same topology. Only the small balls matter.