Show that $d'(x,y)=\min\{1,d(x,y)\}$ induces the same topology as $d$

Let $(M,d)$ be a metric space and define: $d' : M$x$M \rightarrow R$

Show that $d'(x,y)=\min\{1,d(x,y)\}$ induces the same topology as $d$

I know that $d'$ defines a metric on M, since d is a matric (satisfying positivity, symmetry and triangle inequality. But how do I show it induces the same topology?

Solution 1:

Hint: Show that:

1. For any ball defined with the $d$ metric, you can find a ball defined with the $d'$ metric contained in the $d$-ball.
2. 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.

Solution 2:

Since we are in metric spaces, we have nice definition of closed sets, that a limit point is contained in it, so we can try approaching this problem with that fact too. Showing that they induce the same topology is equivalent to showing that any closed set in one metric is closed in the other one also (because complement of open sets is closed)

Let $A$ be a closed set in $(X,d)$ and $x_n$ be any converging sequence in $A$ such that $(x_n) \to x \in A$. Therefore, eventually $d(x_n,x)<\epsilon<1 \implies d'(x_n, x) \leq d(x_n,x)<\epsilon<1$. Hence, $(x_n) \to x \in A$ in the metric $(X,d')$.

For the other way round, let $A$ be a closed set in $(X,d')$ and $x_n$ be any converging sequence in $A$ such that $(x_n) \to x \in A$. Therefore, eventually $d'(x_n,x)<\epsilon<1$. Since, $d'(x,y) < 1 \implies d(x,y) = d'(x,y) \forall x, y\in X$, we have that $d(x_n, x) = d'(x_n,x)<\epsilon<1$ eventually. Hence, $(x_n) \to x \in A$ in the metric $(X,d)$.