Showing the Equivalency of Two Metrics

Question

The following is a problem in Gamelin and Greene's "Introduction to Topology."

Define $\rho$ on $X \times X$ by $$\rho(x,y) = \min(1,d(x,y)), \quad x,y \in X.$$ Show that $\rho$ is a metric that is equivalent to $d$. (Hence every metric is equivalent to a bounded metric.)

The suggested method of proof for this statement is to use the following result (proven in a previous exercise in the same textbook):

Two metrics on $X$ are equivalent if they determine the same open subsets. Show that two metrics $d, \rho$ on $X$ are equivalent if and only if the convergent sequences in $(X,d)$ are the same as the convergent sequences in $(X, \rho)$.

This is used to prove the original statement by stating that $\rho(x_n, x) \xrightarrow{} 0$ if and only if $d(x_n, x) \xrightarrow{} 0$, which shows that the metrics have the same convergent sequences, and hence the metrics must be equivalent by the second result. If I use this approach, I will need to prove the second result before using it in my own solution to the first problem. However, I was wondering if it were possible to prove the first statement directly, using the provided definition of "equivalent metrics." Which would be the best route to use?

(Note that at this point, the textbook has not yet introduced the definition of "topology" or any concepts related to a topology.)

Show that an open set with respect to one metric is also open with respect to the other. — copper.hat, Jan 25 '22 at 03:55
Note that if $\delta \in [0,1]$ then $B_\rho(x,\delta) = B_d(x,\delta)$. — copper.hat, Jan 25 '22 at 04:02
https://math.stackexchange.com/questions/4341941/about-rhox-y-begincases-mathrmdx-y-mathrmdx-y1-1-ma/4341944#4341944 — Mr. Gandalf Sauron, Jan 25 '22 at 07:08

score 0 · Accepted Answer · answered Jan 25 '22 at 07:00

If $O \subseteq X$ is open in $(X,d)$ then it is also open in $(X,\rho)$: let $x \in O$. There is some $r>0$ so that $B_d(x,r) \subseteq O$ by definition of the topology on $(X,d)$. Let $r'=\min(r,1) >0$. Then $B_\rho(x,r') = B_d(x,r') \subseteq B_d(x,r)$ (just by definitions: $y \in B_\rho(x,r')$ iff $\rho(x,y) < r'\le 1$ so that happens iff $d(x,y) < r'$ etc.) and, as $x \in O$ was arbitrary, $O$ is $(x,\rho)$-open too.

The converse is similar.

The intuition is clear: for small radius $d=\rho$ and we only have to consider small balls to check that sets are open anyway: if a ball works, a smaller radius one will also do. So this truncation at $1$ makes no difference for convergence, open sets or continuity etc.

Showing the Equivalency of Two Metrics

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