Showing $e(a,b)=\{\min(1,d(a,b))\}$ is equivalent to $d(a,b)$

Question

Let $d$ be a metric on a non-empty set $X$.

Show that the function $e$ defined by $e(a,b)=\min{1,d(a,b)}$ where $a,b\in X$ and $d$ are equivalent metrics.

My attempted solution:

$\exists c_1\in\mathbb{R}$ such that:

$d(a,b)\leqslant\min\{c_1,c_1d(a,b)\}=c_1 e(a,b)$

In the same way

$\exists c_2\in\mathbb{R}$ such that:

$e(a,b)=\min\{1,d(a,b)\}\leqslant c_2(d(a,b))$ for any $c_2>1$

Question:

1)Is this proof right?

2)If I wanted to prove the topologies induced were the same. How would I do it?

Thanks in advance!

Answer 1

In general for metric space $(X,d)$:

$$U\text{ is open in }(X,d)\iff \forall x\in U\;\exists r>0\text{ such that }B_d(x,r)\subseteq U\tag1$$where $B_d(x,r):=\{y\in X\mid d(x,y)<r\}$.

It is evident that: $$\exists r>0 B_d(x,r)\subseteq U\iff \exists r\in(0,1) B_d(x,r)\subseteq U$$so that we can modify $(1)$ and make it:

$$U\text{ is open in }(X,d)\iff \forall x\in U\;\exists r\in(0,1)\text{ such that }B_d(x,r)\subseteq U\tag2$$

Now realize that for every $r\in(0,1)$ and every $x\in X$ we have: $$B_d(x,r)=B_e(x,r)$$ so that the statements "$U$ is open in $(X,d)$" and "$U$ is open in $(X,e)$" are equivalent.

Answer 2

Use the mutual inclusion of balls criterion instead, there can be no $c_1$ that will work for the usual topology on the reals, as the latter metric is unbounded.