# Inclusions in Metric Topology

Prove that if $\exists L>0 : d_2(x,y)\geq Ld_1(x,y)$ then the topology induced by $d_2$ is finer than the topology induced by $d_1$.

I'll call $T_1$ the topology induced by $d_1$ and $T_2$ the topology induced by $d_2$. Since this problem involves metric space, it would be equivalent to prove that for a ball $B_{d_1}(x,r):=\{y\in X:d_1(x,y)<r\}\; \exists s$ such that for $B_{d_2}(x,s):=\{y\in X:d_2(x,y)<s\}$ we have $B_{d_2}(x,s)\subset B_{d_1}(x,r)$.

But is the inequality in the wrong direction?. My first thought was that if it were $d_2(x,y)\leq Ld_1(x,y)$ then I would have $\frac{1}{L}d_2(x,y)\leq d_1(x,y)<r$ then $B_{d_2}(x,r/L)\subset B_{d_1}(x,r)$ which would prove that $T_1\subset T_2$, but it the question was stated correctly there is something wrong with what I said.

Now, assuming there's nothing wrong with the question, how can I find $s$ such that $B_{d_2}(x,s)\subset B_{d_1}(x,r)$?.

Only hints, please avoid a full answer.

Suppose $d_1(x,y)<r$, then to prove that $B_{d_2}(x,s)\subset B_{d_1}(x,r)$ I have to prove that $d_2(x,y)<s \implies d_1(x,y)<r$.Take $s\in (0,rL)$, then $d_2(x,y)<s\implies Ld_1(x,y)\leq d_2(x,y)<s<rL\implies d_1(x,y)<r$.

• Hint: Add the requirement that L is positive. – Moishe Kohan May 4 '14 at 22:32

## 1 Answer

Hint: all is good, start it over to prove $\exists s:\ B_{d_2}(x,s)\subseteq B_{d_1}(x,r)$.

• Ok, I add a few things, I believe it may be right now, is it?. It seemed easy, I don't know why took me so much to figure it out. I – Cure May 4 '14 at 23:51