Question about open set of a metric space

Question

if $(S,d)$ is a metric space then $\mu(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is also a metric space on $S$ and a set $A$ is open in $(S,d)$ iff it is open in $(S,\mu)$.
I understand the first part that $(S,\mu)$ is also metric space however I am not getting any idea about the rest. can someone give some hints, that would be a great help.

Answer 1

In a metric space, the best way to prove that a topology $\mathcal{T}_d$ is coarser that the topology $\mathcal{T}_\mu$ is to use the bases consisting of open balls i.e. to prove that for every $x\in X$ and every $r>$ there exists an $\epsilon >0$ such that $$B_\epsilon^\mu (x) \subseteq B_r^d (x)$$ in other words every metric open ball in the sense of the metric $d$ centered at $x$ contains another metric open ball in the sense of $\mu$ contained in it. This is done by realizing $$\mu(x,y)$$ $$=\frac{d(x,y)}{1+d(x,y)}$$ $$\le \frac{d(x,y)}{1}$$ $$\le d(x,y)$$ and, hence, $$y\in B_r^{d} (x)$$ $$\Rightarrow d(x,y)<r$$ $$\Rightarrow \mu(x,y)<r \text{ since } \mu(x,y)\le d(x,y)$$ $$\Rightarrow y\in B_r^{\mu} (x)$$ which means that for every $x\in X$ and for every $r>0$, $$B_r^{\mu} (x)\subseteq B_r^{d} (x)$$ so every open set in $\mathcal{T}_{d}$ is also open in $\mathcal{T}_{\mu}$

Answer 2

We'll show that a set $A$ is closed in $(S,d)$ iff $A$ is closed in $(S,\mu)$.

$\implies:$

Let $A$ be closed in $(S,d)$. Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $A$ that converges to some $x \in S$ in the metric $\mu$. That is, $\mu(x_n,x) \to 0$. To show that $x \in A$, the strategy is to show that $d(x_n,x) \to 0$. But we must have that since: $$\mu(x_n,x) = \frac{d(x_n,x)}{1+d(x_n,x)}$$ it must be the case that: $$\mu(x_n,x)(1+d(x_n,x)) = d(x_n,x)$$ $$d(x_n,x) = \frac{\mu(x_n,x)}{1-\mu(x_n,x)}$$ Since $\mu(x_n,x) \to 0$, it follows that $d(x_n,x) \to 0$. Since $A$ is closed in $(S,d)$, this means that $x \in A$.

$\impliedby:$

Now, let $A$ be closed in $(S,\mu)$. We want to show that it is closed in $(S,d)$. Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $A$ that converges to some $x \in S$ in the metric $(S,d)$. That is, $d(x_n,x) \to 0$. I'll leave it to you to show that $\mu(x_n,x) \to 0$. But this will imply that $x \in A$ and you're done. $\Box$

All of this is a consequence of the fact that a subset $A \subseteq X$ of a metric space $(X,d)$ is closed iff every convergent sequence in $A$ converges inside of $A$.