Showing that open subsets for two metrics of same space coincide. 
I am given two metrics d and d'.  I can easily verify d' is a metric from M using the axioms of a metric space(which I have done already)
However, I am stuck on showing that open subsets of (M,d) and (M,d') coincide.   I feel like this proof is most likely related to the concept of continuity in different metrics.

I found this on a website, but I am still confused what is going on and how to show that the open subsets of (M,d) and (M,d') coincide.  Do I have to show that the map is continuous from (M,d) to (M,d')? If so, how would I do this, or I am completely off or confused. Thanks.
 A: You don't need to do anything with continuity. To show that the topologies induced by the metrics $d$ and $d^{\prime}$ are the same you need only show that a metric ball in the topology in induced by $d$ can be written as a union of $d^{\prime}$ balls and vice versa. You might do it in the following way.
Let $x\in M$ and $\epsilon>0$. Then consider the ball $B_{d}(x,\epsilon):=\{y\in M:d(x,y)<\epsilon\}$. Then for $Y\in B_{d}(x,\epsilon)$ we have
$$d(x,y)<\epsilon$$
Then we do the algebra thing.
$$\frac{d(x,y)}{1+d(x,y)}<\frac{\epsilon}{1+d(x,y)}$$
Thus
$$d^{\prime}(x,y)<\frac{\epsilon}{1+d(x,y)}$$
setting $\eta=\frac{\epsilon}{1+d(x,y)}$ we have that $B_{d^{\prime}}(x,\eta)\subseteq B_{d}(x,\epsilon)$. Thus the topology induced by $d^{\prime}$ contains the topology induced by $d$.
Then you show that each $d^{\prime}$ ball can be written as a union of $d$ balls.
Again let $x\in M$ and $\epsilon>0$. As before we consider the $d^{\prime}$ ball $B_{d^{\prime}}(x,\epsilon):=\{y\in M:d^{\prime}(x,y)<\epsilon\}$. Then 
$$d^{\prime}(x,y)=\frac{d(x,y)}{1+d(x,y)}<\epsilon$$
Then you do the algebra thing.
$$\frac{d(x,y)}{1+d(x,y)}<\epsilon\implies d(x,y)<\epsilon+\epsilon d(x,y)$$
$$d(x,y)<\epsilon+\epsilon d(x,y)\implies d(x,y)<\frac{\epsilon}{1-\epsilon}$$
Of course we assume that $0<\epsilon<1$ which we do because balls of radius less than $1$ are contained in balls of radius greater than or equal to $1$. Setting $\eta=\frac{\epsilon}{1-\epsilon}$ we have that $B_{d}(x,\eta)=B_{d^{\prime}}(x,\epsilon)$. Thus the two topologies are equal.
Edit: I didn't do a terribly good job of explaining why this is sufficient. Recall the arbitrary unions of open sets are open. Likewise, every open set in a metric space is a union of open balls with respect to that metric. Thus given a metric ball $B$ with respect to $d$ you show can that $B$ is open in $(M,d^{\prime})$ by showing that for each $y\in B$ there is a $d^{\prime}$ ball, say $B^{\prime}$, containing $y$ that is contained in $B$. Then the union over all of those $B^{\prime}$ is an open set in $(M,d^{\prime})$, thus $B$ is an open set in $(M,d^{\prime})$. Then you do the reverse. You take an open ball $B^{\prime}$ in $(M,d^{\prime})$ and show that for every $z\in B^{\prime}$ there is an open ball $B$ with respect to $d$ that contains $z$ and is contained in $B^{\prime}$. Again, the union over all such balls $B$ is an open set in $(M,d)$ that is equal to $B^{\prime}$. Thus $B^{\prime}$ would be open in $(M,d)$ and the two metric spaces have precisely the same open sets.
