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**Connected** Let $\displaystyle (X,d)$ be a metric space, prove that $\displaystyle \phi(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric over $\displaystyle X.$ Moreover $\displaystyle X$ is bounded with the metric $\displaystyle \phi$ and $\displaystyle \text{diam}\, X<1.$

First part is easy, now I wanted to see how to prove that $\displaystyle X$ is bounded with the metric $\displaystyle \phi,$ is it because $\displaystyle \phi(x,y)<1$ ? Now since $\displaystyle \text{diam}\,X=\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)$ so since $\displaystyle \phi(x,y)<1$ then $\displaystyle \underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)<1,$ then $\displaystyle \text{diam}\,X<1,$ is this correct?