Prove if f: R $\displaystyle \rightarrow$ R is an isometry, then f is a continuous function.
I am just really having trouble understanding the concepts of isometries in general. Thank you for the help.
An isometry is just a mapping from one metric space to another such that the distance between two points in the domain is equal to the distance between the image of those two points in the range. In other words if $\displaystyle f:X\mapsto Y$ is an isometry we have that $\displaystyle d_X\left(x,y\right)=d_Y\left(f(x),f(y)\right)$. In your case we have that $\displaystyle X=Y=\mathbb{R}$ and so, assuming this is the standard case, we have that $\displaystyle d_X(x,y)=d_Y(x,y)=|x-y|$.
So, as Plato mentioned we have that for every $\displaystyle \varepsilon>0$ $\displaystyle |x-y|=|f(x)-f(y)|$ so choosing $\displaystyle \delta=\varepsilon$ we have that $\displaystyle |f(x)-f(y)|=|x-y|<\delta=\varepsilon$
Many neat things can be seen about isometries. They are clearly Lipschitz, and thus uniformly continuous. They are injective. etc.