Definitions
Let $(X, \mathfrak{T}_X), (Y, \mathfrak{T}_Y)$ be two topological spaces and $f: X \rightarrow Y$ be a mapping.
$f$ is called continuous $:\Leftrightarrow \forall U \in \mathfrak{T}_Y: f^{-1}(U) \in \mathfrak{T}_X$
Let $(X,d_X), (Y, d_Y)$ be two metric spaces and $f: X \rightarrow Y$ be a mapping.
$f$ is called an isometry $:\Leftrightarrow \forall x_1, x_2 \in X: d_X(x_1, x_2) = d_Y(f(x_1), f(x_2))$.
Question
Let $(X, \mathfrak{T}_X, d_X), (Y, \mathfrak{T}_Y, d_Y)$ be two topological, metric spaces and $f:X \rightarrow Y$ be an isoemtry.
Is $f$ continuous? According to the German Wikipedia this is "obviously the case, because of the definition". I don't think that is that obvious.
Best Answer
For an isometry, it is easy to see that for every $x\in X$, we have
$$f^{-1}\bigl( B_\varepsilon\left(f(x)\right)\bigr) = B_\varepsilon(x),$$
so $f$ is continuous at $x$, and since $x$ was arbitrary, globally continuous. In this case, the continuity at a point is more evident than the global continuity.