# 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.

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Write down the definition of continuity for metric spaces. Then it's obvious. – Karolis Juodelė Feb 16 '14 at 11:44
@KarolisJuodelė: Thanks (+1), now I understand it :-) (We have shown that "continuous" as it is defined in topology is equivalent to "continuous" as it is defined in analysis.) – Martin Thoma Feb 16 '14 at 11:47
If $f$ is an isometry (=distance preserving) then you can use $\delta = \varepsilon$ to prove $f$ is continuous. – Rudy the Reindeer Feb 16 '14 at 15:04
Like this: $$\|f(x) - f(y)\| = \|x - y\| < \delta = \varepsilon$$ – Rudy the Reindeer Feb 16 '14 at 15:05

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.
I guess $B_\varepsilon(x)$ is the ball with radius $\varepsilon$ around the center $x$? I understand that $f^{-1}\bigl( B_\varepsilon\left(f(x)\right)\bigr) = B_\varepsilon(x)$ is true, but how does it show that $f$ is continuous for arbitrary topologies (I understand that it shows continuity for the standard topology). – Martin Thoma Feb 16 '14 at 11:45