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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 $:\Leftrightarrow \forall x_1, x_2 \in X: d_X(x_1, x_2) = d_Y(f(x_1), f(x_2))$.


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.) – moose 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

1 Answer 1

up vote 3 down vote accepted

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.

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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). – moose Feb 16 '14 at 11:45
@moose: the statement is only true when the topology is that induced by the metric. – Anthony Carapetis Feb 16 '14 at 11:47
@moose I prefer "ball", the sphere is the boundary of the ball. When dealing with metric spaces, the topology is, unless explicitly stated otherwise, always assumed to be the one induced by the metric. – Daniel Fischer Feb 16 '14 at 11:47

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