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Definition of continuous function on a set: Suppose $X$ and $Y$ are metric spaces. Let $f: X \to Y$. We say $f$ is continuous on $X$ if for every $\varepsilon >0$, $\exists \delta >0$ such that $d(x, x_0) < \delta \implies d\left(f(x), f(x_0)\right) < \varepsilon$, $\forall x \in X$.

I learned, not too in depth, that an isometry is a map which preserves distance.

Isn't that what is going on with the definition of a continuous function? What is the difference?

Also, it seems like a continuous function is basically a homomorphism between metric spaces that preserves some sort of distance. Is this correct?

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    $\begingroup$ Here's an example to keep in mind. Let $y \in Y $ and define $f $ so that $f(x)=y $ for all $x \in X$. Then $f $ is continuous, but doesn't preserve distances. $\endgroup$ – littleO Apr 4 '16 at 3:20
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Not quite actually. A continuous function is one where points that are "close" in the domain are guaranteed to be within a certain distance from one another in the codomain. Thus when we say two points are within $\delta>0$ of each other in the domain, we are guaranteed that under a continuous map their images will be within $\epsilon>0$ of each other. The difference between continuity and an isometry is that an isometry will preserve the distance exactly. That is, if $d_X$ is the metric on $X$ and $d_Y$ is the metric on $Y$, an isometry $f:X \to Y$ will satisfy

$$ d_X(p_1, p_2) \;\; =\;\; d_Y(f(p_1), f(p_2)) $$

which is not always guaranteed of continuous maps.

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No. $f$ is a isometry if $$ d(x,y) = d(f(x),f(y)) $$ This is very different from the definition for continuity. You can consider the function $x \mapsto 2x$ on $\mathbb{R}$, it is continuous (choosing $\delta = \epsilon/2$) but not an isometry.

However, an isometry is necessarily continuous, since $f^{-1}[B(f(x);r)] = B(x;r)$. That is, choosing $\delta = \epsilon$.

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Constant functions are always continuous; they are never an isometry if $X$ has more than one point.

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