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Let $X,Y$ be a metric spaces then there is a theorem

$f$ is continuous iff the pre-image of an open set is an open set

On the other hand in topology spaces it seems that the definition is:

Let $X,Y$ be a topology spaces and $f:X\to Y$ then

$f$ is continuous if the pre-image of an open set is an open set

So in metric spaces it is $\iff$ and in topology spaces just in one direction?

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    $\begingroup$ People usually avoid the use of the word "iff" in definitions. Since it is a definition, it is implicit that it's an equivalence despite of not explicitly indicating "only if". Also, do you mean "open set" and not "open group"? $\endgroup$ May 8, 2018 at 14:19
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    $\begingroup$ "group" should be "set". My guess is "group" came from using an on-line translation to English. I often get strange word choices when I use on-line translations to translate a math excerpt from another language to English so that I can read it. Also, I think the Italian word for "group" was originally used for "set" by those in Peano's time. $\endgroup$ May 8, 2018 at 14:20
  • $\begingroup$ @JohnGriffin edited meant set $\endgroup$
    – gbox
    May 8, 2018 at 14:20
  • $\begingroup$ @JohnGriffin so the property of an open set is a definition in topology space which become a theorem in metric space? $\endgroup$
    – gbox
    May 8, 2018 at 14:22
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    $\begingroup$ @gbox The author chose to use a different definition for continuity in the metric space setting. This is likely because there are definitions that directly use the metric and not just the topology it generates. As a result this is a theorem in the metric space setting. For theorems, it's important to indicate "iff" if it meant to be so. In the topology setting, the author chose it to be the definition. This makes it automatically an equivalence. $\endgroup$ May 8, 2018 at 14:25

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So in metric spaces it is $\iff$ and in topology spaces just in one direction?

The metric world has its definition and so does the topological world. It is a theorem that they are equivalent. But this is only since we can generate a topology from a metric but not the other way around. So in a way metric spaces are topological spaces (not literally but they induce a topology) but topological spaces need not be metric in general (in the sense that not every topology arises from a metric).

More generally: there are lots of continuity definitions: Cauchy, uniform, Holder, sequential, $\epsilon-\delta$, Lipshitz, via nets, via preimages, etc. etc. Some of them are equivalent, some are not. Some are only equivalent under the Axiom of Choice.

For metric spaces I assume this is the definition you are referring to: $f:X\to Y$ is continuous if for any $x\in X$ and any $\epsilon\in\mathbb{R}, \epsilon>0$ there exists $\delta\in\mathbb{R}, \delta >0$ such that if $y\in X$ with $d_X(x,y)<\delta$ then $d_Y(f(x),f(y))<\epsilon$. Also known as $\epsilon-\delta$ definition.

Note that this definition does not make sense in the topological world where we don't have the notion of distance.

For topological spaces the definition is a bit shorter: $f:X\to Y$ is continuous if $f^{-1}(U)$ is open whenever $U\subseteq Y$ is.

So these are definitions. And it is a theorem that in the metric world these two definitions are equivalent if we take the usual topology generated by a metric.

But there is at least one more important definition of continuity in topological spaces: $f:X\to Y$ is continuous if for any convergent net $(x_\alpha)\subseteq X$ and its limit point $x\in X$ we have that $f(x_\alpha)$ is a net convergent to $f(x)\in Y$.

So as you can see it is not unusual to have multiple definitions for things.

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  • $\begingroup$ lots of continuity definitions --- Several variations of possible interest are discussed here. $\endgroup$ May 8, 2018 at 17:32

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