In topology we defined a continuous map to be a map between 2 spaces $f: X \mapsto Y$ such that if $U\subset Y$ is open then $f^{-1}(U)$ is open.
Why did we use the preimage; why not say 'a continuous map always maps open sets to open sets'?
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Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there. – Sort of Damocles Feb 02 '19 at 19:56
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1Do you want $f: \mathbb{R} \to \Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition. – Henno Brandsma Feb 02 '19 at 20:02
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Constant functions will also usually be continuous but not open. – Jon Hillery Feb 02 '19 at 20:04
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I think the answer to this question is relevant here. – kimchi lover Feb 02 '19 at 20:05
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"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = \sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it. – fleablood Feb 02 '19 at 20:23
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The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $\mathbb R \rightarrow \mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
Picaud Vincent
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For additional fun, the same map, but viewed as $\Bbb R\to[0,\infty)$ would be continuous in the proposed sense. (But that trick does not work for $x\mapsto x^3-x$.) – Hagen von Eitzen Feb 02 '19 at 20:05
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@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-) – Picaud Vincent Feb 02 '19 at 20:07
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Because with your definition we would have to agree that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $\mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $\epsilon-\delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)\to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $U\subset Y$ the set $f^{-1}(U)\subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
Mark
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