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I am self-studying general topology, and I am curious about the definition of the continuous function. I know that the definition derives from calculus, but why do we define it like that?I mean what kind of property we want to preserve through continuous function?

89085731
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4 Answers4

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Let's formulate the general definition in a somewhat more intuitive way : a function is continuous if under small perturbation in the domain the variation in the range is not too big, or more precisely : we can make the variation in the range as small as we want if we restrict to small enough perturbations in the domain.

In the context of general topology, the "small" sets may be replaced by "well chosen open sets". So the definition should be thought of as : a function $f$ is continuous at $x$ if given any small enough open set $U$ containing $f(x)$, there is an open set $V$ (that we may be forced to choose very small) containing $x$ such that $f(V) \subset U$.

Albert
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    note : if it is not absolutely clear that the definition I gave in the post is equivalent to the one mentionned by OP, note that a union of open sets is open – Albert Mar 07 '13 at 14:35
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    what is wrong with saying that if we have small enough perturbations in X we remains close enough to f(x), wouldn't that be equivalent? – Charlie Parker Jun 18 '18 at 13:56
  • How do you define small enough? In terms of topology, without a metric? – rubikscube09 Jun 19 '18 at 14:51
  • @rubikscube09 I was being purposefully vague. In the context of topology without a metric, instead of talking about small balls, you talk about a basis of neighborhood. An open set $U$ containing $x$ is "small" if "most" open sets containing $x$ contain $U$. So you could define continuity at $x$ by asking that there is a basis of neighborhoods $V_i$ of $f(x)$ such that for all $i$, $f^{-1}(V_i)$ is a neighborhood of $x$. this is equivalent to simply requiring that for ALL neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a nghbd of $x$; – Albert Jun 19 '18 at 16:31
  • (continuation of previous comment) but this last formulation (though more elegant) hides a bit the intuition, I think – Albert Jun 19 '18 at 16:32
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If you mean the definition that $f\colon X\to Y$ is continuous if $f^{-1}(U)$ is open for every open $U\subseteq Y$, then this is because this property is equivalent to continuity in metric spaces, but doesn't refer to the metric. This makes it a natural candidate for the general definition.

It is a little harder to motivate the definition in terms of the preservation of properties, because we have to do it backwards. We are saying that a function $f$ is continuous if $f^{-1}$ (thought of as a function from the set of subsets of $Y$ to the set of subsets of $X$) preserves the property of openness. We run into problems if we try to insist that $f(U)$ is open for every open $U\subseteq X$ (maps with this property are called open), because the image function from subsets of $X$ to subsets of $Y$, taking a subset $V$ to the subset $f(V)$, doesn't respect the operations of union and intersection in the same way that $f^{-1}$ does. For example, $f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)$, but it is not true in general that $f(U\cap V)=f(U)\cap f(V)$. As the definition of a topology involves certain intersections and unions preserving the property of openness, any "structure preserving map" must do so as well, which is why we consider $f^{-1}$.

mdp
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  • i mean why we define continuous like that – 89085731 Mar 07 '13 at 14:08
  • Why do we require $f^{-1}$ to preserve openness? Just knowing the "metric space definition of limit" would seem to imply that the definition is backwards...this must be wrong I know, but why is it wrong? – Charlie Parker Jun 18 '18 at 14:03
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I want to add a bit to Matt Pressland's nice answer:

Sometimes it is useful to think of the notion of a 'continuous' function as more 'primitive' than the notion of a 'topology'. For example:

  • Subspace topology is what makes the inclusion map continuous, i.e. choose the basis of the topology such that the inclusion map is continuous.
  • Product topology is what makes the projection map continuous
  • Quotient topology is what makes the quotient map continuous
Puzzled
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The reason is that $\delta$-neighbourhood and $\epsilon$-neighbourhood in the definition of the continuity of $f$ in Euclidean space respectively correspond to $U$ and $f^{-1}(U)$ in the topological sense. (I'm not explaining the details since you are just wondering the reason not the definition of all these)

Metin Y.
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