# The definition of continuous function in topology

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?

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 possible if we ask for small enough perturbation.

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 $f(V) \subset U$.

• 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 – Glougloubarbaki Mar 7 '13 at 14:35
• 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? – Pinocchio 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 $x$; – Glougloubarbaki Jun 19 '18 at 16:31
• (continuation of previous comment) but this last formulation (though more elegant) hides a bit the intuition, I think – Glougloubarbaki Jun 19 '18 at 16:32

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 $f^{-1}$.

• i mean why we define continuous like that – 89085731 Mar 7 '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? – Pinocchio Jun 18 '18 at 14:03

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)