I was learning the definition of continuous as:
$f\colon X\to Y$ is continuous if $f^{-1}(U)$ is open for every open $U\subseteq Y$
For me this translates to the following implication:
IF $U \subseteq Y$ is open THEN $f^{-1}(U)$ is open
however, I would have expected the definition to be the other way round, i.e. with the 1st implication I defined. The reason for that is that just by looking at the metric space definition of continuous:
$\exists q = f(p) \in Y, \forall \epsilon>0,\exists \delta >0, \forall x \in X, 0 < d(x,p) < \delta \implies d(f(x),q) < \epsilon$
seems to be talking about Balls (i.e. open sets) in X and then has a forward arrow for open sets in Y, so it seems natural to expect the direction of the implication to go in that way round. However, it does not. Why does it not go that way? Whats is wrong with the implication going from open in X to open in Y? And of course, why is the current direction the correct one?
I think conceptually I might be even confused why the topological definition of continuous requires to start from things in the target space Y and then require things in the domain. Can't we just say map things from X to Y and have them be close? Why do we require to posit things about Y first in either definition for the definition of continuous to work properly?
I can't help but point out that this question The definition of continuous function in topology seems to be similar but perhaps lack the detailed discussion on the direction on the implication for me to really understand why the definition is not reversed or what happens if we do reverse it. The second answer there tries to make an attempt at explaining why we require $f^{-1}$ to preserve the property of openness but its not conceptually obvious to me why thats the case or whats going on. Any help?
For whoever suggest to close the question, the question is quite clear:
why is the reverse implication not the "correct" definition of continuous?
As an additional important point I noticed is, pointing out the difference between open mapping and continuous function would be very useful.
Note: I encountered this in baby Rudin, so thats as far as my background in analysis goes, i.e. metric spaces is my place of understanding.
Extra confusion/Appendix:
Conceptually, I think I've managed to nail what my main confusion is. In conceptual terms continuous functions are suppose to map "nearby points to nearby points" so for me its metric space definition makes sense in that sense. However, that doesn't seem obvious to me unless we equate "open sets" to be the definition of "close by". Balls are open but there are plenty of sets that are open but are not "close by", for example the union of two open balls. I think this is what is confusing me most. How is the topological def respecting that conceptual requirement?
Best Answer
The "normal" definition goes like this:
It is claimed that, at fixed point, for any given ball $B_\epsilon$ of radius $\epsilon$ in the image, there exists a ball $B_\delta$, in the preimage, of radius $\delta$ such that $Im (B_\delta) \subset B_\epsilon$. This is the implication $$(...) < \delta \implies (...) < \epsilon $$
Very informally, you could compare the statement, for continuous $f$,
and
and
In topological spaces, the last one is often taken as a definition.
Regarding your interpretation
This is perfectly valid and translates as "IF you give me an $\epsilon$ THEN I can find you a corresponding $\delta$".
Regarding the implication, let me explain in this way, to show what happens with that implication:
Let $U \subset Y$ be open, then for this set you can have its preimage, $f^{-1}(U) \subset X$, which is the set that satisfies: $$x \in f^{-1}(U) \implies f(x) \in U $$ So now you can freely say:
If is just so happens, that $f^{-1}(U)$ is open for any open $U$, then we call $f$ continuous. Translating, this means that if it just so happens that for any given radius $\epsilon$, can find a corresponding $\delta$ such that $$ x\in B_\delta \implies f(x) \in B_\epsilon, $$ then $f$ is continuous.
A few more details:
You have be rather careful when you state exactly what you mean with mapping "nearby points to nearby points".
Given a metric, we can always have balls as subsets of that space. The open sets are precisely those that, for each $x$, have some ball around them completely contained in the open set. This is true regardless of whether the open set is a union of open intervals, the whole space, a single interval, or any other open set.
To say that $f$ maps "nearby points to nearby points" means to say that, if you fix a point $x_0$, and look at what happens to points nearby $x_0$, they will all be mapped to points close to $f(x_0)$. The exact meaning of this is that: for each fixed $x\in f^{-1}(U)$, for any ball $B_\epsilon$ around $f(x)$ (and one exists, and satisies $B_\epsilon \subset U$, by openness), there is a ball $B_\delta$ around the point $x$ that maps into $B_\epsilon$. Since $B_\epsilon \subset U$, we have $B_\delta \subset f^{-1}(U) $, which by definition makes the preimage open. It's a ball around an arbitrary point completely in $f^{-1}(U) $.
Whatever open set you have, all of the points in there will be interior, so continuity (finding matching balls $B_\delta$ and $B_\epsilon$) works at each point at a time, so to speak. And now it almost rolls off the tongue: $$\forall x \ \forall \epsilon \ \exists \delta \ (...) $$
To me, it is somehow intuitively clear that if you want a statement about how some values of $f(x)$ behave, you would start with something about its target set. Maybe that's just me. You sort of start with the question "How close to $f(x_0)$ do you want the outputs of $f$ to be", which is a question about the target set.