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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?

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    $\begingroup$ The implication there is more related to $B_\delta(p) \subseteq f^{-1}(B_\epsilon(q))$ and the part $U$ open $\Rightarrow f^{-1}(U)$ open is more of a "wrapping" argument that allows you to relate the $B_\delta(p) \subseteq f^{-1}(B_\epsilon(q))$ to open sets. Or maybe, the $U$ open $\Rightarrow f^{-1}(U)$ open is more related to the $\forall \epsilon > 0, \exists \delta > 0$ part. $\endgroup$ – Daniel Schepler Jun 18 '18 at 19:00
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    $\begingroup$ Note that if $f : \mathbb{R} \rightarrow \mathbb{R}$ maps $x$ to $x^2$, then the image of any open ball around $x=0$ is not in fact itself an open set, though the metric definition is satisfied because that image is contained in an open ball. $\endgroup$ – aschepler Jun 19 '18 at 0:29
  • $\begingroup$ Topology was originally called " analysis situs" ("position analysis") and was almost entirely about metric spaces. The modern def'n of continuity is a generalization of the "epsilon-delta" def'n, and there are many properties that are equivalent to it, and it is good to learn some of them, as some are more useful than others in some situations... If $f:X\to Y$ maps open sets to open sets then $f$ is called an open mapping (or we say that $f$ is "open"). The real function $f(x)=x^2$ is continuous but not open, as $f((-1,1))=[0,1)$. $\endgroup$ – DanielWainfleet Jun 19 '18 at 14:00
  • $\begingroup$ @DanielWainfleet there seems to be an important but subtle difference between continuous function and open mapping which I do not understand, that must be one of the key issues why my understanding I conjecture. $\endgroup$ – Pinocchio Jun 19 '18 at 15:22
  • $\begingroup$ @aschepler I think I am not understanding your point (sorry I don't understand this concept well yet). What is the concept your trying to emphasizes? $\endgroup$ – Pinocchio Jun 19 '18 at 16:37
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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 $f$,

For any ball $B_\epsilon$ in the image, you can find a ball $B_\delta$ mapping into $B_\epsilon$

and

For any ball $B_\epsilon$ in the image, its preimage contains a ball $B_\delta$

and

The preimages of open sets are open.

In topological spaces, the last one is often taken as a definition.


Regarding your interpretation

IF $U \subseteq Y$ is open THEN $f^{−1}(U)$ is open

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:

For any open $U \subset Y$, there is a set $f^{-1}(U) \subset X.$

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.

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  • $\begingroup$ thanks for helping Julius! I think what confuses me is that yes, I understand intuitively we say given a epsilon ball in the image then we can find an epsilon ball in the pre-image...but the implication starts from the "pre-image" and goes to the image. I find that really confusing. Perhaps thats what's giving me most trouble...isn't the "normal" definition of continuous the reversed one? I can't grasp why the logical implications seems to be written backwards but somehow still say the same thing. Usually the direction of implication is a big deal, so why not here? what makes it work? $\endgroup$ – Pinocchio Jun 19 '18 at 13:50
  • $\begingroup$ do you mind commenting on the difference between open mapping and continuous map? $\endgroup$ – Pinocchio Jun 19 '18 at 16:36
  • $\begingroup$ Here the $\delta - \epsilon$ implication just translates as $x \in f^{-1}(U) \implies f(x) \in U$. Its the existence of such open sets that are at the heart of "pre-images of open sets are open". Sans technical details, this becomes "for any open $U$, (for any $\epsilon$), you can find an open $V$ (find a $\delta$) such that $V =f^{-1}(U)$ and the above impication holds. $\endgroup$ – JuliusL33t Jun 19 '18 at 17:08
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    $\begingroup$ I've edited my answer with some details. $\endgroup$ – JuliusL33t Jun 19 '18 at 19:22
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    $\begingroup$ I made some further edits. Hope it helps. $\endgroup$ – JuliusL33t Jun 20 '18 at 16:38
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The definition of continuity at a point $a$ for a function $f\colon A\to B$ (say between metric spaces) is: for all $\varepsilon >0$ there exists $\delta>0$ such that if $d(x,a)<\delta$, then $d(fx,fa)<\varepsilon$. Now, notice that the $\varepsilon$ is used for a condition in the codomain and the $\delta$ is used for a condition in the domain. So the order of quantification is: for all something in the codomain, there is a something in the domain such that blah blah blah. The topological definition of continuity reads: for all open in the codomain, the inverse image is open in the domain. This shows that in fact the variance in both definitions is the same: continuity of a from $f\colon A\to B$ means you can pull information back from $B$ to $A$. So, the contravariance in the definition of topological continuity is not anything you haven't seen in the metric definition already. You just always thought the metric definition is variant, but it was contravariant all the time. The topological formulation simply makes it unavoidable to notice.

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    $\begingroup$ but the implication is still from balls in X to balls in Y (even though the quantifier is first for balls in Y). I think thats whats confusing me. I don't think I understand how the quantifiers work for the topological definition nor how they relate to the "original" definition of metric spaces. That connection would be really valuable to me, even if its just intuitive. $\endgroup$ – Pinocchio Jun 18 '18 at 14:16
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    $\begingroup$ @Pinocchio The implication $d(x,p) < \delta \Rightarrow d(f(x),q) < \epsilon$ really comes more from an expansion of the notion of $f^{-1}(B_\epsilon(q))$ being open, or more precisely from it being a neighborhood of $p$. Namely, if you expand "$f^{-1}(B_\epsilon(q))$ is a neighborhood of $p$," you get $\exists \delta > 0, B_\delta(p) \subseteq f^{-1}(B_\epsilon(q))$ which is equivalent to the part $\exists \delta > 0, \forall x, |x-p| < \delta \Rightarrow |f(x)-q| < \epsilon$ of the $\epsilon$-$\delta$ formulation of continuity. $\endgroup$ – Daniel Schepler Jun 18 '18 at 20:44
  • $\begingroup$ @DanielSchepler oh so the implication in the $\epsilon-\delta$ is just actually an explanation of the RHS of the topological definition of continuity (as you pointed out), it's not actually a re-write of the topological implication. That makes a lot more sense. I assumed the wrong thing about how the definitions were equivalent. $\endgroup$ – Pinocchio Jun 19 '18 at 15:36
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I think in the translation, it might help to separate out the direct generalization of the notion of "continuity at a point" from the general topological arguments that this generalization being true at every point is equivalent to the condition on inverse images of open sets.

So, recall that for a map $f : X \to Y$ between metric spaces, and $x_0 \in X$, we have $f$ is continuous at $x_0$ if and only if: $$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in X, d(x, x_0) < \delta \rightarrow d(f(x), f(x_0)) < \epsilon. $$ Now let us express what this condition is saying in terms of open balls: first, $d(f(x), f(x_0)) < \epsilon$ is equivalent to $f(x) \in B_\epsilon(f(x_0))$, which is further equivalent to $x \in f^{-1}(B_\epsilon(f(x_0)))$. On the other hand, $d(x, x_0) < \delta$ is equivalent to $x \in B_\delta(x_0)$. Therefore, $f$ is continuous at $x_0$ if and only if: $$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in X, x \in B_\delta(x_0) \rightarrow x \in f^{-1}(B_\epsilon(f(x_0))). $$ Now, the $\forall x \in X$ part is equivalent to a subset condition, so $f$ is continuous at $x_0$ if and only if: $$ \forall \epsilon > 0, \exists \delta > 0, B_\delta(x_0) \subseteq f^{-1}(B_\epsilon(f(x_0))). $$ Now, note that the $\exists \delta > 0, \ldots$ part is precisely equivalent by definition to: "$f^{-1}(B_\epsilon(f(x_0)))$ is a neighborhood of $x_0$." Furthermore, the collection of $B_\epsilon(f(x_0))$ for $\epsilon > 0$ is precisely the neighborhood basis at $f(x_0)$ coming from the metric on $Y$. To summarize, we have seen that more or less directly:

$f$ is continuous at $x_0$ if and only if for all basic neighborhoods $N$ of $f(x_0)$, we have $f^{-1}(N)$ is a neighborhood of $x_0$.


Now, not all topological spaces in general will have a natural system of neighborhood bases, so usually the generalization of continuity at a point to general maps of topological spaces will look something like:

Definition: Let $f : X \to Y$ be a map between topological spaces, and $x_0 \in X$. Then $f$ is continuous at $x_0$ if and only if one of the following equivalent statements is true:

  1. For every neighborhood $N$ of $f(x_0)$, we have that $f^{-1}(N)$ is a neighborhood of $x_0$.
  2. For every open neighborhood $N$ of $f(x_0)$, we have that $f^{-1}(N)$ is a neighborhood of $x_0$.
  3. (In the presence of a given system of neighborhood bases on $Y$:) For every basic neighborhood $N$ of $f(x_0)$, we have that $f^{-1}(N)$ is a neighborhood of $x_0$.

(Of course, I think in practice, most textbooks will likely just choose one of these conditions as the definition - in my experience, usually either (1) or (2) - and then prove the equivalence to the other conditions as separate results.)

Also, we have the general topological fact: "For any subset $U \subseteq X$, $U$ is open if and only if $U$ is a neighborhood of all of its elements." Using this, it is easy to prove the first equivalence in the below revised definition of continuity:

Definition: Let $f : X \to Y$ be a map between topological spaces. Then $f$ is continuous if and only if one of the following equivalent statements is true:

  1. $f$ is continuous at every point of $X$.
  2. For every open subset $V \subseteq Y$, we have that $f^{-1}(V)\subseteq X$ is open.
  3. (In the presence of a given basis for the topology of $Y$:) For every basic open subset $V \subseteq Y$, we have that $f^{-1}(V) \subseteq X$ is open.

(Of course, again most textbooks will present (2) as the definition of continuity, and then prove equivalence to (1) and (3) as separate results.)


Now, according to the translation above, the $\epsilon$-$\delta$ definition of continuity is most closely related to (1) above, with the continuity at a point $x_0 \in X$ being expanded from (3). Looking more closely at the initial expansion, we see that the overall structure "if $V$ is a basic open neighborhood of $f(x_0)$ then $f^{-1}(V)$ is a neighborhood of $x_0$" expands to the $\forall \epsilon > 0, \exists \delta > 0, \ldots$ part. Whereas the part the question is about, the part $d(x, x_0) < \delta \rightarrow d(f(x), f(x_0)) < \epsilon$, is actually part of the expansion of "$f^{-1}(V)$ is a neighborhood of $x_0$."

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The two definitions are equivalent to each other for metric spaces. To see that the first definition implies the second, let $\epsilon>0$ and $y=f(x)$. The open ball $B_\epsilon(y)$ is open in $Y$. Therefore $f^{(-1)}(B_\epsilon(y))$ must be open in $X$. Therefore, it contains the open ball $B_\delta(x)$ for small enough $\delta>0$. Since $B_\delta(x)\subset f^{(-1)}(B_\epsilon(y))$, we have found $\delta>0$ such that $c\in X, d(x,c)<\delta \implies d(f(x),f(c))<\epsilon$.

The reverse implication also uses an argument using open balls.

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I would have expected the definition to be the other way round

I take you to be proposing this:

$f\colon X\to Y$ is continuous if $f(U)$ is open for every open $U\subseteq X$

But that does not serve. In particular, consider constant functions. Constant functions are among those that meet our expectations for continuity, and constant functions over metric spaces are in fact continuous by the metric-space definition of continuity. But if $f\colon X\to Y$ is a constant function and $V \subseteq X$ is nonempty then $f(V) = \{k\}$ for some $k \in Y$, and in many cases we care about, such singleton sets are closed, not open.

On the other hand, consider a constant function $f$ defined as above, and let $U\subseteq Y$ be open. The preimage $f^{-1}(U)$ of $U$ is either $\emptyset$ or $X$, which are both open by definition in every topology over $X$, so the definition you started with serves for this example.

On the third hand, consider $f\colon \mathbb R \to \mathbb R$ defined by $f(x) = -1$ if $x \lt 0$ and $f(x) = 1$ if $x \ge 0$. To demonstrate that it is discontinuous, choose, say, the open interval $\left(\frac{1}{2},\frac{3}{2}\right)$. The preimage of that open set is the closed set $\left[0,\infty\right)$.

More generally, the definition captures the idea of a point of discontinuity in the range of the function, and that should seem natural, because that's what you look for when visually inspecting the graph of a function for discontinuities.

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Perhaps the following paper would be of interest to you:

Velleman, D. J. (1997). Characterizing continuity. The American Mathematical Monthly, 104(4), 318-322. Link.

Here is the beginning:

enter image description here

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    $\begingroup$ Looks like the result in that paper has been published (in greater generality) before: mathoverflow.net/questions/223708/… $\endgroup$ – Daniel McLaury Nov 17 '18 at 2:17
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    $\begingroup$ Theorems 1 and 2 are not the main results of the paper--they are just the motivation. The main result is that the answer to the question at the end of the first paragraph is, indeed, "no". In fact, there do not exists families of sets $\mathcal{F}$ and $\mathcal{G}$ such that a function $f : \mathbb{R} \to \mathbb{R}$ is continuous if and only if for every $X \in \mathcal{F}$, $f(X) \in \mathcal{G}$. $\endgroup$ – Dan Velleman May 10 at 14:25
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I think that understand you.

You have two topological spaces $(X,\tau)$ and $(Y,\tau')$ and a aplication continues f:$X \rightarrow Y$.

For the general definition of continues you can say:

$\forall x \in X , \forall G' \in \tau' : f(x) \in G', \exists G \in \tau : x \in G, f(G) \subseteq G' $. You can prove this using $G=f^{-1}(G') $.

And if you applied this to metric spaces obtains(I suppose that your p verified f(p)=q) your definition of continues function in metric spaces.

You ask why use implication for opens in $Y$ to $X$, and no for opens in $X$ to $Y$.

I give you some reasons:

1-Implication for opens in Y to X is more general because $f^{-1}(G) $ can be $\varnothing$ and you do not contemplate this case for opens in X to Y.

2-Implication for opens in X to Y say that $\exists$ some open that verified... but not say who is this open and for implication for opens in Y to X you know who is this open is $f^{-1}(G')$.

If we change the definition of aplication continues to: $f: X\rightarrow Y$ is continues if $f(U) \in \tau', \forall U \in \tau$.

We have ,for example, that a constant function can be no continue for example:

if we take the constant function 1 for $\mathbb{R}$ in $\mathbb{R}$ we have that $f((0,1))=\{1\}$ that is not open, then $f$ is not continue.

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The notion of topological space and the definition of a continuous function are certainly in the realm of 'abstract' mathematics. To say that a function $f$ is continuous mean that if points are close to other points then they don't get 'ripped away' when applying it - they 'follow the action' of $f$.

Now we can also define a topological space using closed sets. Hmm,

$\quad \text{Open sets: Take open subsets of the codomain back (in some way...)}$

so maybe?!?

$\quad \text{Closed sets: Take closed subsets of the domain forward (in some way)...}$

The OP will find this interesting:

A map is continuous if and only if for every set, the image of closure is contained in the closure of image

$\tag 1 \text{For any } A, \; f(\overline{A})\subseteq \overline{f(A)}$

or intuitively, all points 'close to' $A$ get mapped to points 'close to' $f(A)$.

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