*(Note: in the original version of this, I made an absolutely **huge** error. One of my faults in discussing topology is scrambling when to use forward functions, and when to use inverse functions. Continuity is dependent on properties defined in terms of the *inverse* of the function; I originally wrote it in the other direction. Thanks to commenter Dave Glasser for pointing out my error. I’ll try to be more careful in the future!)*
I’m going to spend a bit more time talking about what continuity means; it’s a really important concept in topology, and I don’t think I did a particularly good job at explaining it in my first attempt.
Continuity is a concept of a certain kind of *smoothness*. In non-topological mathematics, we define continuity with a very straightforward algebraic idea of smoothness. A standard intuitive definition of a *continuous function* in algebra is “a function whose graph can be drawn without lifting your pencil”. The topological idea of continuity is very much the same kind of thing – but since a topological space is just a set with some additional structure, the definition of continuity has to be generalized to the structure of topologies.
The closest we can get to the algebraic intuition is to talk about *neighborhoods*. We’ll define them more precisely in a moment, but first we’ll just talk intuitively. Neighborhoods only exist in topological metric spaces, since they end up being defined in terms of distance; but they’ll give us the intuition that we can build on.
Let’s look at two topological spaces, **S** and **T**, and a function f : **S** → **T** (that is, a function from *points* in **S** to *points* in **T**). What does it mean for f to be continuous? What does *smoothness* mean in this context?
Suppose we’ve got a point, *s* ∈ **S**. Then f(*s*) ∈ **T**. If f is continuous, then for any point p in **T** *close to f(s)*, f-1(p) will be *close to* *s*. What does close to mean? Pick any distance – any *neighborhood* N(f(s)) in **T** – no matter how small; there will be a corresponding neighborhood of M(*s*) around s in **S** so that for all points p in N(f(s)), f-1 will be in M(*s*). If that’s a bit hard to follow, a diagram might help:

To be a bit more precise: let’s define a neighborhood. A neighborhood N(p) of a point p is a set of points that are *close to* p. We’ll leave the precise definition of *close to* open, but you can think of it as being within a real-number distance in a metric space. (*close to* for the sake of continuity is definable for any topological space, but it can be a strange concept of close to.)
The function f is continuous if and only if for all points f(s) ∈ **T**, for all neighborhoods N(f(s)) of f(s), there is some neighborhood M(s) in **S** so that f(M(s)) ⊆ N(f(s)). Note that this is for *all* neighborhoods of *all* points in **T** mapped to by f – so no matter how small you shrink the neighborhood around f(s), the property holds – and it implies that as the neighborhood in **T** shrinks, so does the corresponding neighborhood in **S**, until you reach the single points f(s) and s.
Why does this imply *smoothness*? It means that you can’t find a set of points in the range of f in **T** that are close together, but that weren’t close together in **S** before being mapped by f. f won’t put things together that weren’t together originally. And it won’t pull things apart that weren’t
close together originally. *(This paragraph was corrected to be more clear based on comments from Daniel Martin.)*
For a neat exercise: go back to the category theory articles, where we defined *initial* and *final* objects in a category. There are corresponding notions of *initial* and *final* topologies in a topological space for a set. The definitions are basically the same as in category theory – the arrows from the initial object are the *continuous functions* from the topological space, etc.

1. dave glasser

Um, I’m not so sure about your definition. Choosing M(f(s)) = T always would make any f “continuous”. You really need a definition that talks about “for every neighborhood M of f(s), there’s a neighborhood of s that’s mapped into M”, not the other way round. ie, it’s a universal quantifier over the neighborhood in T and an existential one over the neighborhood in S, not the other way round.

2. ObsessiveMathsFreak

Yes, I think you need to change the function in the definition from f to f^-1. It’s classic that in topology many things are defined by inverse functions. I think a definition might be that for any e > 0 there exists d > 0 with N(s)⊆B(s,d) such that f(N(s)) ⊆ B(s,e). But that’s probably wrong.
Personally I find topology infuriating because of all the potential pitfalls, “between the lines” definitions and of course the extreme abstractness. It’s common to find topology books written with no diagrams at all! Which I’m sure is revealing of certain mental shortcomings with the author.

“Which I’m sure is revealing of certain mental shortcomings with the author.”
As I’ve heard it there are people who can efficiently reason by language (or whatever nonpictural process) alone. Perhaps one may find them concentrated in areas what suit their intelligence better, such as perhaps topology, logic, et cetera. More power to them, its different but doesn’t seem to be a serious shortcoming.
Until perhaps when they want to inform of their work. I’m rather visual oriented myself, and if I can’t assemble some sort of process and/or object image I’m usually quite helpless. But, who has the shortcoming then? 😉 But I agree that books without helpful pictures are inefficient or difficult for many.

4. Daniel Martin

What other folks said.
You’ve gone and reversed your definition from what you want it to be. Let me show you by way of example that your definition is off:
Consider the metric space formed by the surface of the unit sphere, with the metric being the restriction of the normal R3 metric. Now, under your definition, any function from some other metric space to this one is continuous. All you do is choose M(f(s)) to be the set of points within a distance 3 of f(s). Choose this neighborhood regardless of what N(s) is, and the definition is trivially satisfied.
In particular, this bit of your text:

it implies that as the neighborhood in S shrinks, so does the corresponding neighborhood in T, until you reach the single points s and f(s).

Is just at this point wishful thinking – your definition implies no such thing. There’s nothing in there to force M(f(s)) to shrink if N(s) does – the “corresponding neighborhood” could remain identical no matter how small N(s) gets.
One of the best type of problems I remember from one of my undergrad classes was “prove, or disprove and salvage”. They’d give you a statement and you had to either prove it or disprove it, and if you disproved it you then had to tweak it slightly so that it was proveable. So let’s salvage this definition.
Basically, the problem is that you put the “for all” qualifier in the wrong spot. A workable definition is: for all points s ∈ S, and for all neighborhoods M(f(s)), there is a corresponding neighborhood N(s) such that f(N(s)) ⊆ M(f(s)).
Now, because the forall is on M, and not on N, we can say that in order for this to hold, it has to hold as M shrinks down to an arbitrarily small size. This time though, that forces N to get small enough to keep f(N) inside M. I’m not going to go as far as you went above, though, because you only want to count single points as neighborhoods if you want single points to be open sets. Without going into details, you almost certainly don’t want to do that.
This also changes some of the hand-wavy intro text from:

Suppose we’ve got a point, s ∈ S. Then f(s) ∈ T. If f is continuous, then for any point p close to s, f(p) will be close to f(s). What does close to mean? Pick any distance – any neighborhood N(s) no matter how small; there will be a corresponding neighborhood in M(f(s)) of T so that f(p) will be in M(f(s)). If that’s a bit hard to follow, a diagram might help:

To:

Suppose we’ve got a point, s ∈ S. Then f(s) ∈ T. If f is continuous, then for any point p close to s, f(p) will be close to f(s). What does close to mean? To phrase it properly, we have to phrase it in terms of guarantees – p being close enough to s guarantees that f(p) is close enough to f(s). Pick any distance – any neighborhood M(f(s)) no matter how small; there will be a corresponding neighborhood in N(s) of S so that f(p) will be in M(f(s)) for every p ∈ N(s). If that’s a bit hard to follow, a diagram might help:

Fortunately, the diagram still holds – after all, I’ve only changed the order in which you pick the neighborhoods, not where they sit.
I also note that you don’t connect this back with the standard topological definition of continuity, that inverse images of open sets are open.

5. Matt

There’s the classic story of the topologist who 1)draws a diagram, 2)proves a theorem, and then 3)erases the diagram. It’s frustrating for anyone who tries to learn topology by picking up a text, unless you’ve been told, off the record, that part of what you need to learn is just this kind of practice.
About ‘neigborhoods’– I learned that the neighborhood of a point is an open set that contains the point. Sometimes you see the neighborhood of a point defined as a set that contains an open set that contains the point. In any case, the idea of a neighborhood is fundamental and, importantly, topologically invariant.

6. Daniel Martin

Personally I find topology infuriating because of all the potential pitfalls, “between the lines” definitions and of course the extreme abstractness. It’s common to find topology books written with no diagrams at all! Which I’m sure is revealing of certain mental shortcomings with the author.

One of the best topology books I’ve ever seen is called “Counterexamples in Topology” – it lays out where many of those pitfalls are, and what they lead to. Having clearly in your mind where you’d end up if you did follow one of those potential pitfalls helps greatly. It also helps to butress the extreme abstraction to have concrete examples of what happens when the abstractions get tweaked slightly.
As another advantage to someone who can’t justify spending all free cash on math books, it’s a Dover book, so it’s cheap. (~\$11, and very portable) It’s also incredibly dense, and needs to be read in very small doses and digested. (good for me on the train since I then don’t have to read much while in motion)

7. JST

Yes! Neighborhood of x means an open set containing x. The single mainest thing I loved about topology was the simple notion of neighborhoods & continuity. Especially after all those For-every-epsilon-there-is-a-delta-such-that’s I had to struggle with in Real Analysis. It was almost as if Analysis was a weed-out class and they only taught us the simple elegant stuff after the hazing of epsilon-deltas.

8. JST

Daniel: The extreme abstractness is the best thing about Topology! You just have to make sure your diagrams are sufficently blobby and generic. 🙂

9. Mark C. Chu-Carroll

Yeah, you’re right, I blew it. Inverse functions appear all the time in topology, and one of my problems in dealing with topology is that I constantly scramble where the inverses should be.

“It’s frustrating for anyone who tries to learn topology by picking up a text, unless you’ve been told, off the record, that part of what you need to learn is just this kind of practice.”
There’s that too – the usual science culture practice of elimination of all the real work leading up to the result.
I guess I have to learn to draw something topologically equivalent to the proof authors diagrams. 😉

11. Daniel Martin

Some of your explanatory text still appears to be backwards; I’d change this:

The function f is continuous if and only if for all points f(s) ∈ T, for all neighborhoods N(f(s)) of f(s), there is some neighborhood M(s) in S so that f(M(s)) ⊆ N(f(s)). Note that this is for all neighborhoods of all points in S – so no matter how small you shrink the neighborhood around s, the property holds – and it implies that as the neighborhood in S shrinks, so does the corresponding neighborhood in T, until you reach the single points s and f(s).
Why does this imply smoothness? It means that in the mapping from S to T, there is no neighborhood of points that are mapped to be close together in T, but scattered all around in S. You can start with s and f(s), and smoothly expand around it, and you’ll always get mappings from S to T that don’t scatter.

To:

The function f is continuous if and only if for all points f(s) ∈ T, for all neighborhoods N(f(s)) of f(s), there is some neighborhood M(s) in S so that f(M(s)) ⊆ N(f(s)). Note that this is for all neighborhoods of all points in T – so no matter how small you shrink the neighborhood around f(s), the property holds – and it implies that as the neighborhood in T shrinks, so does the corresponding neighborhood in S, for arbitrarily small neighborhoods.
Why does this imply smoothness? It means that in the mapping from S to T, there is no neighborhood of points that are mapped to be close together in S, but scattered all around in T. Look at the definition: it says that you can guarantee f(p) will be close to f(s) so long as p is close enough to s. Finding that M(s) is all about determining what “close enough” means. You can start with s and f(s), and smoothly expand around it, and you’ll always get mappings from S to T that don’t scatter.

12. ParanoidMarvin

Slight nitpick: smoothness is actually a fundamental concept used in differential topology with a concrete definition, so I would be careful with using it here.

“f won’t put things together that weren’t together originally.”
Is that correct, though? According to the definition several points si can be mapped onto the same point f(s), ie f(s1) = f(s2) = … = f(sn) = f(s). Why do we otherwise need to discuss that part of bijectivity or state f-1 continuous for a homeomorphism?
For example, wikipedia mentions “The third requirement, that f−1 be continuous, is essential. Consider for instance the function f : [0, 2π) → S1 defined by f(φ) = (cos(φ), sin(φ)). This function is bijective and continuous, but not a homeomorphism.” Here f puts 0 and points close to 2π close together by continuity. ( http://en.wikipedia.org/wiki/Homeomorphism; I *still* can’t find my old book to steal ideas from…)

14. Darin Brown

I’m glad you’re attempting to have a “math blog” at Seed magazine, Mark. You’re screwing it up so badly, any reasonable mathematician who has the endurance to read through your attempts to relate what is basically bonehead first-year graduate material (if not undergraduate), let alone the equally comic attempts of people to “correct” you, should be so impressed as to come to no other conclusion but that this blog is a joke.
Seeing as it took several HIV apologists TOGETHER with their collective intellect several days or so to arrive at a correct understanding of *continuity*, I wouldn’t expect anyone to take your “AIDS denialist” threads seriously AT ALL, even if they did contain a valid point or two. Besides, anyone who can see how long it took you to understand “continuity”, surely can take the time to figure out the subtle but erroneous reasoning of Duesberg and/or Noble. As well as which error is more important and devastating, as well as which questions are left UNASKED by Noble’s argument.

15. Chris Noble

Besides, anyone who can see how long it took you to understand “continuity”, surely can take the time to figure out the subtle but erroneous reasoning of Duesberg and/or Noble. As well as which error is more important and devastating, as well as which questions are left UNASKED by Noble’s argument.

Why didn’t you correct Duesberg’s error when I pointed it out?
Why was it that no numerate “rethinkers” bothered to take part in the discussions on NAR?
If you are going to respond to any of these points why don’t you respond on the appropriate thread here or even on the new thread at NAR rather than on this unrelated thread?

Darin:
“You’re screwing it up so badly, any reasonable mathematician who has the endurance to read through your attempts to relate what is basically bonehead first-year graduate material (if not undergraduate), let alone the equally comic attempts of people to “correct” you, should be so impressed as to come to no other conclusion but that this blog is a joke.”
You don’t understand this. It is considered to be much harder to write expository and correct educational material than research papers. Any researcher can do papers, fewer can explain ideas simply, and very few can do it correctly.
At the basic level of any theory there are still difficult questions since it also hits the limits of what is known. For example, why is time and energy generally illdefined in general relativity, which purports to be a theory about spacetime and massenergy? Answering such simple questions about the theory basis is often difficult or impossible. But this is what educational material often must demarcate.
Here I’m questioning one sentence, while the exposition is rather simple and correct. This has nothing to do with Mark’s expert math criticism of IDiots and HIVers, which is more general and at times deeper than most.
Criticising ID and HIV denial isn’t rocket science, since the basic ideas and the expositions are so trivially wrong. Tell us what he specifically does wrong in that task, instead of criticising the much more difficult task of writing educational material.

17. Mark C. Chu-Carroll

Darin:
I’ve got only one response for you. One of the marks of a true researcher or scientist is the ability to *admit* when they made a mistake.

18. John Armstrong

If you have trouble remembering that morphisms in the category of topological spaces and continuous maps (I know you like categories) pull back the notion of open sets, you may like to know that there’s an equivalent (but more complicated to describe) definition of a topology on a set: convergence of nets.
As you probably know, the notion of open sets generalizes open balls in metric spaces, which essentially generalizes the notion of epsilon-delta proofs of continuity from calculus.
But there’s another sort of limit in calculus: the limit of a sequence. We say that if as n gets high enough the sequence {xn} eventually stays within any given epsilon-neighborhood of x that the sequence converges to x. The topology of the real numbers is completely determined by the collection of sequences which converge, and to which limits those that do converge, converge to.
Unfortunately, sequences aren’t quite enough in general, which is where nets come in. The point of a sequence is that it’s the image of a mapping from the directed set of natural numbers. Nets just generalize this to any directed set.
First off, a directed set is a nonempty set with a reflexive and transitive binary relation “≤”. Then for any elements a and b, there is a c such that a≤c and b≤c.
So take your favorite directed set A and map the element a to xa. Given an open set U the net is “eventually in U” if there is an a so that for every b≥a the element xb is in U. The net converges to x if and only if for every open neighborhood U of x, the net is eventually in U.
Now, here’s the magic: instead of continuous functions being the ones that pull back the notion of an open set, they’re the ones that push forward the notion of convergence of nets! Given the collection of all nets (/me waves his hands frantically at the set theory) in X and the knowledge of which nets converge and the limits for those that do converge, a function f is continuous if it preserves limits of nets, For an algebraist who’s used to morphisms preserving structure (like group homomorphisms do for group multiplication) this is a much more preferable view.
Of course, it’s a lot harder to work with and prove anything in this view than with the open set view, but you can’t have everything. Personally, I take solace in the notion that directed sets are actually special kinds of categories, and that nets are almost a sort of functor, giving hope that even point-set topology may one day fall to the powers of category theorists.

19. ValJean

“Why does this imply smoothness? It means that you can’t find a set of points in the range of f in T that are close together, but that weren’t close together in S before being mapped by f. f won’t put things together that weren’t together originally. And it won’t pull things apart that weren’t close together originally. (This paragraph was corrected to be more clear based on comments from Daniel Martin.)
Unfortunately, it’s still not quite right, although DM’s correction works. It’s helpful to look at it this way: what we’re trying to say is that continuous functions preserve “closeness”, in the sense that points x and y in S that are close together can’t get mapped to points f(x) and f(y) in T that are far apart. This is half of what you’re saying, but the other half isn’t part of continuity. Certainly, we consider f(x) := |x| to be continuous, but while 100 and -100 are far apart, |100| == 100 == |-100|. Requiring distant points to remain distant is an additional property that is not part of continuity; it leads to the notions of homeomorphism and embedding, and is in essence a requirement that the inverse of f is also continuous.
(I would also like to point out that smooth has a precise meaning in differential geometry, which is not far removed from topology, and that smoothness is a much stronger property than continuity– it states that the function is everywhere differentiable (which requires continuity), and that the derivative is also smooth.)