Thread: A Topological Trifle

So, let's talk about point set topology. First, a point set is, as you might have gathered, a set with no algebraic structure (well, let's say - I don't think that is the most general definition).

OK. Let be a set, and let denote the powerset on . Then one says that is a topology on iff the following are true:

finite intersections of elements in (sets, recall) are in ;

arbitrary union of elements in are in ;



.

The "indivisible" pair is called a topological space.

Before continuing, let me hammer this home. Whereas elements in are points, elements in are sets of points.

We shall that this implies that every element in is likewise a set.

Shortly we will allow this bit of terminological abuse: it is customary to refer to a topological space simply as, say (when is is, of course): one says "let be a topological space", the existence of and the underlying set being assumed. but we shan't doing that just yet.

OK? So the elements in (sets, recall), are called the open sets in . This may seem a bit weird first time around, but I will explain!

Recall we talked about the complement of a set. The closed sets in are those elements in which are the complement in of some set in . Jane would no doubt prefer to say that the closed sets in are elements in the set , and I think on this occasion I would have to agree

Example: Let and suppose that . These are the open sets in

The closed sets (complements in ) are .

From which I hope you can deduce that a set may open, closed, both or neither. Obviously, are both open and closed in any topology.

Which reminds me; I really ought to give some examples of topologies and topological spaces.

Right. Recall that is the set of real numbers Recall also that this set admits of an order. One says that the "standard" (or usual) topology on is given by , where these are the open sets in .

One calls this the topological real line (or real line for short), and interprets this as the union of the open intervals .

Now recall that the union of any arbitrary number of elements in is in . So, . Then and one calls this a closed set in

Does this make sense, or do I need a bucket of cold water throwing over me?

Originally Posted by Guitarist

Jane would no doubt prefer to say that the closed sets in are elements in the set , and I think on this occasion I would have to agree

No, I wouldn’t. I would prefer to say that the closed sets in are elements in the set . Okay, just nitpicking, but that’s what I would do.

Then

More nitpickig, sorry:

Anyway, this is a great thread! I’m looking forward to the next “lesson� already. :-D

So, as I don't have a hot date tonight (any volunteers?), let me continue to harangue you.

Let me first dispose of a couple of very uninteresting topologies on any point set.
Recall I said that a topology on the set is such that .

Then the topology on . where is called the discrete or concrete topology on this set. It is somewhat pathological.

By the same token, I will have that the topology on may be simply . This is called the indiscrete or trivial topology.. This is even less interesting.

Right.

I emphasized that the elements in a topology are sets of points, which is true. It is nonetheless true that the set upon which we are imposing a topology is a set of points.

So, I will make the following assertion: for any open set (i.e. any set ) that contains the point , I will call a neighbourhood of . (Note that some writers require the qualification "open neighbourhood"). Notice that need not be unique, so evidently, if

Now the level of abstraction climbs a little.

You may think it silly, but in any sort of space other than a metric space (and topological spaces may have a metric, but they don't need to), we need to be very sure what we mean when we say that 2 points are the same or different.

This opportunity is provided us by the so-called "separation axioms" of topology. These go by the catchy names of up to 4.

In fact, the only separation axiom of any real interest is the axiom. Any topological space satisfying this separation axiom is called a Hausdorff space, which I will explain another time (since my "hot date" - the bloke next door - has just arrived)

Well, OK, I see now I missed out something important, but since I started wittering about Hausdorff spaces, I had better finish.

A topological space is called a "Hausdorff" space iff, for any , and neighbourhoods (open sets) containing these points, I may find neighbourhoods such that .

Conversely, I will say that, if no such construction exists, then .

Note this well: if and this does not imply that , i.e this property is not transitive.

However, Hausdorff asserts that, nonetheless, there will be neighbourhoods such that .

OK so far?

Let me go back to the bit that should have followed the introduction of closed vs, open sets. Here, our intuition will be a reliable guide, provided we keeps our heads.

Again let be a topological space with (notice I haven't specified whether it is open or closed)

Then the interior of is the largest open set contained in , and is written . Equivalently, we may say that is the union of all open sets in . Evidently is open in , and if is open, , not otherwise..

Conversely (in a manner of speaking), the closure of , written , is the smallest closed set containing . Equivalently, is the intersection of all closed sets of which is a subset. Equally evidently, is a closed set, and if is closed, , not otherwise..

I now define the boundary of as .

This is merely a fancy way of saying that the boundary of a closed set is included in the set, the boundary of an open set is not.

The demonstration that this must be true is easy.

Anyone want a pop? Umm, I have just seen a question looming that I don't quite know how to answer: what if is both open and closed. Hmm... I'd better hit those books!

Originally Posted by Guitarist

This is merely a fancy way of saying that the boundary of a closed set is included in the set, the boundary of an open set is not.

Originally Posted by Guitarist

I have just seen a question looming that I don't quite know how to answer: what if is both open and closed. Hmm... I'd better hit those books!

The boundary of a “clopen� set (a set that’s both open and closed) is empty. (For a clopen set A, and .) So to qualify your first statement: The boundary of a closed set is included in the set; the boundary of an open set that is not closed is not.

Thank you for starting this thread, Guitarist. Unfortunately I'm going to be away for the next week, and without internet access. And then for about 5 days after that, though at that point I may have internet access. Hopefully I'll find time to give everything you've covered so far a good read-through tonight and get any questions I have posted before I leave tomorrow morning. Thanks again for doing this.

Chemboy: You got it! I shall be minimal on this subject for the next 5 days, so you don't get overwhelmed on your return. Have a good trip!

Miss Bennet: I would take it to be the greatest possible delicacy on your part if you never again, ever, used the word "clopen" in my presence; it is regarded in polite circles as being most frightfully vulgar.

Hehe!

Originally Posted by Guitarist

I would take it to be the greatest possible delicacy on your part if you never again, ever, used the word "clopen" in my presence; it is regarded in polite circles as being most frightfully vulgar.

I’m sorry, I haven’t a clue what you’re talking about. :?

Originally Posted by Guitarist

Chemboy: You got it! I shall be minimal on this subject for the next 5 days, so you don't get overwhelmed on your return. Have a good trip!

Thanks. I'm going to print what's been posted so far so I can look over it while I'm away.

I have internet access now and I'll be home Thursday, so feel free to fire up this thread again.

OK, Chemboy, let's try and pick up where we left off.

Recall we talked about the "Hausdorff property". This is one of a set of properties called "topological properties", and I now want to mention just a couple more such properties.

A topological space is said to be connected iff it cannot be written as the union of two non-empty disjoint sets. The intuitive content here should be clear: if , and , then I cannot "move" from a point in A to a point in B without falling into a chasm.

This is, in fact, a rather antiquated (but perfectly serviceable) definition. The better definition is as follows: a topological space is said to be connected iff the only subsets of that are both open and closed are and .

These two definitions are easily brought into register: Let be open. Let , and let (recall this is the the definition of disjointness). Then of necessity, , therefore A and B are both open and closed, and is not connected by either definition.

The other topological property I want to mention is compactness. Again, there are two parallel definitions, and I offer the oldest, and most intuitive, first.

A set is said to be compact iff, for every sequence in that has a limit, that limit is found in .

Again, the content should be clear: a compact set is "self-contained" with respect to limits.

The grown-up version of the same property will require a bit of a detour, which I am going to leave for now.

Let me know, anyone, if the above is less than clear

Definition: If is a sequence of points in a topological space X, a point is called a limit of the sequence iff given any open set U containing x, we can find a natural number N such that for all natural numbers , .

Note that such a limit x may not be unique. It would be unique in a Hausdorff space, but not for non-Hausdorff spaces in general.

Blah...I have questions but I need time to do a really thorough work-through of the material and actually formulate what questions I need to ask. I'm sticking with it though, keep forging ahead. Just wanted you to know I'm into it despite my lack of on-topic posts.

OK, Chemboy, got that. I am reluctant to proceed until I hear your questions, otherwise you will be in over your head and then lose interest.

In order to encourage your questions, let me say this. Some of this stuff is quite hard on first encounter, and you are likely to need some help.

Please don't be shy about asking - I know I am not the best explainer in the world. I am also a terrible show-off, so feed my ego!!!

Originally Posted by Guitarist

A set is said to be compact iff, for every sequence in that has a limit, that limit is found in .

That's interesting. I've never seen compact spaces being defined this way before.

Ok, I just spent ages on it, and I'm all caught up and ready to move on.

Chemboy: I hope you are right, as, I warn you, I tend get a little grumpy if I discover I have been pissing into the wind.

Faldo: I hope to make the connection between the different definitions of compactness clear in what follows.

So. Let's start like this. Let be a topological space. Then a cover for (or covering of ) is simply a collection of sets whose union is all of . Obviously, every space has a cover in this sense.

A sub-cover for is any subset of this collection whose union is still all .

If every set in the the cover of is open in , then the cover is said to be an open cover.

Recall that the axioms of topology allow that the (possibly) infinite union of open sets in is still open. We now suppose that there is some finite subset of our open cover that still covers , and we have our defintion:

is said to be "compact" iff every open cover has a finite sub-cover.

Let's see how this works. We will work with the (topological) real line , with the standard topology i.e. .

This is self-evidently an open cover for . Equally self-evidently, the union, say. is a finite sub-cover for , hence the real line is compact.

Hmm, I had hoped to go a little further, but now I have to fire up the BBQ, cook a few dead creatures, and get drunk.

Later

Originally Posted by Guitarist

This is self-evidently an open cover for . Equally self-evidently, the union, say. is a finite sub-cover for , hence the real line is compact.

No, it definitely is not!

For a space to be compact, every open cover must have a finite subcover. You only considered one open cover and one finite cover. That does not prove compactness. As a matter of fact, is an open cover for which has no finite subcover. Therefore is not compact.

Originally Posted by JaneBennet

Therefore is not compact.

But every convergent sequence of real numbers converges to a real number, doesn’t it?

I agree that is not compact by the finite-subcover definition. Therefore the finite-subcover definition can't be equivalent to the limits definition, can it, since is not compact under the one definition but is under the other? Unless there are some special conditions imposed on those topological spaces.

Well, the finite-subcover definition is the orthodox defintion of compactness for topological spaces in general. If there is any discrepancy with other definitions, the orthodox version stands.

The fact that every convergent sequence of real numbers converges to a real number has nothing to do with compactness – that’s the completeness property of the real numbers. In fact, completeness only makes sense for metric spaces (where the convergent sequences are more properly called Cauchy sequences); it doesn’t make sense for general topological spaces. Guitarist may have his own reason for introducing sequences and limits in topological spaces. Let’s wait and see.

Originally Posted by JaneBennet

The fact that every convergent sequence of real numbers converges to a real number has nothing to do with compactness

There's a definition of a type of compactness that depends on convergence of sequences: in a sequentially compact space, every sequence has a convergence subsequence. Sequential compactness and finite-subcover compactness are equivalent in metric spaces, but not in topological spaces. Actually I don't even think the two are related in topological spaces.

Oh dear, Guitarist, oh dear. What a fool you are!

Jane was completely right - the real line is about as non-compact as they come. Sooooo sorry for my idiocy!

This is because, as a topological space, it is not bounded. The proof of this actually uses the so-called Heine-Borel theorem - the closed and bounded sets of are compact - whose proof is neither easy nor relevant here.

Faldo is right too: the historical definition of a compact space that I gave - contained limits - is now thought of as being a special case called "sequentially compact", which normally mandates a metric space.

I would prefer to leave history aside, and move on, given that the motivation for this thread was a rather superficial survey of the subject.

So. We will agree that the definition of a compact space has to do with finite subcovers of open covers.

However, it might just be worth pointing out the following.

Strictly speaking, the only gadgets allowed to have limits are functions. So when we talk about a limit of a sequence in some space/set , actually we are talking about a limit of the function defined as .

Thus one can define a limit of a sequence in any topological space by this map, which will have a limit if, for every neighborhood , there is some above which all elements of the sequence belong to .

This doesn't require a metric on , nor does it require that have any order. We do have an order, and a metric, on though,

So the notion of a sequence in (or some subset theerof), and of a limit , in a topological sense, is induced by the map

Originally Posted by Guitarist

Strictly speaking, the only gadgets allowed to have limits are functions. So when we talk about a limit of a sequence in some space/set , actually we are talking about a limit of the function defined as .

Thus one can define a limit of a sequence in any topological space by this map, which will have a limit if, for every neighborhood , there is some above which all elements of the sequence belong to .

This doesn't require a metric on , nor does it require that have any order. We do have an order, and a metric, on though,

So the notion of a sequence in (or some subset theerof), and of a limit , in a topological sense, is induced by the map

Ok, I need to make sure I'm clear on this stuff. I don't really have any questions, I just can't really wrap my head around it as is... And could you please explain the type notation so I'm clear on what that's saying? I have a good idea but it can't hurt to ask.

All sequences in general are defined as functions with domain . The image of the natural number n under such a function is denoted with a subscript n and called the nth term of the sequence.

Example: Consider the sequence of real numbers <1, 4, 9, 16, 25, …>. This is primarily a function where . However, we don’t usually write terms of the sequence in this way; instead we write .

So if X is a topological space, a sequence of points in X is the ordered set of points , which is the function where is written . 8)

for every neighborhood , there is some above which all elements of the sequence belong to

There is slight typo here: the “there is some � should just be “there is some �. Hope it didn’t confuse you too much.

Originally Posted by JaneBennet

for every neighborhood , there is some above which all elements of the sequence belong to

There is slight typo here: the “there is some � should just be “there is some �. Hope it didn’t confuse you too much.

Oh, it did. Thanks for the correction. But I still don't understand it. haha.

Originally Posted by Chemboy

Oh, it did. Thanks for the correction. But I still don't understand it. haha.

Did you understand the other definition I gave earlier in the thread, though?

If is a sequence of points in a topological space X, a point is called a limit of the sequence iff given any open set U containing x, we can find a natural number N such that for all natural numbers , .

Guitarist’s defintion is the same as mine, only expressed in different words.

I think what I really need is just to understand the "...we can find a natural number N such that for all natural numbers , ." bit. Maybe I need it to be given to me in non-mathematical terms...though that's probably difficult for you to do. :wink: And I suppose since this is mathematics I should be learning it in mathematics terms anyway.

Right, let’s take , and suppose you have the sequence , so . You can see that the limit of this squence is 0. Now how to interpret this in terms of topological spaces?

First, let’s look for open sets of containing 0. Any open set of is the union of open intervals, so any open set containing 0 must include an open interval containing 0. Hence it suffices to consider open intervals containing 0.

Any open interval containing 0 is of the form where . It’s b we’re interested in. Given any , we can always find a natural number N such that (this is called the Archimedean principle). Then for all natural numbers n greater than N, we have that ; in other words, belongs to the interval for all natural numbers .

In other words, given any open interval I containing 0, is in I for all n greater than some natural number N, which is to say that 0 is a limit of the sequence in .

Hope that makes it a bit clearer.

ok, I get that. But what if it's not a case like and is something like the sequence where or something like that, where larger values don't yield smaller values for ? Could I have an example of something like that?

The sequence doesn’t have a limit. Not every sequence in has a limit.

Try and think of limits in as ordinary limits you encounter in calculus. These familiar limits are actually the same as the limits of considered as a topological space. The latter is used the definition of limits in general topology because not all topological spaces are like .

Originally Posted by JaneBennet

The sequence doesn’t have a limit. Not every sequence in has a limit.

Try and think of limits in as ordinary limits you encounter in calculus. These familiar limits are actually the same as the limits of considered as a topological space. The latter is used the definition of limits in general topology because not all topological spaces are like .

Oh, duh. Right. So will these sequences that we've been talking about always be decreasing, or will they ever increase and approach a finite number? I'm almost there, don't worry.

It doesn’t matter at all. The example I gave happens to be decreasing, but that is just a coincidence. A convergent sequence in can be increasing, decreasing, or neither.

Indeed, you could try this little fun exercise. Let be any function from to ; i.e. is either 1 or −1 for any natural number n. Prove that the sequence where has limit 0.

Originally Posted by JaneBennet

It doesn’t matter at all. The example I gave happens to be decreasing, but that’s just a coincidence. A convergent sequence in can be increasing, decreasing, or neither.

Indeed, you could try this little fun exercise. Let be any function from to ; i.e. is either 1 or −1 for any natural number n. Prove that the sequence where has limit 0.

I think I'll pass on the 'little fun exercise.' I'm all set now. Shall we proceed?

Are we ok to continue, Guitarist? I apologize if I'm being impatient, I'm just very eager to keep learning this, and other things. Just want to make sure we're good to go after that detour.

Sure, gimme a little while

OK, let's assume this little side-show is out of the way. Chemboy wants to "proceed", which I will do with some trepidation.

Let's talk now about maps between topological spaces. But as we all seem a little rusty of some elementary stuff, and moreover, terminology differs sometimes. let me start waaaay back.

Suppose that are sets. Then one calls the map a set function. The set is referred to as the domain of , the set as its co-domain.

Now. Given some the element is called the image in of the element under the function .

The range of is that subset of where we expect to find all elements of the form

Set functions functions can do some weird and wonderful things, but there are 2 things they may not do: they may not "overlook" any element in the domain.

That is, our function must at least try to find an image point in . It may fail to do so, in which case one says that our function is undefined at, say, , which is simply shorthand for saying that the image element .

The other thing that functions of this sort are forbidden from doing is to find multiple image points for a single element .

Moving on. I will define 2 properties of set functions.

The function is said to be surjective if, for all , there is/are some element(s) such that . Notice that this does not exclude the possibly that, say, , in spite if the fact that .

On the other hand, we define an injective function as follows:

If, for any image elements , this implies that , we will say that this function is injective.

Notice that this means that, for any image element there at most one element . Notice also that this does not not imply that for all there is some

A function that is both surjective and injective is said to be bijective. That is, for all there is at least one, and at most one, image point , and for all , there is at least one, and at most one, such that .

Which brings me to my final piece of pedantry, which is, in fact, the main point of this over-long ramble.

We will suppose the set function . Then, using all the above, we will say that every point , for which , is an element in the pre-image set in X.

Notice that this set can never be empty, but it may only have a single member.

There is a truly nasty notation for this: one says that is the pre-image set in X of y under this function.

I say it's nasty notation because has nothing, absolutely nothing to do with inverse functions; in fact it is not a function of any sort - the construction is not "deconstructible", it is merely a label for a set.

Oh fuck, I hear snoring in the background..........

Well, it's not my snoring you're hearing.

Good man. Let me first alert you to a slight mis-speak in my last.

Originally Posted by I

That is, our function must at least try to find an image point in . It may fail to do so, in which case one says that our function is undefined at, say, , which is simply shorthand for saying that the image element .

My last phrase should really be " the image element ".

So in topology, one is mainly concerned with continuous functions.

First I'll tell you why, then give an intuitionist view of what a continuous function is and finally we'll do it properly.

If it is the case that all functions from the space to the space are continuous, and all these functions have continuous inverses, we may think of this as implying that can be continuously "deformed" into and back again.

(This gives rise to the old joke about the topologist who can't tell the difference between a coffee cup and a donut).

Very loosely speaking, the set function will be continuous if, whenever and are arbitrarily "close" to each other in X, then and are arbitrarily "close" to each other in Y.

This view needs to be modified somewhat: first because topological spaces need not understand what the word "close" means - we say there is no metric.

Second, because topology deals in sets of sets, rather than with sets of points. So here's our definition.

The function is said to be continuous if, whenever and is an open neighourhood of , then the pre-image neighbourhood is open in X. In other words, our function maps open sets to open sets.

Notice this implies that continuous functions respect the complement, so one could just as easily say that continuous functions map closed sets to closed sets, which is what Jane did in another thread.

Suppose now there is some .Likewise, if it is the case that whenever is open in X, is open in Y, we will say that is continuous and that, moreover, if and , we will say that are mutual and continuous inverses.

Confusingly, irritatingly, one then writes .

This is the rather tortuous topologist's definition of a continuous bijection, and they say that our spaces are thereby homeomorphic - the topologist's version of isomorphism: for each point there is unique point , with continuous maps between them, and vice versa

Ummm. My next-door-neighbour has been pouring beer down my throat all afternoon, so there are bound to be some errors in the above. Ah well, Jane and Faldo will no doubt be the first to pounce...........

Originally Posted by Guitarist

Ummm. My next-door-neighbour has been pouring beer down my throat all afternoon, so there are bound to be some errors in the above. Ah well, Jane and Faldo will no doubt be the first to pounce...........

Uh, don't rely on me to proofread your posts though. What I know about point-set topology is largely limited to boring definitions and theorems, so there's still a lot in this fascinating topic for me to learn and I'm looking forward to learning more. I'm much better at algebra - that's my strongest subject - but not topology.

I'll be away for the next 4 days without internet access. I'm good with everything in your last post, so feel free to press on, and don't worry about taking it easy, unless of course you feel like it. I wouldn't mind a bunch of fun stuff to read through when I get home.

Know what folks? I'm beginning to think I'm pissing in my own boots here.

I find it impossible to believe that anyone, anyone, could fully absorb the contents of this thread without a single question.

I confess I am not the sharpest knife in the drawer, but I find some of it quite hard to get my head around. Are all of you really that much smarter than me? It's not impossible.........

Look. I am well used to starting threads which die - it is common currency on fora like this. But I will get extremely grumpy if I feel I am being strung along here to no real purpose.

So there.

(I should probably set some exercises and insist they are completed before proceeding - we'll see)

Here are some exercises for anyone who is interested.

1. Let be topological spaces. If and are continuous, prove that is continuous. (Hint: If , .)

2. Suppose is a topological space in which every open set is closed. Show that is a -algebra over .

3. Show that is a disconnected subspace of .

Ooh, wicked. wicked Jane - exercises!

I encourage you (all?) to try ex. 1 - it is a cleverly thought-out test of our understanding of continuity, and it also demonstrates a rather general general set of "phenomena" found in many areas of mathematics which go by the name of "pull-backs". Gwaan, give it a go - it's easy-peasy.....

Anyway, let me apologize for my outburst last night (bad day at work), and then go on to explain how we make new topological spaces from old ones.

This forces me to plug a gap I should have filled, oh, ages ago. Let's start here......

For any structured set, I can define some subset which generates, by repeated applications of the operation that defines our theory, any and all elements in that structured set.

In most cases (monoids, groups, rings, fields, ......) this subset is simply called the "generator". In the theory of vector spaces this set is called a set of basis vectors, or "basis" for short.

In general, for any generator we establish a certain "minimizing criterion" (I invented this term), by which I mean that we require the set to be large enough to do what it says on the tin, but no larger. Again, for example, the set of basis vectors for a vector space is that set of linearly-independent vectors for which, by the usual process of vector addition and scalar multiplication, will give me any vector in the space.

Likewise for topological spaces. Let's forget the shorthand we have been using, and remind ourselves that a topological space is properly written as , where is a point set, and are the open sets in

(Jane uses for this, which is probably more correct - ah well).

So, we now wish to find a generator - a basis - for .

We will define a collection - a family - of open subsets of . Then is a basis (of neighbourhoods) for iff, for any open set (neighbourhood) there is some such that .

It should be easy to see that, for all , any neighbourhood is the union for some (Err, note we may have i = j, here)

It should also be clear that, by the axioms of our theory, that, for any , then there is some .

I hope you can see that this defines a generating set - a basis - for .

So. How do we make new topological spaces from old? Aah, wait and see, as I have train to catch

Originally Posted by JaneBennet

1. Let be topological spaces. If and are continuous, prove that is continuous. (Hint: If , .)

Oh bollocks to it, let's do this

We will suppose that are sets, and that are set functions.

Now let us suppose that . Then by definition of a function, , and therefore . Then it is a notational necessity (not convention, as some would have you believe) we will have the composite function .

We will now suppose that are topological spaces. We are given that is continuous, id est, for some open set , the pre-image set is open.

We are also given that is continuous, which implies that, for the open set then is open in .

Notice that, by my above "necessary notation" this means that the image of, say, under the composite map has a pre-image .

See how our composition is turned around? That a form of pull-back.

So there

Originally Posted by JaneBennet

3. Show that is a disconnected subspace of .

I can certainly say that not everything I've learned has been easy to comprehend the first time around, but I find that if I give it enough thought and stick it out I always get things in the end, and that's why I haven't really been asking any questions. I also have the feeling that I'm just pretty good at grasping concepts like what has been presented so far.

So now, I actually have a question... must the basis of a topological space always contain the singleton sets?

Originally Posted by Chemboy

So now, I actually have a question... must the basis of a topological space always contain the singleton sets?

Hi Chemboy. I emphasized that word in your quote to remind myself that I wasn't perhaps as clear on the following point as I might have been.

In general, there will be many possible bases for the same topology on the same set, so the word "must" isn't quite appropriate. In fact I can think of only one topology where there is no choice. This is the so-called concrete, indiscrete or trivial topology on X i.e. ; the only possible basis here is X itself.

(This free choice of bases is also true for vector spaces, btw)

So the answer to your question is that, if singletons are open in some , i.e. , then they may be in the basis, but they don't need to be.

In facr, it will be illuminating to change the emphasis on this last statement. Let be a set, and let be possible topologies on , Let be bases for respectively.

Then whenever

a) for each and each , there is some such that , and

b) for each with there is some such that .

That is to say, if I know about the basis, then I know pretty much all I need to know about the topology.

Anyhoo... we are now in a position to see how to make new topological spaces from old ones. (Actually I know of 3 ways, but we're only going to do one here)

Recall the Cartesian product of sets. , whose elements are of the form . We want to find a topology for this beast. It is tempting to say that, if are topological spaces, then the topology on is . Sadly, this will not, in general be a topology.

But we can do this. Let be bases for respectively. Let be open neighbourhoods of for all .

Then, by the definition of a basis, there will be such that .

So, the product topolgy on will be generated by the basis for all

Of course, we don't need to restrict ourselves to binary products.

This is about as far as I want to go with this rather sketchy romp over topology. You at least have enough grounding to make some sense of anything you choose to read up on the subject. Beyond this point it gets rather specialized, and, to me, not too interesting.

We also have enough grounding to make sense of the theory of manifolds, which I am tempted to start in an equally superficial way.

Anybody any thoughts? Questions?

Originally Posted by Guitarist

, whose elements are of the form . We want to find a topology for this beast. It is tempting to say that, if are topological spaces, then the topology on is . Sadly, this will not, in general be a topology.

It won’t?

I was under the impression that it does work for finite product spaces at least. In particular, it works for just , the product of just two spaces. It’s only when you try it for “infinite” products that it breaks down.

I’m going to have to think about this for a while. In the mean time, I have to be off somewhere …

Originally Posted by Guitarist

So the answer to your question is that, if singletons are open in some , i.e. , then they may be in the basis, but they don't need to be.

Ok. I thought the "there is some such that " kind of implied that the singleton sets would have to be included... because what if the open set (neighborhood) containing contains only ? Then it seems to me that because of the portion, since can't be larger than since it's a subset or equal to it, then it has to be equal, and has to be simply ... Should've said all that in my first post...ah well.

Originally Posted by Chemboy

Ok. I thought the "there is some such that " kind of implied that the singleton sets would have to be included... because what if the open set (neighborhood) containing contains only ?

Okay, let’s see if I can put it in a different way from Guitarist.

Originally Posted by Guitarist

We will define a collection - a family - of open subsets of . Then is a basis (of neighbourhoods) for iff, for any open set (neighbourhood) there is some such that .

Let be a family of open subsets of . Let . Now suppose is any neighbourhood of . Then, given such a , suppose we can find an open set in the family such that and . Then the family is called a basis for the topology on .

Note that this doesn’t say anything about singleton sets. I am not sure what there is about singleton sets to worry about, really. :?

Originally Posted by JaneBennet

Originally Posted by Guitarist

, whose elements are of the form . We want to find a topology for this beast. It is tempting to say that, if are topological spaces, then the topology on is . Sadly, this will not, in general be a topology.

It won’t?

I was under the impression that it does work for finite product spaces at least. In particular, it works for just , the product of just two spaces. It’s only when you try it for “infinite” products that it breaks down.

I’m going to have to think about this for a while. In the mean time, I have to be off somewhere …

I hope i'm not stealing Guitarists thunder here but is only a base for a topology on the product (the infinite case is different, but more interesting). An interesting example is any open ball in , which is obviously not a member of

Thanks, RR! :-D I was racking my brain over this last evening.

My problem was in not realizing that a subset of a Cartesian product does not have to be a Cartesian product itself. It was a long time before I became aware of that. :?

Welcome back! Please post more.

Originally Posted by river_rat

I was under the impression that it does work for finite product spaces at least. In particular, it works for just , the product of just two spaces. It’s only when you try it for “infinite” products that it breaks down.
quote]

I hope i'm not stealing Guitarists thunder here but is only a base for a topology on the product (the infinite case is different, but more interesting).

The infinite case is only a little different, so newbies need not be too afraid of that case. It can be very interesting indeed.

Since I think more is coming I won't steal anyone's thunder on the subject.

For those who want to learn this subject, and are willing to work I have a recommendation for a book. It is at least available from used book sites. The book is Topology, an Outline for a First Course, by L.E. Ward. It is written for a topology class to be taught using the "Moore Method", which makes it good for self study. In the Moore Method topology is taught by having the students work out all of the proofs and all of the examples on their own -- no references, no consultations with others and most importantly no lectures by the instructor. So the book provides the definitions and statements of the theorems and examples. But it is up to the reader to provide all of the proofs. This is made a bit easier by the way the book is ordered so that by time the student reaches a particular point in the text he has proved the theorems neceessary for the proof of the next one. The book is pretty dense, just over 100 pages goes from prerequisites (elementary set theory) through basic homotopy groups. A formal class from the book takes about a year. And just to make things interesting there are a couple of miinor errors that the student can catch and fix.

Originally Posted by DrRocket

The infinite case is only a little different, so newbies need not be too afraid of that case. It can be very interesting indeed.

That’s a great relief then.

Well, I’m mainly interest in just the case of the product of two spaces. That’s because I’ve been developing a thread on algebraic topology here, and, as you know, the idea of a product space is crucial in the definition of a homotopy.

When I realized that my assumption about the requirements for the topology of a product space was all wrong, I was thrown into consternation.

Fortunately, thanks to the useful notion of topology bases, it looks like my erroneous assumption can be patched up without much difficulty.

I now have a question of mine own. I have a text here that says that a basis for the product topology on the (countably) infinite product of sets is given by where and for all but a finite number of .

Though this is entirely consistent with the construction of the basis for a (countably) infinite product of vector spaces, the reason given (in my text) is that otherwise Tychonoff's thm. (product of compact spaces is compact) would not hold in the infinite case.

This seems completely perverse to me - changing the definition to make some theorem hold - huh?. Surely the correct view would be that, if the theorem doesn't hold in the infinite case, that's just tough tits for the theorem?

Anyway, I agree with Jane - good to see you again river_rat. Try to stick around a bit (so long as you don't mention a certain recent Test series........)

Originally Posted by Guitarist

I now have a question of mine own. I have a text here that says that a basis for the product topology on the (countably) infinite product of sets is given by where and for all but a finite number of .

Though this is entirely consistent with the construction of the basis for a (countably) infinite product of vector spaces, the reason given (in my text) is that otherwise Tychonoff's thm. (product of compact spaces is compact) would not hold in the infinite case.

This seems completely perverse to me - changing the definition to make some theorem hold - huh?. Surely the correct view would be that, if the theorem doesn't hold in the infinite case, that's just tough tits for the theorem?

Anyway, I agree with Jane - good to see you again river_rat. Try to stick around a bit (so long as you don't mention a certain recent Test series........)

Tychonoff's theorem holds for an arbitrary product of compact spaces, not just a countably infinite one.

I don't know which text you are reading (I suspect it might be Hocking and Young), but I am not sure that I agree that the "reason" for the definition is the validity of Tychonoff's theorem. I don't think Hocking and Young actually say that the Tychonoff theorem is the motivation for the definition of the product topology (aka the Tychonoff topology). The product topology in which the inverse image of open sets with respect to the projection maps form a sub-basis is the coarsest topology in which the projection maps are continuous. I think that is the real motivation, and the Tychonoff theoreom on compactness of a product of compact spaces is just a bonus.

Originally Posted by JaneBennet

Originally Posted by DrRocket

The infinite case is only a little different, so newbies need not be too afraid of that case. It can be very interesting indeed.

That’s a great relief then.

...

Fortunately, thanks to the useful notion of topology bases, it looks like my erroneous assumption can be patched up without much difficulty.

An even more useful notion is sub-bases. A sub-base is just a set of subsets of a given space, and it generates the coarsest topology that contains that class. Basically arbitrary unions of finite intersections.

The product topology is then simply the topology generated by inverse images of open sets under the projections onto each "coordinate". It is the coarsest topology that makes all of the projections continuous.

Originally Posted by JaneBennet

Originally Posted by Guitarist

, whose elements are of the form . We want to find a topology for this beast. It is tempting to say that, if are topological spaces, then the topology on is . Sadly, this will not, in general be a topology.

It won’t?

I was under the impression that it does work for finite product spaces at least. In particular, it works for just , the product of just two spaces. It’s only when you try it for “infinite” products that it breaks down.

There's a very obvious reason why can never be a topology for any topological space: it does not contain the empty set! It does contain the ordered pair but that's not the same thing as .

No, no, no. I did not mean the Cartesian product of topologies! If and I was thinking of as a possible topology for . is definitely in this set. would be .

But now that you mentioned it … I see that not only Guitarist but river_rat has written as well.

Originally Posted by river_rat

is only a base for a topology on the product

river_rat, I don’t think you mean , but rather . Can you confirm that this is indeed what you mean? :?

Originally Posted by Faldo_Elrith

There's a very obvious reason why can never be a topology for any topological space: it does not contain the empty set! It does contain the ordered pair but that's not the same thing as .

is not an element of { : is open in space 1 and is open in space 2}. But is. So you do get in your basis. What you don't get are some unions of cross products, and that is what is missing. For instance if you take cross products of open subset of the line to get open sets in the plane, you only get rectangles. To get discs you need to take unions.

Note that is just the set of functions with domain {1,2} and range , which is . Or you can look at it as the set of ordered pairs (x,y) where x and y each belong to which is .

is actually an element of {}X{} where you have to be careful to notice that {} is not empty because it contains .

Originally Posted by DrRocket

Originally Posted by Faldo_Elrith

There's a very obvious reason why can never be a topology for any topological space: it does not contain the empty set! It does contain the ordered pair but that's not the same thing as .

is not an element of { : is open in space 1 and is open in space 2}. But is. So you do get in your basis. What you don't get are some unions of cross products, and that is what is missing. For instance if you take cross products of open subset of the line to get open sets in the plane, you only get rectangles. To get discs you need to take unions.

Note that is just the set of functions with domain {1,2} and range , which is . Or you can look at it as the set of ordered pairs (x,y) where x and y each belong to which is .

is actually an element of {}X{} where you have to be careful to notice that {} is not empty because it contains .

That’s basically what I just said in my previous post.

Originally Posted by JaneBennet

Originally Posted by DrRocket

Originally Posted by Faldo_Elrith

There's a very obvious reason why can never be a topology for any topological space: it does not contain the empty set! It does contain the ordered pair but that's not the same thing as .

is not an element of { : is open in space 1 and is open in space 2}. But is. So you do get in your basis. What you don't get are some unions of cross products, and that is what is missing. For instance if you take cross products of open subset of the line to get open sets in the plane, you only get rectangles. To get discs you need to take unions.

Note that is just the set of functions with domain {1,2} and range , which is . Or you can look at it as the set of ordered pairs (x,y) where x and y each belong to which is .

is actually an element of {}X{} where you have to be careful to notice that {} is not empty because it contains .

That’s basically what I just said in my previous post.

Yes it is. I am a little slow with TEX and was composing my piece when you posted yours. Sorry to have stepped on your post. It was not intentional.

Oh, I wasn’t complaining. In fact, it’s great to have the same problem looked at from different viewpoints (with the same conclusion being reached). :-D

Originally Posted by JaneBennet

river_rat, I don’t think you mean , but rather . Can you confirm that this is indeed what you mean? :?

Yep, its topological shorthand

Some more on the product topology on an infinite family of spaces: There are a number of different topologies you can place on the product of spaces: the box topology, the (normal) product topology etc. The one that got called the product topology was the one that was most useful (as happens with most things in mathematics) and the Tychonoff theorem is a tremendously powerful and useful bit of mathematics. It is the basic element of the stone-cech compactification of a completely regular space and the Banach–Alaoglu theorem for example.

Originally Posted by river_rat

There are a number of different topologies you can place on the product of spaces: the box topology, the (normal) product topology etc. The one that got called the product topology was the one that was most useful (as happens with most things in mathematics) and the Tychonoff theorem is a tremendously powerful and useful bit of mathematics. It is the basic element of the stone-cech compactification of a completely regular space and the Banach–Alaoglu theorem for example.

It would be nice if you could start a new thread on this. Would you like to? :P

Originally Posted by river_rat

Originally Posted by JaneBennet

river_rat, I don’t think you mean , but rather . Can you confirm that this is indeed what you mean? :?

Yep, its topological shorthand

Some more on the product topology on an infinite family of spaces: There are a number of different topologies you can place on the product of spaces: the box topology, the (normal) product topology etc. The one that got called the product topology was the one that was most useful (as happens with most things in mathematics) and the Tychonoff theorem is a tremendously powerful and useful bit of mathematics. It is the basic element of the stone-cech compactification of a completely regular space and the Banach–Alaoglu theorem for example.

It turns out that the product topology provides a topology on a product of a countable family of lines that makes it homeomorphic to a separable Hilbert space. Just one sereidipitous aspect of that topology.

This statement is true, but it is not obvious. Far from it. I don't know the proof, but it is somewhat famous result due to Dick Anderson.

Hi DrRocket, welcome to the forum!

I'm going to take the liberty of expanding on an earlier post of yours, as I think it may have been a little dense for the casual reader - I trust you will forgive me in advance (especially if I mistake your meaning)

Suppose are sets, with their Cartesian product. All such products come equipped with maps of following sort: (this is equally true in the case of countably infinite products).

These maps are called "projections", and their images are sometimes called "coordinates". It is easy to see that these projections are surjective. Then for, say, to be a good topology on the set , there is the requirement that this is the "coarsest" topology for which these projections are all continuous open mappings.

What does this mean? Recall that any topology on any set is a subset of the powerset on , i.e..

Then, if it is the case that admits of the topologies one says that is coarser that whenever [/tex].

It should be self-evident that the coarser topology has the fewer elements, that is has a smaller number of open sets than .

Putting this all together, we see that we will require that the space has just enough open sets for, say, whenever .

And, since it seems obvious that the coarser the topology, the smaller the basis set, this roughly corresponds to my earlier ad hoc assertion that a basis is as large as it needs to be, but no larger.

Forgive my (attempted) intuitive style - but this was the original purpose of the thread.

But I still cannot quite see why the above implies that, for the countably infinite set product we must have that for all but a finite number of in the basis for the topology on

Originally Posted by Guitarist

...

But I still cannot quite see why the above implies that, for the countably infinite set product we must have that for all but a finite number of in the basis for the topology on

If you want those projection maps to be continuous, then the inverse image of any open set in a "coordinate space" under any projection must be open. So each of the is open. Next a topology requires that a finite intersection of open sets must be open so each where for all but a finite number of must also be open. That is enough for a base, since all that remains is arbitrary unions of these guys to get to a topology. Any class of sets (I am using the word class here as a synonym for set) that is closed under finite intersections can be a base for some topology, the topology generated by that class. Here you are starting with what is known as a sub-base and from that taking finite intersections to form a base.