My Topology Questions

**JasonRox** · Jan13-06, 11:06 AM

I have a few questions to ask. They are simple, but the purpose is to make sure I'm staying on track.

First, the definitions that I will use.

T2-Space or Hausdorff Space, find definition here.

Separable Space, find definition here.

Let X be the underlying set.

The questions or statements are...

1 - The Space of Reals is a T2-Space. This seems rather obvious since you can just construct the intervals for x and y (x<>y) such that they do not overlap. Hence, two neighbourhoods that are disjoint.

2 - The Discrete Topology is separable if and only if X itself is countable.

Note: A- is the intersection of the collection of closed subsets in X that contain A. Therefore, if A is closed, A- = A.

Note: A is dense in X if and only if A- = X.

Note: <> means not equal to.

Now, using the notes above.

X can be the only possible solution in this case because every subset is closed, therefore A- <> X, unless A=X. Therefore, if X is countable, it is separable.

Is that right?

3 - The Indiscrete Topology is separable if and only if there is countable subset in X. If X is the set of real numbers, then it is separable because the naturals numbers is a countable subset in R, and it is dense because N- = X, because X is the only set that contains N.

4 - The Space of Reals is separable. Since the rational numbers are countable and it is a subset of R, and R is the only closed set that contains Q then Q- = R. Hence, it is separable.

5 - The Finite Complement Topology is separable. If X is countable, then we are done. If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A.

Therefore, A~N (cardinality) because if this were not true, then A is finite, which is closed and A- <> X, so it is not dense. (A can not have higher cardinality of N because then it wouldn't be countable like we constructed.)

Is my thinking right?

EDIT: I am going to re-read my post later. I might have some mistakes that I don't know yet.

**matt grime** · Jan13-06, 11:27 AM

I can't tell if you are strictly disinguishing between finite and countable, in some cases you are and in some not.

4: in what topology.

It's correct if not written in the shortest way. Eg in cofinite topology, since only finite sets are closed, either X, if X is finite, or any countably infinite set is dense hence X is separable.

Personally I prefer =/= to be not equal rather than<> since that only really is computer speak for 'strictly greater than or strictly less than'

**JasonRox** · Jan13-06, 03:13 PM

What I mean about the Space of Reals, is the topology of intervals, which if you remember from my last question of topology.

Originally Posted by matt grime

I can't tell if you are strictly disinguishing between finite and countable, in some cases you are and in some not.

What do you mean?

**HallsofIvy** · Jan13-06, 03:21 PM

Originally Posted by JasonRox

I have a few questions to ask. They are simple, but the purpose is to make sure I'm staying on track.
First, the definitions that I will use.
T2-Space or Hausdorff Space, find definition here.
Separable Space, find definition here.
Let X be the underlying set.
The questions or statements are...
1 - The Space of Reals is a T2-Space. This seems rather obvious since you can just construct the intervals for x and y (x<>y) such that they do not overlap. Hence, two neighbourhoods that are disjoint.

More specifically, intervals centered on x and y with radius |x-y|/2 will work.

2 - The Discrete Topology is separable if and only if X itself is countable.
Note: A- is the intersection of the collection of closed subsets in X that contain A. Therefore, if A is closed, A- = A.
Note: A is dense in X if and only if A- = X.
Note: <> means not equal to.
Now, using the notes above.
X can be the only possible solution in this case because every subset is closed, therefore A- <> X, unless A=X. Therefore, if X is countable, it is separable.
Is that right?

Actually, it proves the otherway: X is separable only if it is countable.
"If X is countable, then it is separable" is true for any topology.

3 - The Indiscrete Topology is separable if and only if there is countable subset in X. If X is the set of real numbers, then it is separable because the naturals numbers is a countable subset in R, and it is dense because N- = X, because X is the only set that contains N.

?? Are you specifically talking about the real numbers with the indiscrete topology? If so then you should say that. Yes, the set of real numbers contains a countable subset and so is separable with the indiscrete topology. "because X is the only set that contains N" is not correct. I presume you meant "because X is the only set in the topology that contains N" or "because X is the only open set that contains N".

4 - The Space of Reals is separable. Since the rational numbers are countable and it is a subset of R, and R is the only closed set that contains Q then Q- = R. Hence, it is separable.

Yes, that's completely true.
5 - The Finite Complement Topology is separable. If X is countable, then we are done. If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A.
Therefore, A~N (cardinality) because if this were not true, then A is finite, which is closed and A- <> X, so it is not dense. (A can not have higher cardinality of N because then it wouldn't be countable like we constructed.)
Is my thinking right?[/quote]
"Therefore, A~ N"?? Didn't you want to prove "separable"?
How do you get " If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A." Aren't you assuming here that X is separable?

EDIT: I am going to re-read my post later. I might have some mistakes that I don't know yet.

**matt grime** · Jan13-06, 03:24 PM

In my language countable means in bijection with some subset of the naturals, and hence includes finite stuff. Some people disallow finite sets from being countable. You carefully distinguish between countable and finite at some points. I mean in particular the odd point where you use the condition 'iff it has a countable subset' for indiscrete: all non-empty sets have a countable subset in my opinion. Since you distinguish this then apparently you disallow finite countable sets. Then later you say 'let A be countable' and 'prove' that A is inbijection with N, but surely that is your definition of countable, if you're going to condition on something have a countable subset.

**JasonRox** · Jan13-06, 03:36 PM

Originally Posted by HallsofIvy

?? Are you specifically talking about the real numbers with the indiscrete topology? If so then you should say that. Yes, the set of real numbers contains a countable subset and so is separable with the indiscrete topology. "because X is the only set that contains N" is not correct. I presume you meant "because X is the only set in the topology that contains N" or "because X is the only open set that contains N".

Thanks, I see my error now. I was just using the real numbers as an example. So in the topology, X is the only closed set that contains N.

Just like matt grime pointed out before, but he deleted I guess. Every set has a countable subset, hence all Indiscrete Topology's are separable.

"Therefore, A~ N"?? Didn't you want to prove "separable"?
How do you get " If X is not countable, then we can create a countable subset (call it A) within X such that X is the only closed set that contains A." Aren't you assuming here that X is separable?

Let me explain a little bit better.

If X is countable, then it is separable because X is a countable subset, which is dence in X.

If X is not countable, then we can create a countable set (call it A) in X, such that X is the only closed set that contains A.

The reason why we must have A~N is because if it wasn't, then A would be finite because A is countable. Since A is finite, A is also closed, hence A is not dense in X.

This is why A must have A~N (cardinality). This assures us that no finite subset of X will contain A, which leaves us with only X as the closed set that contains A, which makes it dense in X. Since now A is a subset of X, it is countable, and is dense in X, we now have the definition of a separable topology.

**JasonRox** · Jan13-06, 03:37 PM

Originally Posted by matt grime

In my language countable means in bijection with some subset of the naturals, and hence includes finite stuff. Some people disallow finite sets from being countable. You carefully distinguish between countable and finite at some points. I mean in particular the odd point where you use the condition 'iff it has a countable subset' for indiscrete: all non-empty sets have a countable subset in my opinion. Since you distinguish this then apparently you disallow finite countable sets. Then later you say 'let A be countable' and 'prove' that A is inbijection with N, but surely that is your definition of countable, if you're going to condition on something have a countable subset.

That's the definition I use as well.

I apologize for any difficulties in my writing. I'll do my best to improve on my future questions.

**mathwonk** · Jan14-06, 02:26 AM

if you want to understand topology, i recommend you get away from this type of investigation as soon as you can, and start learning about covering spaces, fundamental groups, homology, cohomology, tangent and cotangent bundles, and the properties of specific spaces, like lens spaces, projective spaces, compact 2 manifolds, spheres, tori, lie groups, symmetric products, algebraic varieties, etc.

it is easy to be fooled into thinking topology is about definitions, when really it is about properties of ordinary spaces that occur in practice.

**HallsofIvy** · Jan14-06, 12:36 PM

Spoken like a physicist!

**matt grime** · Jan14-06, 12:45 PM

Well, how many times do really ever actually have to work out if you've just created a T3 space? I don't even know what a T3 space is, and I'm quite proud of that in a perverse way. In real life you only ever need the standard metric topology or some other topology of continuity of functions that you pick a posteri to make things work (eg topology of pointwise convergence, topology of uniform convergence), or the zariski topology, or the compact open topology. Very rarely do you pass outside these things. I would suggest you only need to know about 6 topologies to do maths. And you should learn the properties as needed.

As an example, over the last 6 months off and on I've been dabbling in algebraic geometry so that I can do some mirror symmetry stuff in the near future (interesting conferences in Seoul, San Francisco and Vienna this year for those thinking about working in this area) and finding it too dull for words because the textbooks were written in far too much generality. However by studying curves, ie dimension 2 stuff, alone I've now gotten round to understanding what Riemann Roch is really saying, I've gotten to like divisors since they are just points in dim 2 not bizarre codim 1 things. We only really look at curves anyway, and divisors there are very different from divisors in general.

Here the analogy is: who gives a monkey's about what separable is, in the abstract sense? All that's important is that the Hilbert Spaces we find occuring naturally need to have a countable dense subset to make their study nice, and lo and behold they do.

**mathwonk** · Jan14-06, 03:50 PM

well as a young person, after college and before grad school, i studied kelley's general topology and thought i was learning some topology. he had T1, T2, T3, T4 spaces in there, and I was a little worried i might not remember all the differences, or the right hypotheses for every theorem.
then later i was upset that in studying algebraic topology i did not know much about circles and spheres, and so on, much less lens spaces or projective space.
kelley did admit he meant his book to be titled "what every young analyst should know" and not what every young topologist or geometer should know, or maybe i would have skipped it.
since then i have almost never met a space that was not T2, i.e. in topology, hello, essentially all decent spaces are hausdorff. so maybe kelley was thinking of abstract harmonic analysts and people wanting to study "bornological spaces" and "barrelled spaces" and other far out wierd stuff of infinite dimensions.
normal people (by my definition) study much more mundane spaces.
now i admit that in abstract algebra geometry the "unmentionable" topology, to quote zariski, i.e. the zariski topology is not hausdorff. so what, you just have to understand the correct definition of hausdorff to fix that right up. i.e. affine algebraic varieties ARE hausdorff, they are just not T2. I.e. T2 is not the right definition of hausdorff. what you really want from "hausdorffness" is that two morphisms should agree on a clsed subset, so if they agree on a dense set they agree everywhere, stuff .like that. disjoint neighborhoods is just a tool in a certain setting, to get the more important consequences. when the setting changes, you have to change the definition to get those same consequences.

i.e. a traditional topological space is hausdorff (T2) if and only if the diagonal is closed in the product topology.

hence for an algebraic variety, the right definition of hausdorff, called separated, is that the diagonal be closed, where of course the product is NOT given the product topology, but is given its natural structure in the category of algebraic varieties, i.e. such that algebraic morphisms (not just continuous ones) into the product correspond to pairs of morphisms into the factors.

now suppose you have two morphisms of X into Y. Then you get a morphism of X into YxY, and the pullback of the diagonal is the set where the two morphisms agree, and if the diagonal is closed then the pullback is clsed, so the two morphisms agree on a closed set.

isnt that cool?

then compactness is also defined differently, i.e. a hausdorff topological space X is compact if and only if the projection map XxY-->Y is a closed map for every (hausdorff?) space Y. [I do not want to do the exercise right now.]

anyway, then the right definition of compact, for algebraic varieties, called proper, is that X is proper if and only if X is separated and every projection map XxY-->Y to another (separated?) algebraic variety Y is a closed map.
so the correct way to look at everything is not by what it "is" but what it "does", i.e. the categorical way. via maps, not open sets.

if you do this, as grothendieck emphasized, [and also hausdorff, but not bertrand russell, when hausdorff described numbers by saying he did not care what they were but how they behaved], then everything works.

**mathwonk** · Jan14-06, 03:53 PM

Halls, that is a great compliment, as physicists are intelligent people who are firmly grounded in reality.

Mathematicians also are on safer ground when they stick to the guidance found in observing natural examples.

As someone said about our mathematical forefathers of previous centuries, something like: they may have been somewhat unrigorous in their methods of proof, but their conclusions were supported by extensive experience and by computations with examples of such fundamental importance that they seldom went astray even when their proofs were lacunary.

**matt grime** · Jan14-06, 04:08 PM

that's a good one, the product zariski, for jasonrox.

Consider R, the zariski topology is defined via: K is closed iff K is the zero set of a polynomial - K_f ={x: f(x)=0}

The zariski topology on R^2 is defined as follows: the zariski closed sets are generated by the K_f {(x,y):f(x,y)=0}

By example show that the zariski topology on R^2 is not the product topology from the zariski topology on two copies of R.

It's not hard but is an (the only?) interesting example of a case where the product topology is not what you think it might be.

**mathwonk** · Jan14-06, 04:10 PM

to the OP: I realize this may probably not be the answer that interests you now, but I am like my father, who had me when he was 53 years old: I know you are just starting out, but i don't have time to wait for you to ask the questions I want to answer, I am going to answer them now, and someday when they become the questions you want answered, maybe you will recall the answers.

best wishes,

old poop

**JasonRox** · Jan14-06, 06:51 PM

Originally Posted by mathwonk

to the OP: I realize this may probably not be the answer that interests you now, but I am like my father, who had me when he was 53 years old: I know you are just starting out, but i don't have time to wait for you to ask the questions I want to answer, I am going to answer them now, and someday when they become the questions you want answered, maybe you will recall the answers.
best wishes,
old poop

I think the message is directed towards me.

I'm not sure whether or not this is the direction into topolopy. The author does say the text is aimed at students going in Real Analysis, and that it emphasizes applications to the space of reals (I mentionned earlier).

Topology seems like a very interesting area, but then again, where do I begin. When it comes to learning something new, I'm stuck guiding myself.

When I get home, like tomorrow, I will post the contents of the text.

Although the text might not lead me into topology, but I do think I will benefit from it regardless. The text is short, and only reads 150 pages. I hope to be done shortly, but rightfully.

I just ordered Young's book on Topology from Dover. From the reviews I've read, it sounds like a good text.

Anyways, topology does interest me. I'm most interested in studying infinite-dimensional vector spaces, which I'm still awaiting to do, and topology, but that's hard to say now that I don't really "know" what it is.

My interests to studying infinite-dimensional vector spaces comes from studying finite-dimensional spaces from linear algebra. I'll look into tensors as well when I get there.

Well, I guess that sums up my current interests.

I appreciate that you are all sharing your experiences, which is priceless.

Since, my professor is leading me into the study of infinite-dimensional vector spaces, so I believe he is heading me into the right direction. He has been very helpful on finding a list of books, and in things I should know.

May I ask where to begin for topology?

Maybe if I shared my goal, you can give me better advice.

My current goal is to just garnish enough knowledge to read through articles in math journals that may interest me.

The goal seems pretty straightforward to me, and it can be accomplished. It doesn't have to be a recent article either.

Anyways, that's about it.

Note to matt grime: I had interests in studying elliptic functions, but because the project that I was to follow with my prof, I decide to put it off for now. Since I know nothing about it, it would make no sense to say I'm interested in it. I know nothing about topology, but that's different... I think.

Thanks.

**matt grime** · Jan15-06, 04:19 AM

The thing here is that in some sense you aren't studying what a topologist would call the important bits of topology. Topolgist refers to an algebraic topologist you see, not a pointset top ologist. So the intersting stuff is (co)homology and such ie what you do with the topological space and not separability and those things. It is more the analysts who want to put topologies on strange things so that they can 'do analysis' in odd places. Obviously a topologist does know what pointset topology is, it's just that they almost always study one of very variations on compact, connected, path-connected spaces.