What is the Significance of Completely Regular Spaces?

[WannabeNewton]

The only place where the word line shows up is "real line" and that's just another word for \mathbb{R}...and as mentioned before "for any" is the same thing as "for all".

 

[friend]

The only place where the word line shows up is "real line" and that's just another word for \mathbb{R}...and as mentioned before "for any" is the same thing as "for all".

Thank you. That helps.

I understand now that the definition means the property holds for any and all points x outside F, and for any and all points y inside F. I'm starting to see the global nature of it now. OK, now I'm concerned about the uniqueness of F. What if we wanted to choose x that is inside F. That would mean we would have to choose another F such that this new x is outside the new F. Does this mean that there must be more than one available F to choose from so that we can select x anywhere in the topology and still be gauranteed a closed F somewhere such that x is outside it? How many F's can we choose from? Must there be F's all over the place, as long as the topology is used to select them?

 

[WannabeNewton]

The property must hold for ALL closed subsets of the topological space.

 

[friend]

The property must hold for ALL closed subsets of the topological space.

Yes, but if in the definition, the x outside F can be anywhere in the topology, then it seems necessary that there be at least two closed F's such that they don't intesect. Otherwise there would be some x not outside an F.

I can understand that if the topology is the power set of a connected region, then you could construct an F anywhere of any shape and would guarantee that you could always select an F and an x that fulfills the definition of complete regularity.

But what if the topology is not the power set? Then the topology might restrict the choice of F's. And it may not be guaranteed that you could always select an F for any x to fulfill the definition. In that case, some x in that topology would be excluded from the definition.

 

[micromass]

Yes, but if in the definition, the x outside F can be anywhere in the topology, then it seems necessary that there be at least two closed F's such that they don't intesect. Otherwise there would be some x not outside an F.

I can understand that if the topology is the power set of a connected region, then you could construct an F anywhere of any shape and would guarantee that you could always select an F and an x that fulfills the definition of complete regularity.

But what if the topology is not the power set? Then the topology might restrict the choice of F's. And it may not be guaranteed that you could always select an F for any x to fulfill the definition. In that case, some x would be excluded from the definition.

There is no need to construct an $F$ or to select an $F$.
The definition says something about for all $F$. Nobody cares whether such an $F$ exists or whether there are many of them.

 

[WannabeNewton]

Yes, but if in the definition, the x outside F can be anywhere in the topology, then it seems necessary that there be at least two closed F's such that they don't intesect. Otherwise there would be some x not outside an F.

There are always at least two closed subsets: the empty set and the entire set itself. If these are the only two closed subsets, i.e. the trivial topology, then your over - arching statement is vacuously true.

 

[friend]

There is no need to construct an $F$ or to select an $F$.
The definition says something about for all $F$. Nobody cares whether such an $F$ exists or whether there are many of them.

This seems to contradict your previous statement

We've been trying to tell you this for 3 pages already. $F$ is supposed to be a closed set. Being closed depends crucially on the topology. So yes, the sets $F$ that we consider will be dependent on the topology.

If the F's depends on the topology, are there at least two that do not intersect so that the definition holds for any x in the topology?

 

[friend]

There are always at least two closed subsets: the empty set and the entire set itself. If these are the only two closed subsets, i.e. the trivial topology, then your over - arching statement is vacuously true.

There are no x outside the entire set. There are no y inside the empty set.

 

[micromass]

If the F's depends on the topology, are there at least two that do not intersect so that the definition holds for any x in the topology?

Nobody cares whether there are two that do not intersect. Nothing in the definition of complete regular spaces demands that there must be closed sets that do not intersect. Nothing in the definition of complete regular spaces even asks that there are any closed sets to begin with!

This thread has been going on for 3 pages already, and you seem to have a huge amount of misuderstandings. It is my guess that you never had a formal topology course or that you never read a rigorous topology book. If you want to understand topology, then you need to actually study topology. Just reading wikipedia pages are not going to do it. You will end up with a very superficial understanding.

I advise you to get a topology book and to start working through it. There are many good books out there such as Munkres.

There is no real point in continuing this thread if you don't start studying topology from a actual textbook. Wikipedia contains much useful information, but it can never be the substitute for a textbook and it can never be a replacement for solving exercises. So if you keep using wikipedia as your primary source of learning, then I don't think we should discuss further.

 

[friend]

Nobody cares whether there are two that do not intersect. Nothing in the definition of complete regular spaces demands that there must be closed sets that do not intersect. Nothing in the definition of complete regular spaces even asks that there are any closed sets to begin with!

This thread has been going on for 3 pages already, and you seem to have a huge amount of misuderstandings. It is my guess that you never had a formal topology course or that you never read a rigorous topology book. If you want to understand topology, then you need to actually study topology. Just reading wikipedia pages are not going to do it. You will end up with a very superficial understanding.

I advise you to get a topology book and to start working through it. There are many good books out there such as Munkres.

There is no real point in continuing this thread if you don't start studying topology from a actual textbook. Wikipedia contains much useful information, but it can never be the substitute for a textbook and it can never be a replacement for solving exercises. So if you keep using wikipedia as your primary source of learning, then I don't think we should discuss further.

I'm sorry you're feeling frustrated. I was doing my best to articulate the issues to try and better understand them. And I thought we were making progress. But perhaps we should move on.

I'm trying to understand the continuous nature of the map in the definition, f:X→[0,1]. It seems there is a sharp border between 0 outside and 1 inside, the edge of the closed set F. So how can it be continous?


PS. Next time you feel frustrated, try just simply saying, "I don't know how to make it more clear." And leave it at that. Maybe someone else will be able to explain in terms I might understand.

 

[WannabeNewton]

You keep thinking of this in terms of the euclidean topology on the real line. Continuity is heavily dependent upon the topology. It is not hard to come up with topologies that allow for weird continuous maps, even on the real line. How can you claim something isn't continuous without even knowing what the topology is on the set? Again, I stress that you are stuck on the euclidean topology on the real line but topology is an extremely elegant subject that generalizes continuity to a great extent.

 

[friend]

You keep thinking of this in terms of the euclidean topology on the real line. Continuity is heavily dependent upon the topology. It is not hard to come up with topologies that allow for weird continuous maps, even on the real line. How can you claim something isn't continuous without even knowing what the topology is on the set? Again, I stress that you are stuck on the euclidean topology on the real line but topology is an extremely elegant subject that generalizes continuity to a great extent.

Well I guess it's my turn... What? As I understand it, this definition concerns manifolds which are locally euclidean. So at the point on the boundary between F and not F is a point of locally euclidean geometry, right? Maybe you need to add the Hausdorff property first before if becomes a manifold. But let's assume the easiest case, how is continuity defined for f there?

 

[WannabeNewton]

The general definition of continuity for topological spaces is very simple: f:X\rightarrow Y is continuous if \forall U\subseteq Y open, f^{-1}(U) is open in X.

 

[strangerep]

PS. Next time you feel frustrated, try just simply saying, "I don't know how to make it more clear." And leave it at that. Maybe someone else will be able to explain in terms I might understand.

And maybe when someone tries to explain that you're confused about a basic essential point, perhaps you should be more receptive.

Example: in one of your earlier posts in this thread, we find the following

You are highly mistaken friend,[...] The notion of being open makes no
sense without a topology and the notion of being closed doesn't make sense
without there being a topology. Things like boundary are topological notions.
You are looking at it backwards: you cannot talk about an open set without
there being some topology not the other way around. [...]

Well, perhaps the topology was implicit in those references. I don't want to
get distracted. The question I'm concerned about at the moment is whether a
particular open set belongs to every possible topology constructed on some
background. OK , it can belong to some topology, but does it belong to every
possible topology? I would think that the fact that you can construct
different topologies from the same background means that any particular open
set does not necessarily belong to every possible topology, right?

"Distracted"??

You misunderstand the point that the term "open set" is meaningless in isolation.
Micromass, WannabeNewton, and dx have been trying to explain this to you
over and over again. But you have some kind of mental block about this (probably
because Wiki is slightly misleading on this point), hence no one can de-confuse you.

If you want to continue, please participate in this quiz:

1) What is a "topology"?

2) What is a "topological space"?

3) What is an "open set"?

4) What does "continuous" mean (in the context of general topology)?

(Answers must be given off the top of your head, not merely paraphrased from elsewhere.
"I don't know" is an acceptable answer to any or all of these.)

[BTW, @Micromass: maybe this thread should be moved into the Topology forum?]

 

[micromass]

PS. Next time you feel frustrated, try just simply saying, "I don't know how to make it more clear." And leave it at that. Maybe someone else will be able to explain in terms I might understand.

I'm not frustrated at all. My post contained genuine advice. It was not an attempt to put you down, I am merely trying to help you.
At the moment, I feel that only going through a good topology book will really help you.

 

[Bacle2]

I'm sorry you're feeling frustrated. I was doing my best to articulate the issues to try and better understand them. And I thought we were making progress. But perhaps we should move on.

I'm trying to understand the continuous nature of the map in the definition, f:X→[0,1]. It seems there is a sharp border between 0 outside and 1 inside, the edge of the closed set F. So how can it be continous?PS. Next time you feel frustrated, try just simply saying, "I don't know how to make it more clear." And leave it at that. Maybe someone else will be able to explain in terms I might understand.

I'm sorry, but I think Micromass has a point; given that many have tried unsuccessfully to help, I think is reasonable to believe your background in topology may have something to see with your trouble in understanding.

Like WBN said, there are spaces in which the topology does not allow for a visualization or representation in which the notion of sharp borders is meaningful or illustrative. Maybe the best you have is that, since F is closed , it contains its limit points so that x is not a limit point of F. In this, maybe weak sense, F and x are separated. Maybe it would help for you to consider what happens in a completely-regular space with an open set U and a point x not in U, and, in general, to consider why each part of the hypothesis is necessary in order to understand better.

 

[friend]

The general definition of continuity for topological spaces is very simple: f:X\rightarrow Y is continuous if \forall U\subseteq Y open, f^{-1}(U) is open in X.

Right, that sounds familiar. So the question would then be how is f:X→[0,1] continuous in the definition for complete regularity shown here? I'm understanding that f=0 for every x not in the closed set F, and f=1 for every y in F. I'm not seeing how one can construct open sets with just two values 0 and 1. If the set X were mapped to the continuous interval [0,1], I could see how an open set in one could be mapped to an open set in the other. But for all x in the topology, either x is in F or it is not; it seems to get mapped to either 0 or 1, and not to some continuous value in the interval. So how can a 2 valued map have open sets?

 

[friend]

I'm not frustrated at all. My post contained genuine advice. It was not an attempt to put you down, I am merely trying to help you.
At the moment, I feel that only going through a good topology book will really help you.

I appreciate your efforts. Thank you. But this whole discussion is an example of the trouble I have with topology. The presentations and books I've seen don't really show how the concepts connect to the functions one would see in physics. And I find it hard to keep it in my head because it doesn't seem relevant. However, if you know of a book that provides these connections to everyday functions and their domains and ranges, that would shed a whole new light on the subject.

Other than that, I don't find it useful to answer by basically throwing the book at me... telling me I should go and start from scratch. Instead, it might make it easier to answer if you could cut and pasted from some on-line source. Then I could read it and the context it's in and decide for myself if I need to go back to the very beginning.

 

[WannabeNewton]

Right, that sounds familiar. So the question would then be how is f:X→[0,1] continuous in the definition for complete regularity shown here? I'm understanding that f=0 for every x not in the closed set F, and f=1 for every y in F. I'm not seeing how one can construct open sets with just two values 0 and 1. If the set X were mapped to the continuous interval [0,1], I could see how an open set in one could be mapped to an open set in the other. But for all x in the topology, either x is in F or it is not; it seems to get mapped to either 0 or 1, and not to some continuous value in the interval. So how can a 2 valued map have open sets?

Oh boy. There are a lot of misconceptions floating around here and it isn't going to be easy to deconstruct them. I really think the best thing to do would be to properly learn topology. A forum can only do so much but I'll try to help. No one said the map had to be surjective, just that it had to be continuous and continuity does NOT mean open sets in X are taken to open sets in Y, this is an open map and you are mixing the two up. Open sets are not a property of maps so I don't know what you mean by "how can a 2 valued map have open sets?". If it is what I think it is, you are just asking how can a map take a topological space into an image containing only two values and still be continuous. Well it is trivial to come up with such maps, and even easier you can just look at constant maps which are continuous and whose image is a singleton. For example, it is very easy to show that X is a disconnected topological space if and only if there exists a non - constant continuous map f:X \rightarrow \left \{ 0,1 \right \} where the codomain of course has the discrete topology.

 

[friend]

Right, that sounds familiar. So the question would then be how is f:X→[0,1] continuous in the definition for complete regularity shown here? I'm understanding that f=0 for every x not in the closed set F, and f=1 for every y in F. I'm not seeing how one can construct open sets with just two values 0 and 1. If the set X were mapped to the continuous interval [0,1], I could see how an open set in one could be mapped to an open set in the other. But for all x in the topology, either x is in F or it is not; it seems to get mapped to either 0 or 1, and not to some continuous value in the interval. So how can a 2 valued map have open sets?

Perhaps the set in the codomain has two elements 0 and 1, a discrete set. Yet, IIRC, an open set can be constructed of discrete points, right? Then you can have open sets in [0,1] be mapped to open sets in X.

 

[micromass]

Other than that, I don't find it useful to answer by basically throwing the book at me... telling me I should go and start from scratch.

Everybody here on this forum learned topology this way. It really is the best (and I guess the only) way to learn topology.

Instead, it might make it easier to answer if you could cut and pasted from some on-line source. Then I could read it and the context it's in and decide for myself if I need to go back to the very beginning.

Nobody learned topology from an online source. At least: nobody who has a good grasp on the material. If you want to get good in mathematics, then you have to start reading textbooks and doing exercises. It's the only way. The faster you start, the better.

 

[WannabeNewton]

The presentations and books I've seen don't really show how the concepts connect to the functions one would see in physics.

Just to say one last thing, why in the world would they? They are pure math books so of course there won't be "applications" to physics. The books are meant for you to actually learn the subject properly. If all you want a summary of topology and how it's used in physics but don't properly want to have a deep knowledge of the subject, appreciate its many counter examples and intricacies and elegance, then there are always mathematical physics books like Nakahara.

 

[friend]

Just to say one last thing, why in the world would they? They are pure math books so of course there won't be "applications" to physics. The books are meant for you to actually learn the subject properly. If all you want a summary of topology and how it's used in physics but don't properly want to have a deep knowledge of the subject, appreciate its many counter examples and intricacies and elegance, then there are always mathematical physics books like Nakahara.

"deeper knowledge of the subject"... I should "read a book"... I think this is a cop-out. All I asked is about the definition of completely regular spaces in terms of topologies, open and closed sets, and continuity. These are basic concepts, and I should think that you should be able to easily explain how all these things fit in the definition. But all I'm getting is excuses. The definition is only one paragraph long, and you're not even quoting part of it in any explanation you offer. At this point I'm not confident that you know what you're talking about. If it were so clear to you, you should be able to just cut and paste from some book you like. We don't seem to be communication very well. Maybe you should let someone else reply if they wish.

 

[WannabeNewton]

Yeah good luck with that. I should have read from wikipedia instead of using an actual textbook, then I could at least say I know what I'm talking about. Cheers!

 

[micromass]

Yeah, you're on to us! We don't actually know what we're talking about. I'm sorry for wasting your time. I'll let people reply who know topology now.

 

[friend]

Yeah, you're on to us! We don't actually know what we're talking about. I'm sorry for wasting your time. I'll let people reply who know topology now.

Or at least someone who will not refer me to obscurity. If it's so easy, quote the page. That's better than telling me to go read a book.

 

[jgens]

Maybe you should let someone else reply if they wish.

Others will likely be less patient with you than micromass and WannabeNewton, and despite what you seem to think, they really do know what they are talking about. Instead of complaining, consider taking their advice.

 

[strangerep]

Friend,

Since (apparently) you decline to participate in the quiz I posed in post #64 (which was designed to reveal and correct a basic point that you misunderstand), I'll just mention 2 more books before quitting this thread:

1) Many years ago, I was perplexed about all this "topology" stuff that the experts on sci.physics.research often talked about. Then someone mentioned this book:

Albert Schwarz, "Topology for Physicists",
https://www.amazon.com/dp/3642081312/?tag=pfamazon01-20

Very expensive, but I lashed out and bought a copy. But I couldn't get much out of it, and it still sits gathering dust on my shelf to this day.

2) Semour Lipschutz, Schaum Outline of General Topology,
https://www.amazon.com/dp/0071763473/?tag=pfamazon01-20

It's not a book aimed at physicists, but it is very concise and cheap, with lots of worked examples. When I started reading, I couldn't put it down. Sad, perhaps, but at least that shows it was exactly what I needed, even though it wasn't aimed at physicists.

--------

And BTW, you're deeply wrong about Micromass and WannabeNewton. I'm reminded of a story about when Einstein was asked by a reporter "can you explain your theory of relativity to me", he answered simply "No." When she asked why not, he said "could you explain to someone how to bake a cake if they don't know what flour is?". She was quite taken aback by this, and (probably) offended. But does this mean Einstein didn't understand relativity? Of course not.

 

[friend]

And BTW, you're deeply wrong about Micromass and WannabeNewton. I'm reminded of a story about when Einstein was asked by a reporter "can you explain your theory of relativity to me", he answered simply "No." When she asked why not, he said "could you explain to someone how to bake a cake if they don't know what flour is?". She was quite taken aback by this, and (probably) offended. But does this mean Einstein didn't understand relativity? Of course not.

As I've been told: you don't really understand anything unless you can explain it to someone else.

 

[micromass]

Well, this thread has run its course. So I'm locking it.

If anybody is interested in helping the OP further by answering his questions, then send me a PM and I will open the thread again.

Edit: Thread opened on request of dx.

 

[dx]

friend, here's an intuitive way to think about what a topology is and what an open set is, that might help understanding why you need a topology to speak about open sets:

A 'topology' on a space gives us a way to talk about points "sufficiently close" to members of that space, and the concept of open set can be described as "a set that contains all points sufficiently close to its members". Thus you need a topology to talk about open sets.

For example, the idea of an 'isolated point' can be defined in the language of open sets as a point x such that the set which contains only x is open, i.e. "if there are no points sufficiently close to the point"

 

[friend]

friend, here's an intuitive way to think about what a topology is and what an open set is, that might help understanding why you need a topology to speak about open sets:

A 'topology' on a space gives us a way to talk about points "sufficiently close" to members of that space, and the concept of open set can be described as "a set that contains all points sufficiently close to its members". Thus you need a topology to talk about open sets.

For example, the idea of an 'isolated point' can be defined in the language of open sets as a point x such that the set which contains only x is open, i.e. "if there are no points sufficiently close to the point"

Thank you, dx. But I already accept the need for open sets and unions and intersections thereof to describe topology. And I also accept that open sets cannot be described without topology. For at least any open set and the empty set also describe a topology.

I started this thread to understand the definition of completely regular spaces. The parts I don't get is where the closed set F comes from and if it's unique or arbitrary or constructed from the complement of open sets already in the topology. I'm also not clear about how the function f:X→[0,1] can be continuous when f is either 0 or 1 even up to the sharp boundary of the closed set F. I was hoping these things were easy to answer, but I may have been mistaken.

 

[dx]

The parts I don't get is where the closed set F comes from and if it's unique or arbitrary or constructed from the complement of open sets already in the topology.

A set is a closed if its complement is open, so the open sets determine what the closed sets are. And its not unique. We are not talking about some particular closed set F. For any closed set F, and any point x outside it, there must be a continuous function that separates them.

I'm also not clear about how the function f:X→[0,1] can be continuous when f is either 0 or 1 even up to the sharp boundary of the closed set F.

The definition only says that f must be 0 at x, and 1 on F. It says nothing about what its value must be anywhere else, except that the function must be continuous.

 

[Bacle2]

A set is a closed if its complement is open, so the open sets determine what the closed sets are. And its not unique. We are not talking about some particular closed set F. For any closed set F, and any point x outside it, there must be a continuous function that separates them.



The definition only says that f must be 0 at x, and 1 on F. It says nothing about what its value must be anywhere else, except that the function must be continuous.

But f does not have to be either 0 or 1. And remember that in abstract topological spaces (mostly not R^n and not manifolds) , the concept of borders does not really apply.

 

[dx]

I didn't say it had to be either 0 or 1. I said it has to be 0 on x, and 1 on F. Outside that, it can be anything it wants, as long its continuous.

And friend used the word 'boundary', not border, and boundary is a notion that applies in any topological space.

 

[Bacle2]

I didn't say it had to be either 0 or 1. I said it has to be 0 on x, and 1 on F. Outside that, it can be anything it wants, as long its continuous.

And friend used the word 'boundary', not border, and boundary is a notion that applies in any topological space.

Sorry, I was addressing a coment made by friend, not to your post; in the last paragraph of the most recent post by friend:

"I started...... I'm also not clear about how the function f:X→[0,1] can be continuous when f is either 0 or 1 even up to the sharp boundary of the closed set F. I was hoping these things were easy to answer, but I may have been mistaken. "

 

[dx]

Oh, sorry :) You quoted my post so I assumed you were replying to me.

 

[Bacle2]

Sorry myself for mistakenly quoting your post.

 

[friend]

A set is a closed if its complement is open, so the open sets determine what the closed sets are. And its not unique. We are not talking about some particular closed set F. For any closed set F, and any point x outside it, there must be a continuous function that separates them.

I may have questions about this later.

But first,

The definition only says that f must be 0 at x, and 1 on F. It says nothing about what its value must be anywhere else, except that the function must be continuous.

Yea, I've looked at 3 or 4 definitions on the Web. And they all seem a bit terse. So I'm having trouble visuallizing what the definition means. Does [0,1] mean that f must be somewhere in the closed interval 0≤f≤1?

Am I right in this interpretation:

You have a closed set, F, inside an entire set. (Nevermind for the moment where F comes from.) Chose a point x outside F and chose a point y inside F. Then there must be a continuous function f such that f(x)=0, and f(y)=1. The function f is defined for all points in the entire topology with no discontinuities. Is this right?

 

[dx]

Does [0,1] mean that f must be somewhere in the closed interval 0≤f≤1?

Yes.

Am I right in this interpretation:

You have a closed set, F, inside an entire set. (Nevermind for the moment where F comes from.) Chose a point x outside F and chose a point y inside F. Then there must be a continuous function f such that f(x)=0, and f(y)=1. The function f is defined for all points in the entire topology with no discontinuities. Is this right?

Yes, but you don't choose a particular y in F. The continuous function must be 1 for all y in F.

 

[friend]

Yes, but you don't choose a particular y in F. The continuous function must be 1 for all y in F.

It also seems in the definition you have "for any point x not in F", f(x)=0. So it seems that no matter what point x in the topology not in F, f(x)=0. In other words, the function f must always be zero outside F and 1 inside F, since you can chose x anywhere outside F, and yet must have f(x)=0? Or are we talking about a different continuous function f for each choice of x?

Then if f(x)=0 for all points x outside F, there seems to be a discontinuity where if you approach the edge of F from outside f=0, but if you approach the edge from inside F, then f=1. This sounds like the definition of discontinuous.

 

[dx]

We are talking about different functions for each choice of x and F.

Once you choose x and F, then we must be able to find a continuous function that separates them. This function only needs to be 0 at x, not anywhere else.

 

[WannabeNewton]

Then if f(x)=0 for all points x outside F, there seems to be a discontinuity where if you approach the edge of F from outside f=0, but if you approach the edge from inside F, then f=1. This sounds like the definition of discontinuous.

Your intuitive notions of continuity won't help here so don't use it. Even in \mathbb{R} there are many textbook maps that are continuous in a counter intuitive manner. For example: http://en.wikipedia.org/wiki/Popcorn_function

Try to show this is continuous in the manner specified in the wiki article using the epsilon delta definition of continuity, it will be very instructive in allowing you to get a hold of continuity (but it isn't an easy problem so it might take some time!). In topology, continuity is not as trivial in an intuitive sense as it is for the nice functions one sees in physics.

 

[friend]

We are talking about different functions for each choice of x and F.

Once you choose x and F, then we must be able to find a continuous function that separates them. This function only needs to be 0 at x, not anywhere else.

Even if x approaches the boundary of F?

 

[dx]

Why would that be a problem? x can never be on the boundary of F since closed sets contain their boundaries, but x can be any point outside F.

 

[Bacle2]

Friend, following up upon dx's last post: have you tried to see if the result holds for F open?

Also notice, like dx said, for every choice of F and x there is a function with the given
properties; not for a given F closed and _all x_, but, given F and given a specific x not
in F . If you have a choice of F and x, you can find a function, say, f(F,x) that will have
the given properties; if you choose a different pair F', x' , you will (except maybe in very exceptional situations) a different f'(F'x').

 

[friend]

Why would that be a problem? x can never be on the boundary of F since closed sets contain their boundaries, but x can be any point outside F.

So does continuity say in other words that no matter how close you get to the boundary of F, the open nature of the entire set minus the closed set F will always allow you to find an open set even closer to the boundary that maps to an open interval even closer to 1? Though, I suppose the strict definition would require the inverse map, from the open intervals to the open sets in the topology.

 

[dx]

You can say it this way:

Since f is continuous, and since f is 1 on F, f can be made as close to 1 as you want if you evaluate it at points sufficiently close to F

In fact, in terms of neighborhoods, a map f is continuous if and only if given any point x and any neighborhood M of f(x), there is a neighborhood N of x such that f[N] is a subset of M.

 

[friend]

You can say it this way:

Since f is continuous, and since f is 1 on F, f can be made as close to 1 as you want if you evaluate it at points sufficiently close to F

In fact, in terms of neighborhoods, a map f is continuous if and only if given any point x and any neighborhood M of f(x), there is a neighborhood N of x such that f[N] is a subset of M.

Very good! Thank you. Let's move on...

A set is a closed if its complement is open, so the open sets determine what the closed sets are. And its not unique. We are not talking about some particular closed set F. For any closed set F, and any point x outside it, there must be a continuous function that separates them.

We've talked about what it means to say "any point x not in F". Now I would like to go back and understand "any closed set F". I understand that there is a whole space of points from which to arbitrarily choose any x, well, except perhaps x can not be chosen from F. But I'm not seeing how many choices we have from which to create any F.

I'm trying to avoid any appearance of uniqueness in F so that the property of complete regularity encompasses the whole topology. Otherwise, with x restricted from being chosen in F, it seems there are restricted regions for which the definition does not apply, which means the property is not applicable to the whole topology. So it seems necessary that x can be chosen from any point in the topology and we are guaranteed to be able to find an F and continuous functions f to comply with the definition. So I don't think the definition means draw a circle anywhere you like and call it F. Your comment suggests that F must be constructed from the complement of the available open sets in the topology. So it seem the topology allows more than one F so we can choose x in the previous F and still find a different F that fulfills the definition. Does this sound right?

 

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Yes, a set is closed only if its complement is an open set, i.e. a member of the topology. Given a certain closed set F, then points outside F must be separated from F by a continuous function. The points here are outside F, but they don't have to be outside other closed sets. For a different closed set F', there must be continuous functions that separate F' from points outside F', and again these points are all points outside F', whether they belong to closed sets other than F' or not.