What is the Significance of Completely Regular Spaces?

[friend]

A point x is an accumulation point of a net x_δ if, for any neighborhood N of x, and any δ, there exists a δ' with δ ≤ δ' and x_δ' in N.

A neighborhood of x is a superset of an open set O containing x.

So you can't define an accumulation point without knowing what the open sets are, i.e. without having a topology.

Yes of course. You at least always have the superset, the open set and the empty set is implied. Or perhaps the superset could also be the open set, then since the empty set is always an element of a set, any open set is automatically its own topology. I'm not sure what the point is anymore.

 

[WannabeNewton]

You do know the definition of a limit point itself relies on the concept of a neighborhood right? I'm not sure where this thread is heading but the fact that you need a topology to talk about open sets is like topology 101.

 

[WannabeNewton]

Are you perhaps reading Rudin or some other analysis book like Carothers? It is possible you are looking at notions of open and closed through metric spaces but note that the basis of open balls, of a metric, generate a topology on the set so in the end you are still using topologies.

 

[friend]

We left off in the relevant discussion with:So I think this leaves the question open as to whether F, or rather its compliment belongs to the topology. It doesn't seem obvious to me that every X\F is in the particular topology in the definition so that the property holds for all x in the topology.

Does this mean it is aways possible to construct X\F from the existing open sets in the topology such that y is in F but x is not?

Must every F for every x in F and y not in F be constructed from the allowed open sets in the topology? I have a hard time understanding that one can always arbitrarily construct any closed F anywhere of any size when only the open sets of the topology are allowed.

 

[WannabeNewton]

The complement of an open set is by definition a closed set and a closed set does not necessarily have to belong to the topology although it can depending on the topology in which case it is clopen.

 

[friend]

Are you perhaps reading Rudin or some other analysis book like Carothers? It is possible you are looking at notions of open and closed through metric spaces but note that the basis of open balls, of a metric, generate a topology on the set so in the end you are still using topologies.

Yes, it's been a while since I studied topology. And it seems to go in one ear and out the other because the books I read don't connect the concepts to the properties of the kind of functions I would normally be interested in. If you know of a book that provides this relevance, that would help a lot. Thanks.

 

[friend]

The complement of an open set is by definition a closed set and a closed set does not necessarily have to belong to the topology although it can depending on the topology in which case it is clopen.

F is a closed set, and its complement is open. Does that complement necessarily have to be constructed of the available open sets in the topology? I mean, that is what we have to work with, right? If so, it's not clear we can always construct any size F everywhere from complements of unions of available open sets.

 

[WannabeNewton]

If you are primary interested in studying manifolds, as I suspect you are based on your prior threads, I would recommend "An Introduction to Topological Manifolds - John M. Lee. And to answer your newest question , yes the complement of the closed set has to be in the topology by definition because it has to be an open set.

 

[friend]

If you are primary interested in studying manifolds, as I suspect you are based on your prior threads, I would recommend "An Introduction to Topological Manifolds - John M. Lee. And to answer your newest question , yes the complement of the closed set has to be in the topology by definition because it has to be an open set.

Here we go again. I thought we established that just because you have an open set does not mean it belongs to a particular topology. More than one topology can be constructed from a background superset. They differ in that some topologies have different open sets than other topologies constructed from the same background.

So let me ask again: in the definition of competely regular spaces where F is a closed set with x in and y outside it,... F is a closed set, and its complement is open. Does that complement necessarily have to be constructed of the available open sets in the topology? I mean, that is what we have to work with, right? If so, it's not clear we can always construct any size F everywhere from complements of unions of available open sets.

 

[WannabeNewton]

It doesn't make sense to talk about any of these things unless you first FIX a topology. Then everything regarding topological definitions is with respect to this topology. I don't know how else to explain it, in my head it is obvious so I'm having trouble understanding where your confusion is coming from.

 

[friend]

It doesn't make sense to talk about any of these things unless you first FIX a topology. Then everything regarding topological definitions is with respect to this topology. I don't know how else to explain it, in my head it is obvious so I'm having trouble understanding where your confusion is coming from.

I agree. And I'm asking about a compelely regular space, which is defined on one particular topology defined on a background set. Complete regularity seem to construct closed sets, F, in seemingly arbitrary places of seemingly arbitrary size. I'm asking how this is possible when the particular topology on which complete regularity is defined only has available a certain number of open sets to work with. So it doesn't seem to be possible to construct an arbitrary closed F with the complement of the union of a restrict number of open sets of that particular topology. Are you having difficulty getting out of your comfort zone? :-)

 

[WannabeNewton]

A topological space is completely regular if any subset closed in the topology obeys the usual properties for being completely regular so obviously it only applies to sets that are closed in the topology. What exactly is your confusion with this? It is all rather straightforward definition wise.

 

[friend]

A topological space is completely regular if any subset closed in the topology obeys the usual properties for being completely regular so obviously it only applies to sets that are closed in the topology. What exactly is your confusion with this? It is all rather straightforward definition wise.

"if any subset"... Does that mean you only need one such closed subset somewhere? If so, then wouldn't that mean that complete regularity only applies near that subset and not necessarily everywhere in the topology? I would think in order for the whole topology to be completely regular, then you should be able to find such closed subsets anywhere and everywhere. Thus the question of how to do this with a limited number of open sets available in the topology. Thank you, I think we're getting closer.

 

[dx]

You can't find closed sets anywhere and everywhere. They are determined by the topology.

"if any subset"... Does that mean you only need one such closed subset somewhere?

"given any closed set" does not mean you need only one such set.

Given any closed set whatsoever, and any point x that is not in that closed set, the condition in the definition of completely regular must be satisfied

 

[WannabeNewton]

In addition to what dx said, the phrase "for any" in mathematics is usually denoted by \forall. It is the same thing as for all and for every.

 

[micromass]

"if any subset"... Does that mean you only need one such closed subset somewhere? If so, then wouldn't that mean that complete regularity only applies near that subset and not necessarily everywhere in the topology? I would think in order for the whole topology to be completely regular, then you should be able to find such closed subsets anywhere and everywhere. Thus the question of how to do this with a limited number of open sets available in the topology. Thank you, I think we're getting closer.

You seem to think that complete regularity somehow shows the existence of a closed set. The definition of complete regular is not there exists a closed set and a point outside the closed set such that...

No, the definition is: for any closed set $F$ and for any point $y$ outside the closed set, there must exist a continuous function $f:X\rightarrow [0,1]$ such that $f(F)=\{0\}$ and $f(y)=1$.

So complete regularity shows the existence of a continuous function, and not the existence of closed sets.

And finally, I stress that "size" of a set does not make sense in a topological space. It does not make sense to talk about a closed set being small or large. There is simply no way of comparing the sets.

 

[micromass]

An easy example of a completely regular space is perhaps the indiscrete space. So take $X=\mathbb{R}$ to be any set and take $\mathcal{T}=\{\emptyset,\mathbb{R}\}$. So right now, the only open sets are $\emptyset$ and $\mathbb{R}$. So the only closed sets are $\emptyset$ and $\mathbb{R}$.

This space is completely regular. Indeed, we need to show that for any closed set $F$ and for any $x\notin F$, the condition holds.

So take $F=\mathbb{R}$, then there simply are no $x\notin F$, so the condition is vacuously true.
Take $F=\emptyset$ and take $x\in \mathbb{R}$ arbitrary. The function $f:X\rightarrow [0,1]:x\rightarrow 1$ satisfies $f^{-1}(0) = \emptyset$ and $f(x)=1$. So the condition is satisfied.

The above are the only possibilities of closed sets $F$ and $x\notin F$. So since it works for all those possibilities, the space is completely regular.

Of course, the indiscrete topology is not Hausdorff. So the space is not a Tychonoff space. Tychonoff spaces usually have a very large collection of open sets.

 

[friend]

An easy example of a completely regular space is perhaps the indiscrete space. So take $X=\mathbb{R}$ to be any set and take $\mathcal{T}=\{\emptyset,\mathbb{R}\}$. So right now, the only open sets are $\emptyset$ and $\mathbb{R}$. So the only closed sets are $\emptyset$ and $\mathbb{R}$.

This space is completely regular. Indeed, we need to show that for any closed set $F$ and for any $x\notin F$, the condition holds.

So take $F=\mathbb{R}$, then there simply are no $x\notin F$, so the condition is vacuously true.
Take $F=\emptyset$ and take $x\in \mathbb{R}$ arbitrary. The function $f:X\rightarrow [0,1]:x\rightarrow 1$ satisfies $f^{-1}(0) = \emptyset$ and $f(x)=1$. So the condition is satisfied.

Notice that you are constructing F from the sets in the topology. Are you saying that F is not some arbitrary set independent of the sets in the topology?

Actually, I don't think the empty set will work in the definition because the line is between two points one point inside a closed set, and the other point outside a closed set such that you get a transition from 0 to 1 as you cross some border of a set. The empty set can not contain any points and has no border.

Notice that the empty set can be chosen in any topology and therefore would not be relevant to the definition.

The above are the only possibilities of closed sets $F$ and $x\notin F$. So since it works for all those possibilities, the space is completely regular.

I could understand that the property of complete regularity is global if the point outside the closed set F could be any point in the topology. If they mean that the line can be drawn from each point outside F, and you still get a transition from 0 to 1, then I understand how it could be global. Or if the line could be arbitrarily drawn through any point in the topology, then I could understand how the property could be global. But if the definition only concerns one line from one point to another point through one closed set, then I would think the property is only concerned with those particular objects and is not necessarily relevant to the rest of the topology.

 

[micromass]

Notice that you are constructing F from the sets in the topology. Are you saying that F is not some arbitrary set independent of the sets in the topology?

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.

Actually, I don't think the empty set will work in the definition because the line is between two points one point inside a closed set, and the other point outside a closed set. The empty set can not contain any points.

Notice that the empty set can be chosen in any topology

I could understand that the property of complete regularity is global if the point outside the closed set F. If they mean that the line can be drawn from each point outside F any you still get a transition from 0 to 1, then I understand how it could be global. Or if the line could be arbitrarily through any region in the topology, then I could understand how the property could be global. But if the definition only concerns one line from one point to another point through one closed set, then I would think the property is only concerned with those particular object and does not necessarily relevant to the rest of the topology.

I don't understand what the talk about "lines" is all about. Where do you see a line anywhere in the definition?

If you had a map $f:[0,1]\rightarrow X$, then I could understand why you call this a line (the technical term is "curve" or "path"). But now we have a map $X\rightarrow [0,1]$. So I don't get where the entire talk about "lines" comes from.

 

[friend]

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.



I don't understand what the talk about "lines" is all about. Where do you see a line anywhere in the definition?

If you had a map $f:[0,1]\rightarrow X$, then I could understand why you call this a line (the technical term is "curve" or "path"). But now we have a map $X\rightarrow [0,1]$. So I don't get where the entire talk about "lines" comes from.

On the wiki page, under the Definitions section

Suppose that X is a topological space.

X is a completely regular space if given any closed set F and any point x that does not belong to F, then there is a continuous function f from X to the real line R such that f(x) is 0 and, for every y in F, f(y) is 1.

I'm seeing the word "line" here. I take it x is one end of the line and y is the other end of the line.

Notice is says for "any"... Does that mean that x and y could be any two points in the topology, as long as x is out and y is in F?

 

[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.