What is the Significance of Completely Regular Spaces?

[friend]

Since manifolds are locally compact Hausdorff spaces, manifolds are necessarily Tychonoff spaces. And a Tychonoff space is a topological space that is both Hausdorff and completely regular. This is cut and paste from wikipedia.org. Further,

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.

And so my question is: does this mean that in order for this property of completely regularity to hold for a space that one must be able to construct a set F with a point x in F and y outside F with a function f as described above? And this must hold no matter how close x and y are to each other? Is there any limit to the smallness of the set F? Does this mean that the sets, F, must exist in the topology, or can be constructed from the underlying elments whether in the topology or not? Thanks.

 

[micromass]

The property "There exists a continuous function f:X\rightarrow [0,1] such that f(F)=\{0\} and f(y)=1" must hold for any closed set F and for any point $y$ which is not in $F$. It doesn't matter what $F$ or $y$ is as long as $y\notin F$.

Also, "smallness of $F$" doesn't make sense in topological spaces. You need some kind of metric for that.

 

[friend]

The property "There exists a continuous function f:X\rightarrow [0,1] such that f(F)=\{0\} and f(y)=1" must hold for any closed set F and for any point $y$ which is not in $F$. It doesn't matter what $F$ or $y$ is as long as $y\notin F$.

Also, "smallness of $F$" doesn't make sense in topological spaces. You need some kind of metric for that.

Well, it says X is a topological space. So it makes we wonder if F has to be a part of that topology or not. Thanks.

 

[micromass]

Well, it says X is a topological space. So it makes we wonder if F has to be a part of that topology or not. Thanks.

We just want $F$ to be a closed set. So the complement $X\setminus F$ has to be open (= an element of the topology).

 

[friend]

We just want $F$ to be a closed set. So the complement $X\setminus F$ has to be open (= an element of the topology).

I don't know that every open set is necessarily part of a topology?

 

[micromass]

I don't know that every open set is necessarily part of a topology?

The elements of the topology are by definition the open sets. If (X,\mathcal{T}) is a topological space, then I define U\subseteq X to be open exactly if U\in \mathcal{T}.

 

[friend]

The elements of the topology are by definition the open sets. If (X,\mathcal{T}) is a topological space, then I define U\subseteq X to be open exactly if U\in \mathcal{T}.

I appreciate your efforts. But I'm looking at the wikipedia.org website about topology and reading the definition in terms of open sets, which reads:

Given such a structure, we can define a subset U of X to be open if U is a neighbourhood of all points in U. It is a remarkable fact that the open sets then satisfy the elegant axioms given below, and that, given these axioms, we can recover the neighbourhoods satisfying the above axioms by defining N to be a neighbourhood of x if N contains an open set U such that x ∈ U.[2]

A topological space is then a set X together with a collection of subsets of X, called open sets and satisfying the following axioms:[3]
1.The empty set and X itself are open.
2.Any union of open sets is open.
3.The intersection of any finite number of open sets is open.

The collection τ of open sets is then also called a topology on X, or, if more precision is needed, an open set topology. The sets in τ are called the open sets, and their complements in X are called closed sets. A subset of X may be neither closed nor open, either closed or open, or both. A set that is both closed and open is called a clopen set.

I see where the definition requires that the sets be open. But I'm not reading where it is necessary that every possible open set must be in the topology. Maybe I'm reading it wrong. But the examples given show that some sets that could possibly be constructed are not in the topology, even though they would be called open. Any help in this would be appreciated.

 

[dx]

You can't call a set open if its not in the topology. The word 'open set' is the name that we give to elements of the topology.

But the examples given show that some sets that could possibly be constructed are not in the topology, even though they would be called open.

Which example are you referring to, and how did you decide that those sets which are not in the topology should be called open? i.e. what definition of 'open set' did you use to determine that the set you refer to should be called open?

 

[WannabeNewton]

A set is open if its an element of the topology by definition of a topological space...

 

[friend]

You can't call a set open if its not in the topology. The word 'open set' is the name that we give to elements of the topology.
Which example are you referring to, and how did you decide that those sets which are not in the topology should be called open? i.e. what definition of 'open set' did you use to determine that the set you refer to should be called open?

A set is open if its an element of the topology by definition of a topological space...

Yes, of course, if it is in the topology, then it is open. But that does not mean that if it is open, then it is in the topology.

In the first figure on this wiki webpage, there is a topology with the empy set, the whole set and element 1 circled. Elements 2 and 3 are not in the topology. One could just as easily constructed a set from element 2 and called it open. The mere fact that in the figure it is not circled and called an open set does not mean that someone else could have come along and constructed a different topology using 2 and not 1. Here in, elements 2 and 3 are in the underlying set but not separately identified as sets in the topology. Thus the question: is every open set constructed from the underlying whole set necessarily in the topology? I think the answer is no. Thanks.

 

[dx]

Which sets are called open depends on which topology you are using. A certain set could be open in one topology and not open in a different topology.

 

[dextercioby]

Yes, of course, if it is in the topology, then it is open. But that does not mean that if it is open, then it is in the topology.
[...]

No, it means. The open set notion is not primitive, the primitive one is topology (topological space). You then DEFINE the open sets as the (sub)sets in the 3 conditions which define the topology. You can't have open sets in the absence of a topology and viceversa.

 

[micromass]

Yes, of course, if it is in the topology, then it is open. But that does not mean that if it is open, then it is in the topology.

It actually does mean that. By definition, a set is open if and only if it is an element of the topology

In the first figure on this wiki webpage, there is a topology with the empy set, the whole set and element 1 circled. Elements 2 and 3 are not in the topology. One could just as easily constructed a set from element 2 and called it open. The mere fact that in the figure it is not circled and called an open set does not mean that someone else could have come along and constructed a different topology using 2 and not 1. Here in, elements 2 and 3 are in the underlying set but not in the topology. Thus the question: is every open set constructed from the underlying whole set necessarily in the topology? I think the answer is no. Thanks.

Of course, if you change the topology, then you're going to change the open sets! The notion open sets depends on which topology you are considering.

If $X=\{0,1\}$ and if $\mathcal{T}=\{\emptyset, X,\}$, then $\{0\}$ is not an open set for the topology $\mathcal{T}$.
But if I put $\mathcal{T}^\prime=\{\emptyset, X, \{0\}\}$, then $\{0\}$ is an open set for the topology $\mathcal{T}^\prime$.

You can't talk about open sets without first defining a topology. And you can't change the topology and expect to keep the same open sets.

 

[friend]

It actually does mean that. By definition, a set is open if and only if it is an element of the topology



Of course, if you change the topology, then you're going to change the open sets! The notion open sets depends on which topology you are considering.

If $X=\{0,1\}$ and if $\mathcal{T}=\{\emptyset, X,\}$, then $\{0\}$ is not an open set for the topology $\mathcal{T}$.
But if I put $\mathcal{T}^\prime=\{\emptyset, X, \{0\}\}$, then $\{0\}$ is an open set for the topology $\mathcal{T}^\prime$.

You can't talk about open sets without first defining a topology. And you can't change the topology and expect to keep the same open sets.

Very good! But that does not answer the question: If you can construct an open set from any part of the whole set, X, does that automatically mean it is in the topology one may be given to start with? I think you are saying that it necessarily is an open set is some topology but not necessarily the topology you started with, right?

 

[dx]

Yes, any subset that you choose could be an 'open set', if you choose the right topology. For example, the discrete topology makes all subsets open sets.

 

[micromass]

If you can construct an open set from any part of the whole set, X

What do you mean with this??

If you merely have a set $X$, then you can't talk about open sets. You need a set $X$ and a topology on $X$. So if we talk about open sets, then we have (or should have) specified a specific set and a specific topology on that set.

 

[friend]

What do you mean with this??

If you merely have a set $X$, then you can't talk about open sets. You need a set $X$ and a topology on $X$. So if we talk about open sets, then we have (or should have) specified a specific set and a specific topology on that set.

As I understand it, an open set is defined independently of a topology. You can talk about open sets without talking about a topology. And open set can be constructed with the elements of a superset. For example an open set is a closed set minus its boundary - at least I'm talking about it.

A topology is defined as consisting of open sets of some background superset and unions and intersections of those open sets. You can not talk about a topology without talking about open sets.

Since you can talk about constructing an open set from a subset of some background superset, apart from talking about topologies, the question remains: is every open set constructed from the overall superset necessarily part of every topology on that superset?

 

[WannabeNewton]

As I understand it, an open set is defined independently of a topology. You can talk about open sets without talking about a topology. And open set can be constructed with the elements of a superset. For example an open set is a closed set minus its boundary - at least I'm talking about it.

A topology is defined as consisting of open sets of some background superset and unions and intersections of those open sets. You can not talk about a topology without talking about open sets.

You are highly mistaken friend, may I ask where you saw this? Maybe you got the wrong information from someone. 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. Dexter already elucidated this.

 

[micromass]

As I understand it, an open set is defined independently of a topology. You can talk about open sets without talking about a topology. And open set can be constructed with the elements of a superset.

No, these three statements are wrong. You need a topology to be able to talk about open sets.

For example an open set is a closed set minus its boundary - at least I'm talking about it.

How would you define "closed set" or "boundary" without topology?

 

[friend]

You are highly mistaken friend, may I ask where you saw this? Maybe you got the wrong information from someone. 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. Dexter already elucidated this.

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?

 

[WannabeNewton]

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?

You are indeed correct that a subset of a set does not need to open in every topology on that set EXCEPT for the set itself and the empty set. So for example \left \{ a \right \}\subset \mathbb{R} is closed and NOT open for \mathbb{R} with the euclidean topology, which is the topology generated by the 2 - norm, but \left \{ a \right \}\subset \mathbb{R} is both closed AND open when \mathbb{R} is equipped with the discrete topology.

 

[dx]

You keep talking about 'open sets' as if they are defined independent of a topology. Whether a particular set is open or not only has an answer once you choose a topology, unless that set is the space itself or the empty set since they are open in all topologies, as wannabeNewton mentioned.

 

[friend]

You keep talking about 'open sets' as if they are defined independent of a topology. Whether a particular set is open or not only has an answer once you choose a topology, unless that set is the space itself or the empty set since they are open in all topologies, as wannabeNewton mentioned.

This is a bit of an aside for the thread. And I don't want to pretent to be an expert. Let me just say that every definition I've seen for a topology was in terms of open sets. There may be other definitions. But consider, if open sets are defined in terms of topologies, then that makes for a circular definition for a topology in terms of a topology. This goes against my expectation that complicated math terms are defined in terms of more premitive terms. And topology seems to be a more complicated thing as unions and intersection of open sets, which seem more primitive. I think I'm ready to move on to the main topic of this thread.

 

[dx]

Let me just say that every definition I've seen for a topology was in terms of open sets.

But 'topology' is not defined in terms of a pre-defined notion of open sets. 'open set' is simply the name that you give to elements of the topology.

 

[friend]

We left off in the relevant discussion with:

The property "There exists a continuous function f:X\rightarrow [0,1] such that f(F)=\{0\} and f(y)=1" must hold for any closed set F and for any point $y$ which is not in $F$. It doesn't matter what $F$ or $y$ is as long as $y\notin F$.

Well, it says X is a topological space. So it makes we wonder if F has to be a part of that topology or not. Thanks.

We just want $F$ to be a closed set. So the complement $X\setminus F$ has to be open (= an element of the topology).

I don't know that every open set is necessarily part of a topology?

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?

 

[friend]

But 'topology' is not defined in terms of a pre-defined notion of open sets. 'open set' is simply the name that you give to elements of the topology.

[0,1) is an "open" set.

 

[dx]

In the space R of real numbers, with the standard topology, that set is not an open set.

 

[dx]

If, however, you choose the discrete topology, then that set is an open set.

Which again illustrates that the topology determines which sets are open. You cannot decide whether a set is open or not if you don't have a topology.

 

[friend]

If, however, you choose the discrete topology, then that set is an open set.

Which again illustrates that the topology determines which sets are open. You cannot decide whether a set is open or not if you don't have a topology.

IIRC, I think I remember an open set being defined as a set not containing all its accumulation points. That doesn't seem to involve a topology.

 

[dx]

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.

 

[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?