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.