# Completely regular spaces

1. Mar 12, 2013

### 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.

Last edited: Mar 12, 2013
2. Mar 12, 2013

### 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.

3. Mar 12, 2013

### friend

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.

4. Mar 12, 2013

### micromass

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

5. Mar 12, 2013

### friend

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

6. Mar 12, 2013

### micromass

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 $U\in \mathcal{T}$.

7. Mar 12, 2013

### friend

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:

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.

8. Mar 12, 2013

### 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.

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?

9. Mar 12, 2013

### WannabeNewton

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

10. Mar 12, 2013

### friend

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.

11. Mar 12, 2013

### 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.

12. Mar 12, 2013

### dextercioby

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.

13. Mar 12, 2013

### micromass

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.

14. Mar 12, 2013

### friend

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?

15. Mar 12, 2013

### 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.

16. Mar 12, 2013

### micromass

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.

17. Mar 12, 2013

### friend

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?

18. Mar 12, 2013

### WannabeNewton

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.

19. Mar 12, 2013

### micromass

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

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

20. Mar 12, 2013

### friend

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?

21. Mar 12, 2013

### WannabeNewton

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.

22. Mar 12, 2013

### 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.

Last edited: Mar 12, 2013
23. Mar 12, 2013

### friend

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.

24. Mar 12, 2013

### dx

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.

25. Mar 12, 2013

### 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?

Last edited: Mar 12, 2013