# Topological indistinquisable points and set theory.

1. Jan 7, 2014

### center o bass

In set theory a set is defined to be a collection of distinct objects (see http://en.wikipedia.org/wiki/Set_(mathematics)), i.e. we must have some way of distinguishing any one element from a set, from any other element.

Now a topological space is defined as a set X together with a topology T. And one might run into so called 'coarse' topologies (ex. the trivial topology T = {Ø,X}) which are not able to distinguish between the points of the topological space. I.e. suppose we have two points p and q in X. If we had a 'fine topology', so that T contains a lot of sets one could distinguish the points p and q by sets in T to which p belongs and not q. However in the 'coarsest' topologies (as for example the trivial topology above), p and q can not be distinguished in this way.

So my question is; doesn't this contradict the axiom of set theory, that a set must be a collection of element which are distinct in some way? Have I missed some way of distinguishing the point p and q in the example above? Or do we just define the points of a topological to be different elements?

2. Jan 7, 2014

### Office_Shredder

Staff Emeritus
The points p and q can be distinguished, just not in the topology. For example we can take the real numbers, and give it the trivial topology with only the empty set and $\mathbb{R}$ as open sets. The numbers 1 and 0 are indistinguishable topologically, in the sense that if 0 is contained in an open set, 1 is contained in it as well. But 0 and 1 are distinguished on a set-theoretic level, as they are different elements of $\mathbb{R}$. We have more information about the set $\mathbb{R}$ than just the topology we place on it.

3. Jan 7, 2014

### HallsofIvy

Staff Emeritus
That's not what "distinguished" means in topology. Two points, p and q, can be "distinguished" if there are distinct open sets such that p is in one and q is in the other. Of course, if p and q are "two" points then they are two different points in set X.

4. Jan 9, 2014

### center o bass

Yes, but that presupposes some additional structure: In the case with the real line the extra structure is the arithmetical axioms of numbers etc. But a topological space, as it is defined, does not presuppose any additional structure, only a set X and a topology T.

5. Jan 9, 2014

### center o bass

But then that is a definition. If two mathematical objects are to be different, they must have some way of actually being distinguished. In a coarse topology, purely as a topological space, they have no way of being distinguished. So it seems like there must be some additional structure that is able to, or one must identify the points.

6. Jan 9, 2014

### jgens

The topology might not be able to distinguish between the two points, but the points are distinguished in the underlying set.

As above they are distinguished in the underlying set.

7. Jan 9, 2014

### center o bass

Which presuppose some additional structure which must exist for it to be a set.

8. Jan 9, 2014

### jgens

I have no idea what your issue is here. The set-level data is more elementary than the topological data. It makes no sense to talk about topological spaces without their underlying sets. So if you have two distinct points which are topologically indistinguishable, then they are always distinguishable on the set-level. There is no additional structure (on top of what is already needed to define a topological space) used here.

9. Jan 9, 2014

### center o bass

What do you mean by 'being distinguishable on the set-level'? If the set is $\mathbb{N}$ the only reason that, say, 1 and 2 are distinguishable as mathematical objects is because of their arithmetical properties. (I.e. the arithmetical structure.) The collection of symbols \mathbb{N} is not a set if the elements did not somehow have different properties. The fact that we assume a topological space (X, T) to consist of a set X implies that there must be some structure which can distinguish the elements of X. It's not an issue any longer, it has become clear to me.

10. Jan 9, 2014

### jgens

This is actually not true.

Again this is not true.

The elements of X are distinguished by definition. This is all information contained in the structure of X as a set.

Evidently not. This last post of yours was all crazy talk.

11. Jan 9, 2014

### center o bass

Could you please state a reason? I might have left out some properties of the natural numbers, but my only purpose was to illustrate that you need some structure to distinguish between the elements.
A definition is certainly also a implication. If ${a,b}$ is a set, this implies that a and b is distinct objects which again implies that there is some way of distinguishing them.

Thanks.

12. Jan 9, 2014

### jgens

The problem with your thinking is that it is backwards. You are starting with N as a monoid and then saying that these arithmetical properties allow us to distinguish the elements of N as a set. But really N starts life as a set and then is endowed with all of the nice arithmetical properties that we like. Elements of N are distinguishable by definition.

As long as being distinct members of a set counts as "some way of distinguishing them" then this works. But "a" and "b" need no additional structure a priori to distinguish between them.

13. Jan 9, 2014

### center o bass

Well I think that your thinking is backwards. The natural thing to do is surely to define what a natural number is by defining it's properties and then consider the set of all natural numbers. But this again boils down to a matter of opinion, as it seem to do with you. I explain what properties the objects have that distinguish them from one another and then consider the set of all such objects. You define a set of objects, and later justify why it is a set by defining the properties of it's elements.

You can go on doing that as much as you like, but it does not justify your accusation of me 'talking crazy'.
Alright, so how then would you distinguish between them without any additional structure? In what sense is $a$ different from $b$?

14. Jan 9, 2014

### jgens

From a set-theoretical/foundations of mathematics standpoint it really is not.

This procedure is natural from a certain standpoint yes. We can axiomatically characterize the natural numbers in this manner for example. Unfortunately it still does not mean that the arithmetical properties are what distinguish the elements of N. Once we have decided what are natural numbers are, the elements of N are distinguished by virtue of N being a set. Not by the arithmetical properties.

This particular claim of mine is definitely not a matter of opinion.

What you have described here actually cannot give you the natural numbers. For one it is not well-defined. Anyway it is not like there is some god given set of natural numbers. We have to construct them manually and then define operations on this set that satisfy all the nice arithmetical operations that we like.

Nope.

The fact that they are distinct elements of a set is all the structure needed.

15. Jan 9, 2014

### economicsnerd

I think the confusion is about what it means to "distinguish" two elements of a set.

Informally, a set is a collection $X$ of objects.
- We write "$a \in X$" if $a$ is one of the objects in $X$. This isn't an opinion of mine; it's a notational convention.
- If $a,b\in X$, we say "$a=b$" if $a$ and $b$ are the same. That doesn't mean they have "the same properties" or anything like that. It just means they're labels for the same object.
- If we write $a\neq b$, it just means they're labels for two different objects. Nothing about "distinguishing properties", whatever that means in this abstract setting.

A perhaps way too silly analogy:
-Say I see listings for two courses at my university. One is a course called "Teutonic Literature 101" and the other is a course called "Germanic Literature 101". I know nothing that would help me pick between these classes. I couldn't even venture a guess of a topic that might show up in one class but not the other. That said, they are different courses. If $C$ is the set of classes, there are elements $TL101,GL101\in C$ such that $TL101\neq GL101$.

16. Jan 27, 2014

### center o bass

So if I come to you with the following proposition: $a$ is a member of $\mathbb{N}$.
What proof would I have to give you to convince you of it's truth?

Would you not appeal to the properties of the natural numbers?

More generally, if I gave you two set $A$ and $X$ I claim that $A=X$. Then by definition of set equality I would have to show you that $a$ is a member of $A$ if and only if $a$ is $a$ member of $X$. But I can only do this if I know what it means to be a member of $A$ and of $X$. In other words I must appeal to properties $a$ must have in order to be a member a set.

Wikipedia on the axiom of extensionality (http://en.wikipedia.org/wiki/Axiom_of_extensionality): [Broken]

The axiom of extensionality can be used with any statement of the form $\exists A \ \forall X \ (X \in A \iff P(X) \ )$ where P is any unary predicate that does not mention A, to define a unique set A whose members are precisely the sets satisfying the predicate P. We can then introduce a new symbol for A; it's in this way that definitions in ordinary mathematics ultimately work when their statements are reduced to purely set-theoretic terms.

Last edited by a moderator: May 6, 2017
17. Jan 27, 2014

### jgens

Depends on how you defined N. If we go with the usual definition (first countable ordinal), then there are easy ways to check that claim which do not appeal to the arithmetic properties of the natural numbers.

In general no! The arithmetic properties of natural numbers are not sufficient to determine N uniquely as a set (i.e. there is more than one set with the arithmetic properties of N).

This all checks out, but what "properties" needs to mean here is different than you have been using the term in this thread.

I am aware of all this. Unless you assume a priori a collection of numbers with all the desired arithmetic properties, however, you cannot use these sorts of predicates to weasel out the natural numbers.

Last edited by a moderator: May 6, 2017
18. Feb 13, 2014

### FactChecker

You are thinking that adding a topology to a set destroys the already existing way of distinguishing elements in the set. That is not true. If the elements of a set were distinguishable without the topology, they remain distinguishable with the topology.

19. Feb 13, 2014

### Number Nine

The problem (one of them) with your thinking is that you're equivocating with the word "distinguishable". The word has a very specific meaning in topology which is different from its meaning in set theory. Saying that two points are or are not "topologically distinguishable" has precisely nothing to do with the use of the word "distinguishable" in set theory.