Topological space help

1. Sep 19, 2008

Unassuming

1. The problem statement, all variables and given/known data[/b]

U is T1 open iff $$\Re$$/U is countable, U=$$\oslash$$, or U=$$\Re$$.

What is the bd(0,1), the cl(rationals), and the int(rationals).
2. Relevant equations

3. The attempt at a solution
Could somebody please explain this problem to me? I feel like I could try it once I understand what it is asking for. It is for my Topology class. I am confused to the fact of whether U is any set that we pick, or U is in this case bd(0,1), or Q.

2. Sep 19, 2008

Dick

I think what they are saying is that the set of all such subsets U defines a topology. In that particular topology, what is the boundary of (0,1), the closure of Q and interior of Q? Does that sound right?

3. Sep 19, 2008

Unassuming

You clarified a bit for me but I still am unsure of myself now. When is R/U countable? I have only one thing in mind, and that is R/P with P=irrationals? I'm hoping that is what is meant by this topology?

So then, T1 is the topology of all open sets an element of P, or U=empty sets, or U=empty sets because R/R=empty sets. Am I right?

...bd(0,1)=empty set, because I feel that there is no closure.

...clQ=empty set, because every neighborhood around a rational is not guaranteed to hit another rational.

...intQ=P, because Q is dense in R.

I don't know. This is the best I think do for now. What does anybody think?

4. Sep 19, 2008

Dick

Hint: Is R/Q in U? If it is then the complement of Q is open in this topology. What does that say about Q? What is cl(Q)? Work that out and keep thinking about this. Look up the EXACT definitions of int, cl and bdy.

Last edited: Sep 19, 2008
5. Sep 19, 2008

Dick

BTW I shouldn't have said "R\Q in U", that's confusing. What I meant is "Is R\Q T1 open". T1 open is the name the problem is using for the set of all such U's.

6. Sep 19, 2008

Unassuming

Okay...

(R/Q) is T1 open.

(R/Q) is the irrationals.

Q is closed.

cl(Q) = R/Q

That's my thought bubble, is it right?

I am thinking that the closure is every point in which the smallest neighborhood intersects the set. Since Q is dense, any neighborhood around an irrational will intersect a piece of Q. But a neighborhood around a rational will not necessarily intersect Q. Thus, cl(Q)=R/Q.

7. Sep 19, 2008

Dick

Almost. The closure of Q is the intersection of all closed sets that contain Q. If Q is closed, the conclusion is pretty easy. cl(Q)=Q. Isn't that right? Be careful with statements like "Q is dense". That's in the standard topology. The T1 open topology is NOT the standard topology. Try another one of the questions.

8. Sep 20, 2008

Unassuming

I see now why I shouldn't say dense. It is hard to think about these topologies. I think that the int(Q)= empty set, and bd(0,1)={0,1}.

It sounds like the irrationals are open and Q is closed. I can't imagine there being an int(Q) because you would have to be able to find an open N in Q, which I can't see.

Also, bd(0,1). 0 and 1 are in Q, and Q is closed. All of the intermediate points are intertwined with irrationals and so I don't think I can include them.

9. Sep 20, 2008

Dick

It's fairly easy to think about topology if it's the standard topology. It's really hard to think about nonstandard ones if you let your intuitions about the standard one color your thinking. As I said, you have to look up the exact definitions. I think int(Q) is empty as well. Can you prove it? When you say bd(0,1) do you mean the boundary of the open interval (0,1) or the boundary of the set {0,1}?

10. Sep 21, 2008

Unassuming

Well, I've given up on this problem for now. This is just too much. Thanks Dick, maybe I will repost again after I try to learn more fundamentals.

If you think it possible, ... is there a certain section of topology that additional study time would increase my understanding in this type of problem? I just can't get my mind focused on this!!

11. Sep 21, 2008

Unassuming

Also, is this type of problem a common theme in Topology?

12. Sep 21, 2008

morphism

This type of exercise is common in the sense that it should be something you ask yourself when you first meet a topology. It's always good to have examples in mind when you're working with any topology. It's also good to get as much information out of the definition of a given topology as you can.

Looking at the definition of topology T_1, we can see that its closed sets are either countable or all of R. Thus the closure of (0,1) is all of R, since (0,1) is uncountable (and since a proper closed set can't contain an uncountable set in this topology). The same can be said for the complement of (0,1). Thus the boundary of (0,1) is R, and not {0,1}.

You were right in thinking that cl(Q)=Q. This is precisely because Q is closed. Make sure you completely understand this.

Now, as for int(Q), this is an open set contained in Q (in fact, the largest open set contained in Q). Since Q is countable, so too are its subsets. What does this say about the open sets contained in Q?

By the way, using R/U to denote set difference isn't standard notation; you should use R\U instead.

13. Sep 21, 2008

Unassuming

How do you say that U is T_1 closed if U=Q or U= all of R? I am changing your words but this is how I interpret this problem...any help for me? I believe the U=Q, since R\U is countable implies irrationals and the complement of irrationals are Q.

14. Sep 21, 2008

morphism

I can't understand your question at all!

15. Sep 21, 2008

Unassuming

I meant that I cannot understand this part of your response. Will you please elaborate? I am having a difficult time with these type of problems.

16. Sep 21, 2008

Unassuming

Am I correct if I say that there are only 3 open sets in T_1. U=empty, U=irrationals, U=R ?

17. Sep 21, 2008

morphism

- A topology T on a set X is a special collection of subsets of X. (This is where you look up the actual definition and make sure you understand it. Having done this, make sure you understand why the following two are topologies on an arbitrary set X: (1) the collection of all subsets of X; (2) {$\varnothing$, X}.)

- We call the sets that belong to the topology T "open sets." Remember: members of T are subsets of X. (So in example (1) above, every subset is open, while in example (2), the only open sets are $\varnothing$ and X.)

- The complements of open sets are called "closed sets." Another way to say this is: A set F is closed (with respect to T) iff its complement X\F is in T. (In example (1) every open set is closed, and in example (2) the closed sets are $\varnothing$ and X. This should show you that "open" and "closed" aren't antonyms.)

In our case, X=R with the topology T_1. Members of T_1 (the "open sets") are those subsets of R whose complement is countable or all of R. Since the "closed sets" are precisely the complements of the elements of T_1, we see that the closed sets are precisely the countable subsets of R, together with R itself.

Does this make sense now?

18. Sep 21, 2008

Unassuming

You blew my mind. First, I am going to work out some examples and list the open sets and the closed sets. I feel that this is my duty since I need to erase my image of closed and replace it with this new one. Then I will come back to T_1, maybe one hour.

Sound good?

19. Sep 21, 2008

morphism

Yup. Good luck!

20. Sep 21, 2008

Unassuming

A set F is closed iff its compliment X\F is in T.

Does F have to be an element of T?