Infinite number of open intervals

1. Feb 1, 2006

Kocur

I am a computer scientist that only recently got interested in topology. I have got the following question:

How to prove, in an elegant way, that an arbitrary sum of open intervals (a, b), where a is in R and b is in R, is open?

I have got some kind of a proof, but I am not sure it is right. Moreover, it is very bulky and I do not like it.

2. Feb 1, 2006

matt grime

Pick a point in the union, it lies in one of these intervals, and thus the union (not sum) must be open. Write down the definition of open if you need to.

Last edited: Feb 1, 2006
3. Feb 1, 2006

Kocur

Matt, I think I that I do not quite get it.

It is clear that any point of the union belongs to at least one of the contributing intervals (by the definition). Unfortunately, the second part escapes me.

4. Feb 1, 2006

matt grime

Write down the definition of what it means for a set to be open. If you do not know what you're trying to prove you will never be able to prove it, or at least you won't know when you have proved it and when to stop.

5. Feb 1, 2006

Kocur

Matt, I am clearly very resistant to the reasoning .

I think it goes like that:

By the definition, any point of a union belongs to at least one of the contributing intervals.

An open interval is an interval which does not contain its borders.

Combining these two above we get:

The union does not contain the borders of the respective intervals (so, it is open), unless these intervals overlap.

And this is the overlapping that poses a problem for me. How can I check what happens when an infinite number of intervals overlap?

I am worried about the overlapping, because I know that an infinite intersection of open intervals might yield a closed interval. How can I prove that nothing like that happens in the case of a union of open intervals?

It might sound like a really stupid question, by I am just used to operating on discrete values .

6. Feb 1, 2006

arildno

Recheck your textbook on the definition of an open interval.

7. Feb 1, 2006

Kocur

I think I got it:

By the definition, any point x of a union belongs to at least one of the contributing intervals.

Since the respective intervals are open, each interval containing x must also contain a neigbourhood of x.

By the definition of union of sets, the union contains some neighbourhood of x and, hence, is an open set.

Is this correct?

8. Feb 1, 2006

matt grime

Definition: X is open (in this topology) if for any point x in X there is an open interval I with x in I and I contained in X. Now, that is trivially true, here.

9. Feb 1, 2006

HallsofIvy

Staff Emeritus
That's one definition (probably the most common) of "open set" in a metric topology.

Another, equivalent, definition of "open set" that I have seen (that corresponds to the statement "An open interval is an interval which does not contain its borders." that arildno objected to) is this:

A point, p, is said to be an "interior point" of set A if and only if there exist some d> 0 such that Nd(p) is a subset of A (usual definition of "interior point").
A point, p, is said to be an "exterior point" of a set A if and only if it is an interior point of the complement of A.
A point, p, is said to be a "boundary point" of set A if and only if it is neither an interior point nor an exterior point of a.

Now, one can, indeed, define a set to be "open" if and only if it contains none of its boundary points and "closed" if and only if contains all of its boundary points.

My experience has been that for students who are just beginning topology, but who have experience with "open" and "closed" intervals, this makes more sense than the standard definitions.

Last edited: Feb 2, 2006
10. Feb 2, 2006

Kamataat

A union of sets is composed exactly of the elements of those sets (no more, no less). Let's say we have the sets A=(a,b) and B=(c,d), where a<b<c<d. If their union were closed, then it would have to

a) include "more" elements than there are in A and B, for example their endpoints
OR
b) include "less" elements than there are in A and B, thus excluding some of their elements. For example, the "left" endpoint of the union would have to be a point e so that a<e<b.

hence the union must be open.

PS: Don't know anything about topology though, so take this with a grain of salt!

- Kamataat

Last edited: Feb 2, 2006
11. Feb 2, 2006

matt grime

Well, thanks, but surely you can see that since that in no way depends on the nature of 'open' surely you can see it's nonsense. Insert the word closed instead and look, your argument is just as 'valid', and since the union of arbitrary closed sets is not in general closed....

Now, how about actually checking the definitions before trying to answer a question?

12. Feb 2, 2006

JasonRox

This isn't Topology yet.

13. Feb 2, 2006

Kamataat

OK, I apologize for this mess. Feel free to delete that posting of mine. More careful next time.

- Kamataat

14. Feb 3, 2006

Kocur

Thank you for help guys. I think I got it at last .

By the definition of union of sets, any point x of a union belongs to at least one of the contributing intervals.

Since the respective intervals are open, each interval containing x must also contain a neigbourhood of x.

A union contains, by the definition, all elements of the contributing sets. Thus, a union of open intervals contains, for each point x, a neighbourhood of x.

Kamataat, we all make mistakes. Do not worry. You tried to help.

15. Feb 3, 2006

JasonRox

Is it that obvious?

The infinite union of closed intervals isn't always closed.

16. Feb 3, 2006

matt grime

Yes it is that obvious.

It uses a particular property of each part of the union, one that fails for closed sets, so the argument says nothing about what the union of closed sets is, as it shouldn't.

Last edited: Feb 3, 2006
17. Feb 3, 2006

JasonRox

Yes, it really is.

I just don't think it's good idea that a beginner should just jump to a conclusion.

Because the infinite intersection of open intervals is not always an open interval.

18. Feb 3, 2006

matt grime

And nothing they reasoned made them suspect that it was.

19. Feb 4, 2006

HallsofIvy

Staff Emeritus
What do you mean by "jump to that conclusion"? Certainly nobody here, including Kocur, did.