# Homework Help: Is the uncountably infinite union of open sets is open?

1. Jan 26, 2010

### Johnson04

This is not a homework problem, just a question from a discussion with my classmates about the Cantor set. The original goal is to prove Cantor set is closed. My earlier attempt is to show the complement of the Cantor set is open. Since when construct the Cantor set each time the sets removed are open, I thought the complement of the Cantor set contains uncountably infinite many open sets, and the union of all of these open sets is still open. So the Cantor set is closed. But first, one of my classmate disagree with this proof. He thought the complement of the Cantor set only contains countably infinite many open sets. Second, he thought only countably infinite union of open sets is open, uncountably infinite union of open sets may not be open. But he could not give me convincing proof. So I posted my question here. Any suggestions are highly appreciated!

2. Jan 26, 2010

### Dick

ANY union of open sets is open, countable or not. That's an easy proof. Try it! Prove it to your classmate.

3. Jan 26, 2010

### Johnson04

Thanks a lot. I see, since any points in a open set are interior points, the points in the any union of the open sets are still interior points, accordingly, the union of the open sets is still open, no matter if the union is finite, countably infinite or uncountably infinite. Well, I was wondering if the complement of the Cantor set consists of uncountably infinitely many open sets?

4. Jan 26, 2010

### Dick

Depends on what you mean. ANY open set can be written as an uncountable union of open sets. You can write it as an uncountable union of copies of itself. On the other hand, it's also true that any open set on the real line can be written as a countable union of disjoint intervals, if that's what you really mean. That's a bit harder to prove. But not super hard.

5. Jan 26, 2010

### Johnson04

I think that was my question: if the complement of the Cantor set on [0,1] is the set of uncountably infinitely many disjoint open sets. I think it is, because the end points of the open sets form the Cantor set. It can be proved that the Cantor set is uncountable set by using the Cantor's diagonal process. So there are uncountably infinitely many open sets in the complement of the Cantor set. Did I miss anything important?

6. Jan 26, 2010

### Dick

Yes. You are missing the deep weirdness of the Cantor set. You can't have an uncountable number of disjoint open intervals on the real line. Each interval contains a rational number which is not in any other interval. Right? That means you can associate each interval with a unique rational number. Also, right? That means the cardinality of intervals is equivalent the cardinality of a subset of the rationals. Also, also, right? Can't be denied so far, right? Hence the intervals are countable. The Cantor set is uncountable. The picture of a string of intervals with single points between them is WRONG. Boo!

7. Jan 26, 2010

### Dick

Look at it this way. If you pick x=1/2, then what's the next rational number that is larger than 1/2? I think that kind of sums the problem up.

8. Jan 27, 2010

### Johnson04

Thanks Dick! I think I see what you mean. Each interval contains a unique rational number, because the set of all rational numbers is dense on real line. On real line there are only at most countably infinite many open intervals. Actually this conclusion make the proof of the Cantor set is closed a little easier.

9. Jan 28, 2010

### Johnson04

Well, I believe I totally understand the related properties of the Cantor set now. First, not all points in Cantor set are endpoints of the removed open sets. Second, the complement of the Cantor set only contains countably infinitely many open sets, that means the Cantor set does contain only countably infinitely many endpoints of the removed open sets (since they only have countably infinitely many endpoints). With this very clear picture, I have got proofs of several properties of the Cantor sets. Thanks!