Distinct open set B, non-open set A: Is A U B open or not?

[pisrationalhahaha]

Hi
If we have two sets A and B where A is not an open set and B is open, then AUB is open or not ?
Note that A and B are distinct ( their intersection is empty)
I am sure that the union is not open (logically it's true )
But how can we prove this on an exam ?
I'm stuck on how should I explain my idea and how to start

 

[Dr.Peterson]

Hi
If we have two sets A and B where A is not an open set and B is open, then AUB is open or not ?
Note that A and B are distinct ( their intersection is empty)
I am sure that the union is not open (logically it's true )
But how can we prove this on an exam ?
I'm stuck on how should I explain my idea and how to start

I think you meant to say "disjoint", not "distinct" (which just means they aren't the same set). So the problem really is,

If we have two disjoint sets A and B, where A is not an open set and B is open, then AUB is open or not?

(Actually, it seems to me that the answer does not have to be "yes" or "no"; it could be "sometimes"! Since we already know that you didn't quote the question exactly, please do so.)

Tell me your thinking as to why, "logically", the union is not open.

Then tell me the definition of "open set", and what it would take to prove that a set is not open.

These two things, together, should make it easy to answer your question.

 

[pisrationalhahaha]

I think you meant to say "disjoint", not "distinct" (which just means they aren't the same set). So the problem really is,

If we have two disjoint sets A and B, where A is not an open set and B is open, then AUB is open or not?

(Actually, it seems to me that the answer does not have to be "yes" or "no"; it could be "sometimes"! Since we already know that you didn't quote the question exactly, please do so.)

Tell me your thinking as to why, "logically", the union is not open.

Then tell me the definition of "open set", and what it would take to prove that a set is not open.

These two things, together, should make it easy to answer your question.

A set is not open if we can find at least one element of it such that for any open ball centered at this element , this open ball is not a subset of the main set
If we let an element "e" of AUB then e belongs to A or e belongs to B
Then there is two cases
If e belongs to B then we can prove easily that AUB is open
But if e belongs to A ...... (then what)??

 

[Dr.Peterson]

A set is not open if we can find at least one element of it such that for any open ball centered at this element , this open ball is not a subset of the main set
If we let an element "e" of AUB then e belongs to A or e belongs to B
Then there is two cases
If e belongs to B then we can prove easily that AUB is open
But if e belongs to A ...... (then what)??

Ah! That is the question!

If you sketch a picture of disjoint sets, one non-open and the other open, you will probably draw them in such a way that it is clear that the union is not open. You might think about what it would take to prove that there can never be an open ball around e, or (really) that there would always be some e with no open ball around it.

Perhaps a more fruitful thing is to play "devil's advocate", and try to find a counterexample, in which the union is open. What would it take to make that happen? You will either find your counterexample, or have an idea how to prove there is no counterexample. I find this strategy is often useful. I won't tell you yet which way it works out ...

 

[pisrationalhahaha]

Ah! That is the question!

If you sketch a picture of disjoint sets, one non-open and the other open, you will probably draw them in such a way that it is clear that the union is not open. You might think about what it would take to prove that there can never be an open ball around e, or (really) that there would always be some e with no open ball around it.

Perhaps a more fruitful thing is to play "devil's advocate", and try to find a counterexample, in which the union is open. What would it take to make that happen? You will either find your counterexample, or have an idea how to prove there is no counterexample. I find this strategy is often useful. I won't tell you yet which way it works out ...

Okay, if we suppose that "e" is on the bounadary of the set then we can't find an open ball
Thus there is a counterexample such that the union is not open
Is this it ?

 

[Dr.Peterson]

Okay, if we suppose that "e" is on the boundary of the set then we can't find an open ball
Thus there is a counterexample such that the union is not open
Is this it ?

I was talking about a counterexample in the sense of a pair of sets A and B that contradicts our guess that the union will not be open. You're just supposing we have a pair of sets. But we don't know yet that all such pairs will have a non-open union; that's what we want to prove or disprove.

Think about what it would take to find non-open set A and open set B whose union is open. What would have to be true of the boundary of A?

This does stretch your mind a bit; when I used the word "easy" a while ago, I meant that only relatively ...

By the way, you haven't told me the exact wording of the problem. Does it at least use the word "disjoint"? Does it demand a yes or no answer?

 

[pisrationalhahaha]

I was talking about a counterexample in the sense of a pair of sets A and B that contradicts our guess that the union will not be open. You're just supposing we have a pair of sets. But we don't know yet that all such pairs will have a non-open union; that's what we want to prove or disprove.

Think about what it would take to find non-open set A and open set B whose union is open. What would have to be true of the boundary of A?

This does stretch your mind a bit; when I used the word "easy" a while ago, I meant that only relatively ...

By the way, you haven't told me the exact wording of the problem. Does it at least use the word "disjoint"? Does it demand a yes or no answer?

The sets are like that according to the exercise ,

Q: Is AUB open or not ? Why ?
AUB is open if there is no any boundary points which is not true because they exist
Is this it ?

 

[Dr.Peterson]

The sets are like that according to the exercise ,
View attachment 9430
Q: Is AUB open or not ? Why ?
AUB is open if there is no any boundary points which is not true because they exist
Is this it ?

This totally changes the problem.

I wish you had said that you were shown a particular pair of sets on a plane, rather than asking as if you had to prove a general fact, "If we have (any) two sets A and B where A is not an open set and B is open, then AUB is open or not?" The answer does not follow from some general rule about unions, but from the details of the picture, which you withheld.

As drawn, evidently the lower boundary of A is part of A, making it not open. Since that is a boundary of the union as well, the union is also not open. It is not that the boundary points "exist", as you say, but that they are part of the set.

I'll tell you where I was headed in the problem as I perceived it, so that I don't completely waste the effort: If A was a closed interval on the real number line, and B its complement, then A would not be open, B would be open, and their union, the entire line, would be open. (This is the counterexample I had in mind.) The condition for this is that the boundary of A is (part of) the boundary of B.

Please, in the future, obey the instructions in the READ BEFORE POSTING announcement, "Post the complete text of the exercise. This would include the full statement of the exercise and its instructions, so the tutors will know what you are working on. If there is a graphic or table or some other non-textual information necessary to the exercise, include a detailed description." This would save a lot of wasted time and effort for both of us.

 

[pisrationalhahaha]

This totally changes the problem.

I wish you had said that you were shown a particular pair of sets on a plane, rather than asking as if you had to prove a general fact, "If we have (any) two sets A and B where A is not an open set and B is open, then AUB is open or not?" The answer does not follow from some general rule about unions, but from the details of the picture, which you withheld.

As drawn, evidently the lower boundary of A is part of A, making it not open. Since that is a boundary of the union as well, the union is also not open. It is not that the boundary points "exist", as you say, but that they are part of the set.

I'll tell you where I was headed in the problem as I perceived it, so that I don't completely waste the effort: If A was a closed interval on the real number line, and B its complement, then A would not be open, B would be open, and their union, the entire line, would be open. (This is the counterexample I had in mind.) The condition for this is that the boundary of A is (part of) the boundary of B.

Please, in the future, obey the instructions in the READ BEFORE POSTING announcement, "Post the complete text of the exercise. This would include the full statement of the exercise and its instructions, so the tutors will know what you are working on. If there is a graphic or table or some other non-textual information necessary to the exercise, include a detailed description." This would save a lot of wasted time and effort for both of us.

Thanks