Is the Intersection of Intervals Empty?

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  • #1
mathmari
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Hey! :eek:

I want to determine the following sets:


I have done the following:
  1. Let . This means

    To be able to divide by we have to consider three cases: , , . Case 1:

    We get which is true for every . .
    Case 2:

    We get

    Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection .
    Case 3:

    We get

    Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection . Therefore we get that the intersection contains only the element and so we have .

    Is everything correct and complete? (Wondering)


  2. We have that for all . So it follows that .

    Let . Now we have to prove that there is a such that and , or not? But how can we show that? (Wondering)


  3. Let . This means

    We see that it must hold that . Now we want to solve for .

    From the first inequality we have .

    From the second inequality we have . From that we get , since is positiv and so the case is rejected, right?

    So we have Since is unbounded we can find always a value for that is bigger than , or not?

    Would that means that the intersection can't contain any element, and so ? (Wondering)
 
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  • #2
mathmari said:
1.

I have done the following:

Let . This means

To be able to divide by we have to consider three cases: , , . Case 1:

We get which is true for every . .
Case 2:

We get

Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection .
Case 3:

We get

Since is unbounded, we can find a value for that is bigger than . So this inequality is not true for every . This means that if the element cannot belong to the intersection . Therefore we get that the intersection contains only the element and so we have .

Is everything correct and complete?

Looks fine to me. (Nod)

mathmari said:
2.

We have that for all . So it follows that .

Let . Now we have to prove that there is a such that and , or not? But how can we show that?

Can we pick ? (Wondering)
mathmari said:
3.

Let . This means

We see that it must hold that . Now we want to solve for .

From the first inequality we have .

From the second inequality we have . From that we get , since is positiv and so the case is rejected, right?

So we have Since is unbounded we can find always a value for that is bigger than , or not?

Would that means that the intersection can't contain any element, and so ?

Yep.

I think we can do it a bit quicker.
The intervals have a lower bound of .
If some would be in the intersection, then must be in all intervals. But will not be in any interval where , so it won't be in the intersection. This is a contradiction, so the intersection must be the empty set. (Nerd)
 
  • #3
Klaas van Aarsen said:
But will not be in any interval where , so it won't be in the intersection.

Could you explain that to me further?

Do we consider two intervals and then we show that cannot belong to both?

(Wondering)
 
  • #4
mathmari said:
Could you explain that to me further?

Do we consider two intervals and then we show that cannot belong to both?

The intersection is of the form:

For an to be in this intersection, it must be in all intervals.
But there will always be an interval that starts with an that is higher than is. (Thinking)
 
  • #5
Klaas van Aarsen said:
The intersection is of the form:

For an to be in this intersection, it must be in all intervals.
But there will always be an interval that starts with an that is higher than is. (Thinking)

Ahh I got it now! (Malthe) As for 2. if we pick then we have that . But how do we get the union? I got stuck right now. (Wondering)
 
  • #6
mathmari said:
As for 2. if we pick then we have that . But how do we get the union? I got stuck right now.

For an to be in the union, it must be in at least one of the intervals.
And we just found an interval that it is in, didn't we? (Wondering)
 
  • #7
Klaas van Aarsen said:
For an to be in the union, it must be in at least one of the intervals.
And we just found an interval that it is in, didn't we? (Wondering)

Ahh yes!

So we have the following:

We have that for all . So it follows that .

Let . For it holds that . Therefore it holds that and so we get .

Therefore . Is everything correct? (Wondering)
 
  • #8
mathmari said:
Ahh yes!

So we have the following:

We have that for all . So it follows that .

Let . For it holds that . Therefore it holds that and so we get .

Therefore . Is everything correct?

Yep. (Nod)
 
  • #9
Klaas van Aarsen said:
Yep. (Nod)

Thank you very much! (Sun)
 
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