Proving something is not the empty set

• l888l888l888
In summary, the problem at hand is to prove that the intersection of all Vn's from n=1 to infinity, where Vn is a nonempty closed subset of a compact space T, is not equal to the empty set. This can be proven by contradiction, assuming that the intersection is empty and using the fact that T is compact. By constructing an open cover of T with no finite subcover, it can be shown that the assumption leads to a contradiction, proving that the intersection must be nonempty.
l888l888l888

Homework Statement

for every n in the natural numbers, Vn is a nonempty closed subset of a compact space T. Vn+1 is contained in Vn. Prove the Intersection of all the Vn's from n=1 to infinity does not equal the empty set.

The Attempt at a Solution

This question seems rather easy, actually a little too easy for an advanced topology class. but this is what i did. I tried to prove by contradiction and assume that the Intersection of all the Vn's from n=1 to infinity EQUALS the empty set. By this assumption this would imply that there exists an n in the natural numbers such that that Vn is the empty set. I say this because... for any set A, A n {empty set} = {empty set}. so choosing a Vn to be the empty set would guarantee that the intersection of all the Vn's is the empty set. But this is a contradiction because in the original problem it says for every n in the natural numbers, Vn is NONEMPTY. Is this enough?

No. Your proof doesn't use that the space T is compact. Take the space to be T=(0,1]. Take Vn=(0,1/n]. The intersection of the Vn's is empty, but none of them are empty. You need to use that T is compact.

OH I SEE. UR RITE. DONT KNOW WHAT I WAS THINKING... ANYWAY I THINK I HAVE IT NOW. THERE WAS A PREVIOUS PROBLEM WITH THE END RESULT BEING THAT... A NECESSARY AND SUFFICIENT CONDITION FOR A SPACE TO TO BE COMPACT IS : IF {Vi : i in I} is an indexed family of subsets of T such that the intersection of the Vi's does not equal the empty set with the i's in a finite subset J of I, then the intersection of the Vi's for all the i's in I is not the empty set. So using this result and taking a finite subset of the natural numbers, if the intersection of the Vn's with the n's being in the subset is not equal to the empty set the the intersection of all the Vn's does not equal the empty set. I hope this is making sense. It is hard to explain without all the mathematical symbols.

l888l888l888 said:
OH I SEE. UR RITE. DONT KNOW WHAT I WAS THINKING... ANYWAY I THINK I HAVE IT NOW. THERE WAS A PREVIOUS PROBLEM WITH THE END RESULT BEING THAT... A NECESSARY AND SUFFICIENT CONDITION FOR A SPACE TO TO BE COMPACT IS : IF {Vi : i in I} is an indexed family of subsets of T such that the intersection of the Vi's does not equal the empty set with the i's in a finite subset J of I, then the intersection of the Vi's for all the i's in I is not the empty set. So using this result and taking a finite subset of the natural numbers, if the intersection of the Vn's with the n's being in the subset is not equal to the empty set the the intersection of all the Vn's does not equal the empty set. I hope this is making sense. It is hard to explain without all the mathematical symbols.

I'm not sure I even understand that without trying harder. Which I'm not going to do because there is a much simpler and instructive approach. Why don't you try to show if the intersection of all of Vn's is empty then you can construct an open cover of T with no finite subcover?

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What does it mean to prove something is not the empty set?

Proving that something is not the empty set means showing that it contains at least one element. In other words, the set is not empty and has some value or element in it.

How do you prove that something is not the empty set?

One way to prove that something is not the empty set is by providing a concrete example of an element that is in the set. This can be done through direct observation or by using logical reasoning.

Why is it important to prove that something is not the empty set?

Proving that something is not the empty set is important because it establishes the existence of at least one element in the set. This is necessary for further analysis and understanding of the set and its properties.

Can a set be both empty and not empty?

No, a set can only be either empty or not empty. If a set is empty, it has no elements and therefore cannot be not empty. If a set is not empty, it must have at least one element and therefore cannot be empty.

How does one prove that a set is not empty using a proof by contradiction?

To prove that a set is not empty using a proof by contradiction, assume that the set is empty and then show that this leads to a contradiction. This contradiction would prove that the set cannot be empty and therefore must contain at least one element.

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