Proving something is not the empty set

[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.

Homework Equations

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?

 

[Dick]

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.

 

[l888l888l888]

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.

 

[Dick]

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?