# Infinite Unions and Intersections

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I know this isn't the forum for this, but I am just stuck dead in the water. This book gives an excercise to find an infinite collection of open sets whose intersection is closed. They then proceed to state that a union of open sets is always open and only an intersection of a finite number of open sets is certain to be open. They just don't give any justification for that statement. I don't get it. All I can do is say, ok, they said it, it must be true. It just wasn't intuitively obvious to me. I just don't know how to find something that would explain it rather than simply state it.

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Well, a topology is *defined* to be a collection of subsets of a set X, which we'll call "open", such that:
*X is open
*An arbitrary union of open sets is open
*The intersection of two open sets is open

It so happens that if we consider the real numbers as our X and we call a set open if it contains a little interval around each of its points, those properties are satisfied, which means that the usual topology on R is indeed a topology. Try to prove this yourself; if you need help, ask again.

A set is said to be "closed", by definition, if its complement is open. The intersection of all the intervals (-epsilon,+epsilon), where epsilon takes all positive values is an infinite collection of open intervals whose interserction is {0}, which is closed because R-{0} is open. An easier [and less interesting] example is the intersection of all open sets, which is the empty set, which is closed, since its complementary, R, is open.

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There is where my problem lies. I just don't see how the infinite intersection of the open sets (-e,+e) for all real numbers e contains {0}. It seems to me at infinity you have an open interval on the real number line of width zero. That seems to me to say that collection of sets contains the empty set and their intersection is empty. Specifically the summation of 1/2^n for n a natural number going from 1 to infinity is 1, but no finite number of terms ever equals one. If that collection of sets does contain {0} then that seems two conflicting views of what taking something to infinity means. I can accept that it does contain {0}. I'm not arguing that it doesn't. Rather I'm saying I don't understand how to take into account that infinity part. Why doesn't that collection of sets include the empty set?

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The idea of "taking something to infinity" is very vague and kind of meaningless. The good news is that the intersection of a possibly infinite family of sets is well defined. The intersection of a family of sets is the set of elements that belong to all the sets in the family.

The only real number that belongs to all the intervals (-epsilon, +epsilon), where epsilon takes all positive values, is 0. If you think that another number x is in the intersection, take epsilon=abs(x)/2 and you'll see that x can't possibly be in all of the intervals. If you think that 0 shouldn't be in the intersection, you need to find a value of epsilon for which 0 is not in (-epsilon, +epsilon). Convinced yet?

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Thanks, that was exactly what I was looking for. Thinking about it that makes perfect sense. It is certainly consistant with how you prove the limit of a function using an epsilon/delta arguement. It was the cantor set that got me all screwed up. I just got started thinking that at infinity the irrationals in it were endpoints of intervals since the width of the intervals goes to zero. That makes no sense. I couldn't figure out exactly what it was wrong with that though. What was wrong is nothing happened at infinity. They were not eliminated in any finite stage so they must be part of the set.

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review question =):

take the sequence x_n = (1-1/n) in the topological space (R, U) where U is the set of all halfopen sets (a,b] (trivially defining a topology on R).

Prove that x_n does not converge to 1 in this space.

A topology on R is more or less a definition on what is open and what is not open. Given that some constraints on U (given by alvaro).

hint: how do you define convergence using open sets...

regards

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