Consider the subsets of R defined as follows: An is the interval (0,1/n) for all natural numbers. Show that:
U (union) n=1 to infinity of An = (0,1)
Here's what I tried as a proof:
Show that (0,1) is a subset of U n=1 to infinity An
Take an arbitrary real number a in (0,1) so a is in the set {x such that there exists some n such that x is in (0,1/n) for all natural numbers n}
Show that U n=1 to infinity is a subset of (0,1)
Take x in U (0,1/n)
By definition of union this means there exists some n such that x is in (0,1/n) for all natural numbers n
I'm not sure if this is the right way to go? Or should I prove if the set is countable?
You are on the right track with your proof. To show that the union of all the subsets An is equal to (0,1), you need to prove two parts:
1. Show that (0,1) is a subset of the union of all the subsets An.
2. Show that the union of all the subsets An is a subset of (0,1).
Let's go through each part step by step:
1. To show that (0,1) is a subset of the union of all the subsets An, you correctly started by taking an arbitrary real number "a" in (0,1). Now, you want to show that "a" is also in the union of all the subsets An.
Since "a" is in (0,1), it follows that for any positive integer n, 1/n > 0. Therefore, 1/(n+1) > 0 and (0,1/(n+1)) ⊂ (0,1/n) for all natural numbers n.
This means that "a" is in each of the intervals (0,1/n) for every natural number n. Thus, "a" is in the union of all the subsets An.
2. To show that the union of all the subsets An is a subset of (0,1), you started by taking an arbitrary number "x" in the union of all the subsets An. By the definition of a union, this means there exists some natural number n such that "x" is in (0,1/n).
For this part, you can argue that since "x" is in (0,1/n) for some positive integer n, it is also in the interval (0,1/(n+1)). Therefore, "x" is in the interval (0,1), which means the union of all the subsets An is a subset of (0,1).
Putting these two parts together, you have shown that (0,1) is a subset of the union of all the subsets An, and the union of all the subsets An is a subset of (0,1). Therefore, you can conclude that the union of all the subsets An is equal to (0,1).
Note: Countability is not necessary to prove this result. The proof you've outlined is sufficient.