I'm in introduction to analysis right now and I'm having a hard time proving intersections and unions of a family of sets.
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Chances are, you're "making it harder than it really is", as is often said.
Your goal is to show that 2 given sets are equal:
intersection over all n of (-1/n, 1/n) = {0}.
For convenience, let's denote the infinite intersection of all those intervals by the letter "A".
In general, to show 2 sets are equal, you can show that each is a subset of the other.
Equivalently, you can show that the number 0 is in A, and that NO other numbers are in A.
So, why is 0 in A?
Well, to show that 0 belongs to the intersection of all the intervals (-1/n,1/n), you must show that 0 is in
every single one of the intervals of the form (-1/n, 1/n). This is obviously true, since for every positive integer n, it's true that -1/n < 0 < 1/n.
Now why are no other numbers in A?
Pick an arbitrary nonzero number y.
For some value of N, 0 < 1/N < |y|.
This is the "Archimedean property" of the set of natural numbers at work.
So we have 1/N < |y|.
This is equivalent to saying that either y > 1/N or y < -1/N,
which means y isn't in the interval (-1/N, 1/N) for our particular value of N.
Hence y fails to be in at least one interval of the form (-1/n, 1/n).
Hence y fails to be in the intersection of all such intervals, namely, A.
y was an arbitrary nonzero number, so all nonzero numbers are not in A.
Overall, we've shown that for any real number z, z is in A if and only if z is 0.
The strategy was to show that for any number other than 0, the intervals we're intersecting get so small that they can't "hold on to" any numbers that aren't 0. Their endpoints keep shrinking closer to 0.
Your goal is to show that 2 given sets are equal:
intersection over all n of (-1/n, 1/n) = {0}.
For convenience, let's denote the infinite intersection of all those intervals by the letter "A".
In general, to show 2 sets are equal, you can show that each is a subset of the other.
Equivalently, you can show that the number 0 is in A, and that NO other numbers are in A.
So, why is 0 in A?
Well, to show that 0 belongs to the intersection of all the intervals (-1/n,1/n), you must show that 0 is in
every single one of the intervals of the form (-1/n, 1/n). This is obviously true, since for every positive integer n, it's true that -1/n < 0 < 1/n.
Now why are no other numbers in A?
Pick an arbitrary nonzero number y.
For some value of N, 0 < 1/N < |y|.
This is the "Archimedean property" of the set of natural numbers at work.
So we have 1/N < |y|.
This is equivalent to saying that either y > 1/N or y < -1/N,
which means y isn't in the interval (-1/N, 1/N) for our particular value of N.
Hence y fails to be in at least one interval of the form (-1/n, 1/n).
Hence y fails to be in the intersection of all such intervals, namely, A.
y was an arbitrary nonzero number, so all nonzero numbers are not in A.
Overall, we've shown that for any real number z, z is in A if and only if z is 0.
The strategy was to show that for any number other than 0, the intervals we're intersecting get so small that they can't "hold on to" any numbers that aren't 0. Their endpoints keep shrinking closer to 0.