sol1 - Math 235 Assignment 1 Solutions 1.1 For n N not zero...

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Math 235: Assignment 1 Solutions 1.1 : For n N not zero, let A n = [ - n, n ] (The closed interval in R containing all real numbers x satisfying - n x n ). It is easy to see that we have the chain of inclusion A 1 A 2 A 3 ... , and therefore N [ n =1 A n = A N = [ - N, N ] and N \ n =1 A n = A 1 = [ - 1 , 1] . 1.2 : Notice that the consecutive closed intervals, [ n, n + 1] and [ n + 1 , n + 2] do in fact overlap (This is because we have taken the closed interval. What if we had taken the open intervals instead?) We conclude that the finite union is N [ n =1 [ n, n + 1] = [1 , N + 1] . Since any real number x 1 will eventually lie inside the interval [1 , N + 1] for some sufficiently large N N , we conclude [ n =1 [ n, n + 1] = [1 , ) . (Note we have the open bracket ”)” on the right side of the above interval. Why?). Similarly, the consecutive open intervals ( n, n + 2) and ( n + 1 , n + 3) also overlap, so N [ n =1 ( n, n + 2) = (1 , N + 2) [ n =1 ( n, n + 2) = (1 , ) . ( means ”therefore”) 1.3 : It is helpful to say in words what each A n is. A 1 is the set of all natural numbers, N . A 2 is the set of all squares of natural numbers. A 3 is all cubes , etc. Of course, a square of a natural number is in particular a natural number, so A 2 A 1 . Likewise for cubes, 4 th powers, etc. We conclude A 1 A n for all n N , [ n =1 A n = A 1 = N . Again it is helpful to ask in words, what does T n =1 A n mean? Since each A n is the subset of natural numbers consisting of n th powers, their intersection must be all natural numbers x such that x is a square, a cube, a 4 th power, a 5 th , etc. There are only 2 such numbers: 1
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2 \ n =1 A n = { 0 , 1 } . (Challenge: Can you find the smallest natural number x 6 = 0 , 1 such that x is a square, a cube, a 4 th power, ..., and a 10 th power? What if we replace 10 with some arbitrarily large N th power?) 2.1 : Consider the sets A = B = C = { 1 } . Then ( A \ B ) \ C is the empty set φ , while A \ ( B \ C ) = A \ φ = A . This counterexample is enough to show that the formula in 2.1 does not hold (There are of course many more interesting counterexamples that show this as well, but any one is sufficient). 2.2: Note that no amount of examples will suffice to prove that this formula does hold, hence we need to get down to the nitty-gritty and construct a proof. In general, when proving two sets are equal (say X and Y ), a good place to start is to try to show that if x X then x must also be in Y and vice versa. This will show that X is a subset of Y and Y is a subset of X , which is possible only if X = Y .
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