how do i show that sup A is 1 for the set {cos(X)} from 1 to infinity
thanks for any help
Well, you start by knowing or looking up the definition of "sup"!
The "supremum" or "sup" of a set of numbers (also called "least upper bound" or "lub") is, as "least upper bound" explicitely says, the smallest of all possible upper bounds. While a general set of real numbers may not have a "smallest" or "largest number (the set of all real numbers, x, such that 0< x< 1, has neither) the set of all upper bounds for a set of real numbers always contains smallest member.
So, what are the "upper bounds" for this set. What numbers are larger than any possible value for cos(x)? What is the smallest of those?
(You know, I hope, that $\displaystyle -1\le cos(x)\le 1$ for all x.)
well i am aware that the supremum of this set is 1, and that this is not a maximum as -1<= cos(x)<=1 only over the real numbers and not the integers cos(2k *pi) = 0 however pi is irrational and so for this set we have -1< cos(x) < 1 however i was told that to prove that 1 is the supremum (although being obvious) is difficult and beyond the scope of the course (our lecturer said that he would give a proof after the lecture but i had another lecture so could not stay to see the proof) and i was wondering if anyone could give me one/point me in the right direction because im afraid i do not think that i will be able to do it myself,
thanks for any help
I did not understand that "x" was to be a positive integer! That does make the problem a lot harder than I a had thought.
Given any integer, n, write $\displaystyle 2\pi$ to n decimal places. Then we can say that $\displaystyle 2\pi$ is equal to $\displaystyle \frac{6283185307179586476925286766559...}{10^n}$ where the numerator is $\displaystyle 2\pi$ to n decimal places with the decimal point dropped (i.e. multiplied by $\displaystyle 10^n$). Take k to be the integer $\displaystyle 10^n$. For that $\displaystyle 2k\pi$ is the integer 6283185307179586476925286766559 yet is very close to a multiple of $\displaystyle 2\pi$ so the cosine of that integer is very very close to 1. Taking more and more decimal places of $\displaystyle 2\pi$ we get higher and higher integers whose cosine is closer and closer to 1. Given any number, $\displaystyle \delta< 1$, we can take enough decimal places to get cosine of that integer closer to 1 than $\displaystyle \delta$. Thus, no number less than 1 can be an upper bound on the set and since 1 is an upper bound, it is the least upper bound.